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1

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are exited to states with principal quantum number n = 6 ? The value of X is ______________.

## Explanation

No. of different wavelengths = $${{n(n - 1)} \over 2}$$

$$= {{6 \times (6 - 1)} \over 2} = {{6 \times 5} \over 2} = 15$$
2

### JEE Main 2021 (Online) 27th July Morning Shift

Numerical
A particle of mass 9.1 $$\times$$ 10$$-$$31 kg travels in a medium with a speed of 106 m/s and a photon of a radiation of linear momentum 10$$-$$27 kg m/s travels in vacuum. The wavelength of photon is __________ times the wavelength of the particle.

## Explanation

For photon $${\lambda _1} = {h \over P} = {{6.6 \times {{10}^{ - 34}}} \over {{{10}^{ - 27}}}}$$

For particle $${\lambda _2} = {h \over {mv}} = {{6.6 \times {{10}^{ - 34}}} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$$

$$\therefore$$ $${{{\lambda _1}} \over {{\lambda _2}}} = 910$$
3

### JEE Main 2021 (Online) 27th July Morning Shift

Numerical
In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 $$\mathop A\limits^o$$. If the speed of electron is 2.2 $$\times$$ 166 m/s, then the current associated with the electron will be _____________ $$\times$$ 10$$-$$2 mA. [Take $$\pi$$ as $${{22} \over 7}$$]

## Explanation

$$I = {e \over T} = {{e\omega } \over {2\pi }} = {{eV} \over {2\pi r}}$$

$$I = {{1.6 \times {{10}^{ - 19}} \times 2.2 \times {{10}^6} \times 7} \over {2 \times 22 \times 0.5 \times {{10}^{ - 10}}}}$$

= 1.12 mA

= 112 $$\times$$ 10$$-$$2 mA
4

### JEE Main 2021 (Online) 25th July Evening Shift

Numerical
The nuclear activity of a radioactive element becomes $${\left( {{1 \over 8}} \right)^{th}}$$ of its initial value in 30 years. The half-life of radioactive element is _____________ years.

## Explanation

We know, $$A = {A_0}{e^{ - \lambda t}}$$

For half life

$${{{A_0}} \over 2} = {e^{ - \lambda {t_{1/2}}}}$$

$$\Rightarrow$$ $${\lambda {t_{1/2}}}$$ = ln 2 .....(1)

And when radioactive element becomes $${\left( {{1 \over 8}} \right)^{th}}$$ of its initial value in 30 years

$${{{A_0}} \over 8} = {A_0}{e^{ - \lambda \times 30}} \Rightarrow \lambda \times 30 = \ln 8$$

$$\Rightarrow$$ 30$$\lambda = 3\ln 2$$

$$\Rightarrow$$ $$\lambda = {{3\ln 2} \over {30}}$$ .....(2)

Putting value of $$\lambda$$ in (1), we get

$${{3\ln 2} \over {30}} \times {t_{1/2}}$$ = ln 2

$$\Rightarrow$$ $${t_{1/2}}$$ = 10 years

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