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# Pulsed High Power LED

I am going to design power drivers for a high power LED that is capable of high currents (up to several amps). I am using a power MOSFET for fast switching. I have designed a PCB for this circuit, but I have some problems with it.

I connected a dimmer circuit with the pcb. When I give 5V to the circuit the LED doesn't glow. I can check voltage switching between 5V at the gate of mosfet but there are also little voltages switching at drain and source and mosfet gets hot after some time.

Can anyone tell me the reason why there is no current at load, that is, why the LED doesn't light? Why is there some voltage at the drain and source of the mosfet, and why does it get hot very quickly?

Vcc=5 V, Vs=12 V, LD1=ordinary LED.

Datasheets: MOSFET, MOSFET driver.

## Update

I have removed the diode from the circuit but very low voltage is present there on the output that cannot drive my led. I am facing the same problem. the supply voltage is vcc=10V and Vs= 3V. and the resistor r1 is 80 ohm. can anyone tell me what could be the problem...

• I'm guessing you're not driving the FET correctly, but of course we can't tell what you are doing and how things are hooked up without a schematic. – Olin Lathrop May 3 '12 at 12:56
• You are obviously connecting things incorrectly in some way. You MUST provide a circuit diagram of what you are trying to do. Please also provide part numbers (LED, MOSFET, ...) and datasheet web links. || Trouble shooting: Apply power. Connect a 100 ohm resistor from drain to source on the MOSFET. If the MOSFET is the ONLY switch in the LED circuit then the LED must now light slightly if all is connected correctly. If not, check the circuit against what you have built until you find the mistake or accidental short circuit. – Russell McMahon May 3 '12 at 13:00
• A normal LED does not stand 12 V. It may be fried. That's one thing. Then, T1 may be hot because either D1 or T1 are wrongly connected. Just remove D1, and tell us if T1 gets hot. – Telaclavo May 3 '12 at 13:54
• i have connected these T1 and D1 correctly on the pcb. there are also little voltage switching at drain and source of mosfet – dani May 3 '12 at 13:56
• @scorpdaddy should i give 10-15V at Vcc (means gate of the mosfet) instead of 5V. and what about Vs. is it okk. – dani May 3 '12 at 14:10

If you used an ordinary LED for your your first test it's probably gone to LED heaven: you only have 1$\Omega$ current limiting resistor, that's good for 12A (if your power supply can deliver that). Now if the LED broke shorting that's also the current through the FET. It has a very low $R_{DS(ON)}$, so it shouldn't get hot, even at 12A. That probably means the gate voltage isn't high enough. Increase the supply voltage of your driver.

Try again with a higher value for R1, like 470$\Omega$. That should get you 25mA. Good enough for an ordinary LED.

BTW, D1 is not necessary. Get rid of it.

edit
On second thought it seems unlikely to me that an indicator LED, when broken, will conduct 12A. The thin bonding wire from the anode post to the die will break and leave an interruption. Then the next suspect is D1. If you mounted it wrong it would explain the high current, and why the LED doesn't light. One more reason to remove D1.

• D1 is only in the ciruit for protecting the led (i will use power led later). and also i want to drive the current in several amps upto 40amp or more – dani May 3 '12 at 14:00
• @dani You only need D1 if you have the LED far away from your circuit, so that its wires add up a non-negligible inductance. – Telaclavo May 3 '12 at 14:08
• @Telaclavo if i remove D1 from the ciruit then what you think current will be switched to the led pin in pcb. Please tell me what this D1 causing problem in the circuit? – dani May 3 '12 at 14:17
• @dani I don't understand your first sentence. Regarding the second one, no, D1 would be causing problems only if it was connected upside down. – Telaclavo May 3 '12 at 14:19
• @Telaclavo do you think the whole problem is because of this D1? if i remove it from circuit then current will be switched to the load i.e led. and also there are some voltage at source and drain of mosfet. – dani May 3 '12 at 14:23

VGS must be >4.5 Volts at 40A. Remove R2 and test again.

If the FET is getting warm, there is current flow through the FET. If the LED is not turning on, there is insufficient current flow through the LED. The two most likely possibilities are:

1) Your LED needs more current to visibly light than the FET can deliver in this circuit. 2) Your circuit topology is wrong. Possibly you have confused the drain and the gate, and the FET is shorting across the power supply.


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# Mortivation for the definition of measurable function?

1) We say that $f:\mathbb R^n\to \mathbb R$ is measurable if $$\{x\in\mathbb R^n\mid f(x)<\alpha \},\tag{D}$$ is measurable for all $\alpha$. What is the motivation for such definition ?

We can define for example continuity as : $f^{-1}(O)$ is open in $\mathbb R^n$ for all open set in $\mathbb R$. A possible definition for measurability could be for example $f^{-1}(O)$ is measurable for all open of $\mathbb R$, no ? So what is the motivation for the definition (D) ?

2) In more abstract measurable spaces, $f:(X,\mathcal F)\to (Y, \mathcal G)$ we say that $f$ is measurable if $$f^{-1}(U)\in \mathcal F\tag{D'}$$ for al $U\in \mathcal G$. Is there a correlation with the definition (D) ? For example, if $Y$ is a space with an order $\mathcal R$, would it be equivalent to $$\{x\in X\mid f(x)\mathcal R\alpha \}\in \mathcal F,$$ for all $\alpha \in Y$ ? Because I unfortunately don't really see the relation between (D) and (D').

• By combining (D) and the properties of the inverse image you get that the inverse images of the elements of the algebra generated by the sets $(-\infty,\alpha)$ also need to be measurable. – user574380 Jul 13 '18 at 10:26
• I would practicize (D') as definition and (D) as statement that can be shown to be equivalent with the definition in a particular case. – drhab Jul 13 '18 at 10:44
• The connection between $D$ and $D'$ is more clear if you write $D$ as $f^{-1}((-\infty,a)) \in \mathcal F$ for all $a$. – littleO Jul 13 '18 at 11:43
• A good motivation is indeed to extend the topological definition of continuity by saying that $f^{-1}(\mathcal O)$ is measurable for all open set $\mathcal O$. In fact : the function $f:(\mathbb R^n,\mathcal M)\longrightarrow (\mathbb R,\mathcal B)$ is measurable $\iff$ $f^{-1}(\mathcal O)$ is measurable for all open $\mathcal O\subset \mathbb R$. – Surb Jul 13 '18 at 11:47

$(D)$ is actually a necessary and sufficient condition for $f:\mathbb R^n\to\mathbb R$ to be measurable if $\mathbb R$ is equipped with the Borel-$\sigma$-algebra.

Especially the fact that it is sufficient is useful, and makes it easy to prove that functions are measurable.

In general if $X$ is a topological space with topology $\tau$ then by definition the corresponding Borel-$\sigma$-algebra is the $\sigma$-algebra generated by $\tau$ and denoted by $\sigma(\tau)$.

Here $\mathbb R$ is looked at as equipped with its usual topology.

If $\tau$ denotes this topology and $\mathcal M$ denotes the $\sigma$-algebra on $\mathbb R^n$ then it can be shown that $$f^{-1}(\sigma(\tau))\subseteq \mathcal M\text{ if and only if }f^{-1}((-\infty,\alpha))\in\mathcal M\text{ for every }\alpha\in\mathbb R$$

Here $f^{-1}(\sigma(\tau))$ serves as a notation for $\{f^{-1}(A)\mid A\in\sigma(\tau)\}$.

Note that $f^{-1}((-\infty,\alpha))=\{x\in\mathbb R^n\mid f(x)<\alpha\}$.

• Are you sure that $(D)$ is necessary and sufficient for $f$ to be measurable with $\mathbb R$ and $\mathbb R^n$ are both equipped with Borel $\sigma -$algebra ? Take for example $n=1$. If $B$ is a Borel set and $f:\mathbb R\longrightarrow \mathbb R$ is continuous, there is no reason for $f^{-1}(B)$ to be Borel although it's measurable... – Peter Jul 13 '18 at 10:53
• Yes, I am sure. In the situation you sketch continuity of $f$ means exactly that $f^{-1}(\tau)\subseteq\tau$. From this it follows that $\sigma(f^{-1}(\tau))\subseteq\sigma(\tau)$. But it can be shown in general that $\sigma(f^{-1}(\tau))=f^{-1}(\sigma(\tau))$ (for a proof of that see here) so we get $f^{-1}(\sigma(\tau))\subseteq\sigma(\tau)$. That means that - if $f$ is continuous - $f^{-1}(B)$ is Borel whenever $B$ is Borel. – drhab Jul 13 '18 at 11:04
• Ok for $f$ continuous, but for $f$ not continuous ? Because my teacher said that measurability if for $f:(\mathbb R,\mathcal M)\to (\mathbb R, \mathcal B)$ (and not $f:(\mathbb R,\mathcal B)\to (\mathbb R, \mathcal B)$, i.e. $f^{-1}(B)\in \mathcal M$ for all $B\in \mathcal B$ (where $\mathcal M$ is the lebesgue measurable function $\sigma -$algebra). See here – Peter Jul 13 '18 at 11:11
• Define $\mathcal A$ as collection of open sets $(-\infty,a)$. Let $(X,\mathcal M)$ be measurable space and let $f:X\to\mathbb R$ be a function that satisfies $f^{-1}(\mathcal A)\subseteq\mathcal M$. Then $\sigma(f^{-1}(\mathcal A))\subseteq\mathcal M$. But as I said: we have $\sigma(f^{-1}(\mathcal A))=f^{-1}(\sigma(\mathcal A))$ so that $f^{-1}(\sigma(\mathcal A))= \sigma(f^{-1}(\mathcal A))\subseteq\mathcal M$. Next to that it can be shown that $\tau\subseteq\sigma(\mathcal A)$ and consequently $\sigma(\tau)=\sigma(\mathcal A)$. So we end up with $f^{-1}(\sigma(\tau))\subseteq\mathcal M$. – drhab Jul 13 '18 at 11:25
• Under $15$? Not any more :-) Glad to help. – drhab Jul 13 '18 at 12:28

The two definitions are equivalent with the standard Borel sigma algebra.

The main trick is as follows: when checking the second definition, you don't need to check all $U\in\mathcal{G}$ - it is sufficient to check a generating set. For example, the open sets do generate the Borel sigma algebra so your suggested definition of measurability is also equivalent. But we can find an even smaller generating set, which is what the definition (D) is based on: all rays of the form $(-\infty,a)$.

And that's how you should read $(D)$. $f$ is measurable if $f^{-1}((-\infty,a))$ is measurable for each $a\in\mathbb{R}$.

format: plain


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# Cartan subalgebra and the decomposition of eigenspaces

Let $\frak g$ be a semisimple Lie algebra over a finite dimensional field $F$, and let $\frak h$ be a Cartan subalgebra in $\frak g$. I need actually some explanation on why $$\mathfrak g = \bigoplus_{\alpha\in \frak h^{*}} \mathfrak g_{\alpha}$$ Where $\mathfrak g_\alpha= \{x\in \mathfrak g; \ [x,h]=\alpha(h)x\text{ for all } h\in \mathfrak h\}$.

The proof relies on $F$ being algebraically closed. There are examples of Lie algebras which don't decompose like this when $F$ is not algebraically closed.

This is the viewpoint of Humphrey's Intro to Lie Algebras and Rep Theory. The big picture is: The set of matrices $\{\text{ad}(h):\ h\in\mathfrak h\}$ is a bunch of diagonalizable (aka semisimple) matrices which all commute with each other. When diagonalizable matrices commute then they can be simultaneously diagonalized.

This is basically because if $A$ and $B$ are commuting matrices, then the action of $B$ preserves the $\lambda$-eigenspace for $A$. So you pick out a basis of eigenvectors for $B$ in the space of $\lambda$-eigenvectors for $A$ for every $\lambda$, then both $A$ and $B$ are diagonal with respect to this basis.

So if you are with me so far, then you are basically done — there is a basis of $\mathfrak g$ so that for any $h\in \mathfrak h$, the matrix of $\text{ad}(h)$ is diagonal. That is exactly what that decomposition means — each root space is simultaneously an eigenspace for each $h$ in $\mathfrak h$.

The fact that $\text{ad}(h)$ for $h\in\mathfrak h$ always semisimple is little bit subtle. The crux of it is that you can define what it means for an element $s$ in a semisimple Lie algebra to be `ad-semisimple', meaning that $\text{ad}(s)$ is a semisimple endomorphism (i.e. it is diagonalizable). In fact, every element in a semisimple Lie algebra can be decomposed into the sum of an ad-semisimple element with a ad-nilpotent element such that the two commute — and this is unique. This is called the abstract Jordan decomposition.

A toral subalgebra is defined to be an algebra containing only semisimple elements. A Cartan subalgebra is defined to be a maximal toral subalgebra. These are proven to exist and be nonzero in semisimple Lie algebras and all the properties of Cartan subalgebras can be derived from there.

It should be noted that this is just one approach to Cartan subalgebras. There are some equivalent definitions and different approaches to the theory.


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# Table cross-chapter reference works fine but shows “??”

When I reference a table in a posterior chapter in my thesis, instead of the correct table number it puts ??. However, the link works correctly, and the table appears in the List of Tables fine.

I reference it as:

\ref{tab:datasets}

The table is:

\begin{table}[h] \label{tab:datasets} ...

The log shows

Reference 'tab:datasets' on page 16 undefined

How can I make Latex show the proper table number?

• Put the label behind the caption. – Ulrike Fischer Jun 24 '15 at 14:26
• And you need to compile at least twice. The label number is written to the .aux file and read at the beginning (\begin{document}) of the next LaTeX run. – Heiko Oberdiek Jun 24 '15 at 14:31
• @HeikoOberdiek: As the table appears in the list of table and links work I think one can assume that the document has been compiled twice. – Ulrike Fischer Jun 24 '15 at 14:54
• @UlrikeFischer That does not explain the question marks. The \label could be added at a later time, the the auxiliary files already contain the caption, but not yet the label. – Heiko Oberdiek Jun 24 '15 at 15:01
• @UlrikeFischer gave the correct answer. – CrystalCalculator Jun 24 '15 at 15:33

You need to run the program twice, one after another. And before that, be sure you have used the standard \label-\ref system: \label{labelname} and \ref{labelname}. In your case \label{tab:datasets} and \ref{tab:datasets}. Try to avoid using the same "label name" more times.


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# Kernel approximation

A page where I document what I don’t know about kernel approximation. A page about what I do know would be empty.

What I mean is: approximating implicit Mercer kernel feature maps with explicit features; Equivalently, approximating the Gram matrix, which is also related to mixture model inference and clustering. I’m especially interested in how it might be done using random linear algebra via random embeddings such as random projections.

Also I guess we need to know trade-offs of computational time/space cost versus approximation bounds, so that we can decide when to bother. When is it enough to reduce computational cost merely with support vectors, or to evaluate the kernels efficiently using, e.g. the Fast Gauss Transform and related methods, rather than coming up with alternative kernel bases? (e.g. you don’t want to do the fiddly coding for the Faust Gauss transform)

Not the related but distinct (terminological collision) of approximating functions from [mixtures of kernels](./mixlow-dimensional random projections. while commonly used to approximate convolution kernels, can also approximate certain Mercer kernels but it’s not what I mean here - fast multipole methods gives you an approximation to the field generated by all kernels and all the support points In kernel approximation we look for approximations, in some sense, to the component kernels themselves.

## Stationary kernels

I am mostly interested in translation-invariant kernels, so assume I’m talking about those unless I say otherwise.

Short introduction at scikit-learn kernel approximation page.

The approximations might be random projection, or random sampling based, e.g. the Nyström method, which is reportedly effectively Monte Carlo integration, although I understand there is an optimisation step too?

I need to work out the difference between random Fourier features, random kitchen sinks, Nyström methods and whatever Smola et al call their special case Gaussian approximation. I think Random Fourier features are the same as random kitchen sinks, and Nyström is different . When we can exploit (data- or kernel-) structure to do better? (say, ) Quasi Monte Carlo can improve on random Monte Carlo? (update: someone already had that idea: ) Or better matrix factorisations?

Recently, Maji and Berg and Vedaldi and Zisserman proposed explicit feature maps to approximate the additive kernels (intersection, $$\chi^2$$, etc.) by linear ones, thus enabling the use of fast machine learning technique in a non-linear context. An analogous technique was proposed by Rahimi and Recht for the translation invariant RBF kernels. In this paper, we complete the construction and combine the two techniques to obtain explicit feature maps for the generalized RBF kernels.

## Inner product kernels

See e.g. . These have some elegant relations to low-dimensional random projections. How well do these work with NN kernels?

## Implementations

This is the LIBASKIT set of scalable machine learning and data analysis tools. Currently we provide codes for kernel sums, nearest-neighbors, kmeans clustering, kernel regression, and multiclass kernel logistic regression. All codes use OpenMP and MPI for shared memory and distributed memory parallelism.

[…] PNYSTR : (Parallel Nystrom method) Code for kernel summation using the Nystrom method.

## Connections

MDS and Kernel PCA and mixture models are all related in a way I should try to understand when I have a moment.

For the connection with kernel PCA, see and for metric multidimensional scaling.

use projection tricks to calculate Sinkhorn divergence.

## References

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Le, Quoc, Tamás Sarlós, and Alex Smola. 2013. “Fastfood-Approximating Kernel Expansions in Loglinear Time.” In Proceedings of the International Conference on Machine Learning. http://www.jmlr.org/proceedings/papers/v28/le13-supp.pdf.
Minh, Ha Quang, Partha Niyogi, and Yuan Yao. 2006. “Mercer’s Theorem, Feature Maps, and Smoothing.” In International Conference on Computational Learning Theory, 154–68. Lecture Notes in Computer Science. Springer. https://doi.org/10.1007/11776420_14.
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# Find the area of the region enclosed by one loop of the curve. r = 7 cos 8{eq}\theta {/eq}

## Question:

Find the area of the region enclosed by one loop of the curve.

r = 7 cos 8{eq}\theta {/eq}

## Area Under Curves:

Area of the region enclosed by the curve is given by the definite integration. For a curve having farthest point from the origin is r, then the area enclosed by the curve is:

{eq}A = \frac{1}{2} \int r^2 d \theta {/eq}

Become a Study.com member to unlock this answer!

Here, we use some trigonometric formulas:

{eq}sin^2 \theta + cos^2 \theta = 1 \\ sin^2 \theta = \frac{1-cos2\theta}{2} {/eq}

One loop is created by...

Evaluating Definite Integrals Using the Fundamental Theorem

from

Chapter 16 / Lesson 2
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In calculus, the fundamental theorem is an essential tool that helps explain the relationship between integration and differentiation. Learn about evaluating definite integrals using the fundamental theorem, and work examples to gain understanding.


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# Finding the Equation of Tangent line on cos wave

1. Aug 31, 2010

### slash123

Finding the Equation of Tangent line on cos wave!!!

Find the equation of the normal line to y = 2cos ( 4x) at x = $$\pi$$ / 3
I dont even know where to start with this question, i have searched the text book and internet, help would be appreciated.

2. Aug 31, 2010

### Staff: Mentor

Re: Finding the Equation of Tangent line on cos wave!!!

So tangent or normal?

Tangent should be quite easy - what do you know about value of derivative at a point?

What do you know about normal and tangent? Any specific angle between them?

3. Aug 31, 2010

### slash123

Re: Finding the Equation of Tangent line on cos wave!!!

its finding the normal sorry, should of said that instead,, what ive put in as the question is all the questions asks.

4. Aug 31, 2010

### Staff: Mentor

Re: Finding the Equation of Tangent line on cos wave!!!

It doesn't change much - try to answer questions from my previous post.

Last edited by a moderator: Aug 13, 2013
5. Aug 31, 2010

### HallsofIvy

Staff Emeritus
Re: Finding the Equation of Tangent line on cos wave!!!

Since this problem doesn't have anything to do with differential equations, I am moving it to "Calculus".

6. Aug 31, 2010


text:
# Totally different adsorption-degradation habits of methylene blue and Congo pink in nanoceria/H 2 O 2 system below alkaline situations

### Characterization of nanoceria

Nanoceria was synthesized by way of a hydrothermal course of. The morphologies and sizes of nanoceria had been studied by transmission electron microscopy (TEM). As in our earlier research30,31, the CeO2 nanocubes had uniform cubic shapes with a dimension of 20–30 nm (Fig. 1a), whereas the CeO2 nanorods had diameters of roughly 15–20 nm and lengths of 100–200 nm (Fig. 1b). Dynamic mild scattering evaluation was carried out to detect the agglomeration dimension of nanoceria in pH 9.zero of aqueous resolution. As proven in Fig. S1, each of nanocubes and nanorods are inclined to agglomerate into bigger particles with related dimension.

Determine 1

TEM photos of CeO2 nanocubes (a) and nanorods (b).

The section purity and crystal construction of CeO2 nanocubes and nanorods had been recognized by XRD (Fig. 2a). All of the diffraction peaks of each samples may be listed to the pure fluorite construction of CeO2 (JCPDS 34–0394). In the meantime, the nanocubes exhibited sharper XRD diffraction peaks than these of the nanorods, indicating the next crystallinity and larger crystallite dimension of the nanocubes32. The floor chemistry of nanoceria was analyzed by XPS. The XPS survey scan is given in Fig. S2. The potential for look of sodium on the floor of nanoceria may be dominated out as a result of no peaks belonging to sodium had been detected. The Ce(3d) XPS spectra are proven in Fig. 2b. Peak vo, v’, uo and u’ belong to Ce3+ species, whereas v, v′′, v′′′, u, u′′ and u′′′ are derived from Ce4+. The relative focus of Ce3+ on the floor of nanoceria was calculated as follows33:

$$[C^3+]=fracA_+A_+A_+A_+A_v+A_v^ +A_+A_u+A_u^ +A_occasions 100 %$$

(1)

the place Ai is the built-in space of peak ‘i’.

Determine 2

XRD patterns (a), Ce3d XPS spectra (b) and the zeta potential as a operate of pH of CeO2 nanorods and nanocubes (c).

The relative focus of Ce3+ on the floor of nanorods was 42.9%, which was a lot bigger than that of nanocubes (36.5%). The O(1 s) spectra for each samples are proven in Fig. S3. The height with a binding vitality of round 529 eV may be ascribed to the lattice oxygen species of bulk CeO2 (Olat); the height with a binding vitality of round 531 eV is attributed to floor chemisorbed oxygen (Osur) and the height at round 533 eV is assigned to molecular water adsorbed on the floor34,35,36. In keeping with the literature34,35,36, the relative focus of Osur is estimated from the relative areas of peaks. The ratios of Osur/(Olat + Osur) are roughly 50.6% and 59.four% for nanocubes and nanorods, respectively. The upper proportion of chemisorbed oxygen in nanorods could as a consequence of their larger Ce3+ focus. As a result of Ce3+ in nanoceria can generate oxygen vacancies after which facilitate oxygen adsorption37.

The zeta potential of nanoceria was decided as a operate of pH, and the isoelectric factors (PI) of nanocubes and nanorods was at roughly pH four.5 and 5.eight, respectively (Fig. 2c). The particular floor areas of nanoceria had been measured by nitrogen fuel adsorption/desorption isotherm and calculated by BET methodology. The nanorods exhibit a bigger floor space (89.9 m2 g−1) than that of nanocubes (39.7 m2 g−1).

### Adsorption of MB and CR on nanoceria within the absence of H2O2

To grasp the decolorization of MB and CR in nanoceria/H2O2 system, the adsorption of dyes on nanoceria floor was surveyed within the absence of H2O2 at first, and the adsorbed quantity of dye was decided after 30 min of equilibrium time. The quantity of MB adsorbed on CeO2 nanorods and nanocubes was solely zero.01 mg m−2 and zero.11 mg m−2, respectively. The outcomes confirmed that the adsorption of MB on CeO2 floor was very weak. Curiously, a considerable amount of negatively charged CR was adsorbed: zero.78 mg m−2 on the nanorods floor and zero.39 mg m−2 on the nanocubes floor (Fig. S4a).

### Decomposition of H2O2 over CeO2

The catalytic H2O2 decomposition over CeO2 nanorods and nanocubes was investigated at 25 °C. As proven in Fig. three, 82.7% of 20 mM H2O2 was decomposed after 120 min within the presence of nanorods, whereas solely 38.eight% within the presence of nanocubes below the identical situations. Usually, the catalytic exercise of nanoceria is straight associated to its floor chemical states and particular floor space11. The excessive focus of Ce3+ facilitates the adsorption and decomposition of H2O24,5,42. Therefore, the excessive exercise of nanorods was owing to the presence of extra Ce3+, as proved by XPS evaluation.

Determine three

H2O2 decomposition over CeO2 nanorods and nanocubes (1.zero g L−1 CeO2, 20 mM H2O2, 25 °C).

To probe the oxidative species in the course of the catalytic decomposition of H2O2 over nanoceria, EPR spectroscopy measurement was carried out utilizing DMPO as a spin entice. As proven in Fig. S5, a typical sign of DMPO-OH adducts (1:2:2:1 quartet) was detected for a suspension of nanorods upon the addition of H2O2, which recommended the era of HO· within the catalytic decomposition of H2O2. Related outcomes had been obtained for a suspension of nanocubes, besides the low depth of the DMPO-OH peaks. Fig. S6 confirmed the Raman spectra of CeO2 nanorods earlier than and after H2O2 remedy at pH 9.zero. The looks of band at 830 cm−1 signifies the presence of the η2-peroxide (O22-) species on the floor of CeO2 nanorods after H2O2 treatment8,43,44. These outcomes recommend that the presence of HO· and peroxide-like intermediates in nanoceria/H2O2 system at pH 9.zero. Hamoud et al.13 recommended the connection between floor Ce(IV) peroxo species and HO· radicals manufacturing:

$$C^+fracthree__to C^four+(_)^2-+_O+^+$$

(2)

$$C^four+(_)^2-+2^+to C^four++2HOcdot$$

(three)

### Decolorization of MB in nanoceria/H2O2 system

#### Impact of H2O2 focus on decolorization of MB

Pre-experiments confirmed that little change in MB focus was noticed with H2O2, nonetheless, upon introduction of H2O2 into suspension of nanoceria with MB, the colour of nanocubes and nanorods grew to become deep blue and inexperienced in 5–10 min, respectively (Fig. S7). Clearly, the adsorption of MB on nanoceria was enhanced within the presence of H2O2. And the decolorization of MB in nanoceria/H2O2 system may be ascribed to each adsorption and oxidative degradation. Then the whole quantity of MB (labeled as MBt) in suspension was divided into three components: (1) free MB (labeled as MBf) in supernatant; (2) adsorbed MB (labeled as MBa) on nanoceria floor; and (three) degraded MB (labeled as MBd) in nanoceria/H2O2 system, which was decided by the distinction within the quantity of MBt, MBf and MBa:

$$M_=M_-M_-M_a$$

(four)

When the focus of H2O2 was 20 mM, the decolorization of MB after 30 min in each nanoceria/H2O2 programs was investigated. Within the case of nanorods, 80.three% of MB was decolorized; within the case of nanocubes, the decolorization was 76.2% (Fig. four). Nevertheless, when the nanoceria was remoted from each programs and added into pH three.zero HCl aqueous resolution, desorption occurred and the quantity of desorbed MB was nearly equal to the quantity of decolorization. The outcomes clearly recommended that the decolorization of MB in nanoceria/H2O2 system was primarily as a consequence of adsorption relatively than degradation at 30 min. In the meantime, the improved adsorption of MB on nanoceria floor attributable to the presence of H2O2 was reversible and could possibly be completely desorbed utilizing pH three.zero HCl aqueous resolution because the eluting agent.

Determine four

Distribution of MB in nanoceria/H2O2 system at 30 min.

The affect of H2O2 focus on MB adsorption on nanoceria floor was additional investigated. The contact time was restricted to 30 min to keep away from the interference of degradation of MB. As proven in Fig. 5, upon the introduction of 5 mM H2O2, the adsorbed quantity of MB on CeO2 nanorods and nanocubes elevated by 6.6 and 1.three occasions in 30 min, respectively. Within the case of nanorods, the adsorbed quantity of MB elevated with growing H2O2 focus from 5 mM to 20 mM after which remained nearly unchanged. Nevertheless, additional improve of the H2O2 focus didn’t have a big affect on the adsorption of MB within the presence of nanocubes.

Determine 5

The impact of H2O2 focus on the adsorption of MB on nanoceria floor (15 mg L−1 MB, 1.zero g L−1 CeO2, 25 °C, pH 9.zero).

The impact of H2O2 on the zeta potential of nanoceria at pH 9.zero was investigated. It’s clear that the addition of H2O2 made the zeta potential of CeO2 nanorods extra damaging, and absolutely the worth of the zeta potential elevated from 5 mM to 20 mM of H2O2 (Fig. 2c). Nevertheless, the zeta-potential of CeO2 nanocubes didn’t change clearly within the presence of H2O2 (information not present). The elevated damaging zeta-potential of CeO2 nanorods could also be attributed to the formation of damaging floor peroxide species by way of Ce3+–H2O2 interactions43,45. The era of floor peroxide species elevated with growing H2O2 focus till the Ce3+ websites of nanoceria floor had been utterly occupied5,13. The unchanged zeta-potential of CeO2 nanocubes could also be defined by the presence of much less Ce3+ websites and better unique zeta-potential. As a result of the adsorption of MB on nanoceria floor is by way of electrostatic adsorption, we speculate that the negatively charged floor peroxide species on the floor of nanoceria could also be an necessary motive for the improved adsorption of MB on nanoceria, particularly on the nanorods.

#### Impact of contact time on MB decolorization

To additional examine the catalytic oxidation of MB in nanoceria/H2O2 system, the impact of contact time on the decolorization of MB was investigated. Determine 6a exhibits the UV–vis spectral modifications of the supernatant as a operate of time. Upon the addition of H2O2 to the suspension of nanorods/MB, the attribute band (centered at 664 nm) of MB decreased instantly, after which it continued to lower as much as 30 min, after which started to extend as much as 12 h. The UV–vis spectra confirmed that the focus of MB in supernatant had a particular change: it first decreased after which elevated.

Determine 6

Impact of contact time on MB decolorization in nanoceria/H2O2 system (15 mg L−1 MB, 1.zero g L−1 CeO2, 20 mM H2O2). (a) CeO2 nanorods/H2O2 system (The samples had been diluted 1:2 in pH 9.zero aqueous resolution earlier than examined). (b) CeO2 nanocubes/H2O2 system (the samples marked * had been diluted 1:2 in pH 9.zero aqueous resolution earlier than examined).

To grasp the decolorization of MB in CeO2 nanorods/H2O2 system, the distribution of MB in suspension at 12 h was analyzed and is proven in Fig. 7; most of MB was dissolved in supernatant, 15.eight% was degraded and solely four.eight% was adsorbed. These outcomes indicated that MB could possibly be degraded in suspension of nanocubes/H2O2 at pH 9.zero, though the speed of degradation was relatively sluggish. The particular change of MB focus in supernatant was the results of adsorption, desorption and degradation. However the change of MB focus in supernatant was primarily decided by adsorption and desorption as a result of the degradation fee of MB was low. In the course of the 12 h lengthy MB degradation response, adsorption dominated and the content material of MB in supernatant shortly decreased in the course of the first 30 min due to the formation of a considerable amount of floor peroxide species, after which MB was desorbed from nanoceria within the subsequent 11.5 h because of the decomposition of floor peroxide species with lengthy contact time. That was in accord with the remaining H2O2 focus in system. The 240–290 nm spectral vary, which corresponds to the H2O2 focus, confirmed a continuous lower in absorbance with growing response time (Fig. 6a).

Determine 7

Distribution of MB in nanoceria/H2O2 system (*: the addition of H2O2 was repeated at 12 h).

The sluggish oxidative means of MB was probably as a consequence of their stability and troublesome degradation. In the meantime, few floor peroxide species on nanorods, which was attributable to the excessive decomposition exercise for H2O2, was additionally one necessary motive. To additional affirm this assumption, we repeated this experiment and added H2O2 into the system once more at 12 h (the ultimate H2O2 focus at 12 h was nonetheless 20 mM). As anticipated, the proportion of degraded and adsorbed MB elevated to 41.four% and 13.four%, respectively (Fig. 7).

Nevertheless, a special phenomenon was noticed in nanocubes/H2O2 system. The absorbance band at 664 nm decreased promptly upon the addition of H2O2, after which it continued to lower steadily over the subsequent 32 h (Fig. 6b). The distribution of MB in suspension at 12 and 32 h had been proven in Fig. 7. The proportion of degraded MB elevated from 38.four% to 50.zero%, whereas that of the adsorbed MB decreased from 56.three% to 47.6%, respectively. It could possibly be noticed that a small quantity of MB dissolved in resolution. Thus, the change of MB focus in supernatant was the results of adsorption and degradation. This may be defined as a result of the catalytic exercise of nanocubes in direction of H2O2 decomposition was decrease than that of nanorods, and numerous H2O2 remained within the system (Fig. 6b). The excessive focus of H2O2 in suspension may repeatedly present the floor peroxide species, which induced the adsorption and degradation of MB over nanocubes.

The outcomes recommended that the adsorption and degradation of MB was tightly associated to the content material of floor peroxide species by way of Ce3+-H2O2 interplay. A excessive focus of H2O2 within the resolution would improve the adsorption and degradation of MB over the nanoceria floor. However the desorption of MB grew to become dominant after the focus of H2O2 was dramatically decreased.

### CR decolorization in nanoceria/H2O2 system

The decolorization means of CR in nanoceria/H2O2 system was additionally investigated at 25 °C. Management experiments confirmed that CR resolution was secure within the presence of H2O2. The UV−vis spectral modifications of the supernatant as a operate of time had been proven in Fig. S8. In presence of CeO2 nanorods, the depth of the attribute band of CR centered at 497 nm considerably decreased as a result of numerous CR adsorbed on the floor of nanorods. After the addition of H2O2, a fast and apparent improve (zero < t < 2 min) of the depth of the attribute band of CR was evident firstly of degradation. This function was additionally noticed for CeO2 nanocubes (Fig. S4b and S8). As beforehand reported, these options are largely assigned to the desorption of CR from the floor of CeO2 due to the adsorption competitors between CR and H2O210,13.

Determine eight

Take away of CR in nanoceria/H2O2 system (70 mg L−1 CR, 1.zero g L−1 CeO2, 20 mM H2O2).

### Reusability of CeO2 nanorods

The recyclability of CeO2 nanorods was evaluated by CR degradation on the situations of pH 9.zero, 70 mg L−1 of CR, 20 mM of H2O2 and 1.zero g L−1 of CeO2. As seen in Fig. S9, the degradation proportion of CR nearly stored unchanged throughout 5 successive runs after 2 h of response, indicating a superb reusability of CeO2 nanorods.

### Response mechanism dialogue

Within the nanoceria/H2O2 system, the character of the oxidative species performs a key position in oxidative degradation of natural compounds. In keeping with the literature8,9,10,47,48, Ce3+ on the floor of nanoceria may advanced with H2O2 and generate floor peroxide species13, which might decompose into hydroxyl radicals at low pH or straight act as an oxidative species below alkaline situations. We speculated that the floor peroxide species must also be the primary oxidative species within the present nanoceria/H2O2 system as a result of the experiments had been carried out at pH of 9.zero. Primarily based on the experimental observations, the adsorption-degradation processes of MB and CR are schematically illustrated in Fig. 9.

Determine 9

Schematic illustration for the doable mechanism of substrate-dependent Fenton-like exercise of nanoceria.

When MB was current in nanoceria/H2O2 system, MB could possibly be adsorbed on the floor of nanoceria by way of electrostatic attraction with floor peroxide species, which subsequently act because the catalytic lively websites for MB oxidation47,48. With respect to CR, the adsorption of CR on the nanoceria floor is thru Lewis acid–base interactions, and the adsorption competitors exists between CR and H2O213,14. The degradation of CR would happen by assault from the adjoining peroxide species, which might induce an intermolecular rearrangement of CR8,10 or degrade CR by way of a homolysis of the O-O bond to kind HO5.

In contrast with CeO2 nanocubes, CeO2 nanorods displayed the next catalytic exercise for the decomposition of H2O2 and generated extra floor peroxide species below the present situations. Subsequently, environment friendly degradation of CR was achieved in CeO2 nanorods/H2O2 system. Nevertheless, MB couldn’t be effectively degraded by CeO2 nanorods within the presence of H2O2. The totally different chemical construction of the natural dyes could also be one motive for the distinction within the degree of degradation as a result of the degradation of MB is mostly very troublesome14,49. Moreover, the excessive fee of H2O2 decomposition over the ceria nanorods would considerably lower the focus of floor peroxide species and the adsorbed MB, and therefore the oxidation fee of MB.


text:
Whole Numbers

Ex 2.3 Question 1.

Which of the following will not represent zero:
(a) 1 + 0
(b) 0 x 0
(c) 0/2
(d) 10 – 10/2
Solution:
(a) 1 + 0 = 1 ≠ 0, (Does not represent zero.)
(b) 0 x 0 = 0, (Represents zero)
(c) 0/2= 0, (Represents zero.)
(d) 10 – 10 / 2 = 0/2= 0 (Represents zero.)

Ex 2.3 Question 2.

If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Solution:
Yes, If we multiply any number with zero the resultant product will be zero.

Examples-
8x 0 = 0
0 x 5= 0

If both numbers are zero, then the result also be zero.

Example-

0 x 0 = 0

Ex 2.3  Question 3.

If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.
Solution:
This is only true when each of the numbers is 1.
Example- 1 x 1 = 1

If there is only one number is 1 then the product cannot be 1

Examples- 5 x 1 = 5 , 6 x 1 = 6

Ex 2.3  Question 4.

Find using distributive property:
(а) 728 x 101
(b) 5437 x 1001
(c) 824 x 25
(d) 4275 x 125
(e) 504 x 35
Solution:
(a) 728 x 101

= 728 x (100 + 1)
= 728 x 100 + 728 x 1
= 72800 + 728
= 73528

(b) 5437 x 1001

= 5437 x (1000 + 1)
= 5437 x 1000 + 5437 x 1
= 5437000 + 5437
= 5442437

(c) 824 x 25

= 824 x (20 + 5)
= 824 x 20 + 824 x 5
= 16480 + 4120
= 20600

(d) 4275 x 125

= 4275 x (100 + 20 + 5)
= 4275 x 100 + 4275 x 20 + 4275 x 5
= 427500 + 85500 + 21375
= 534375

(e) 504 x 35

= (500 + 4) x 35
= 500 x 35 + 4 x 35
= 17500 + 140
= 17640

Ex 2.3 Question 5.

Study the pattern:
1 x 8 + 1= 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
Solution:
Step I: 123456 x 8 + 6 = 987654
Step II: 1234567 x 8 + 7 = 9876543

Working pattern:
(1) x 8 + 1 = 9
(12) x 8 + 2 = (11 + 1) x 8 + 2 = 98
(123) x 8 + 3 = (111 + 11 + 1) x 8 + 3 = 987
(1234) x 8 + 4 = (1111 + 111 + 11 + 1) x 8 + 4 = 9876
(12345) x 8 + 5 = (11111 + 1111 + 111 + 11 + 1) x 8 + 5 = 98765

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