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A monochromatic beam of light ...

Updated On: 20-06-2022

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Text Solution

may be in the first excited statemay be in the second excited statemay be in both first and second excited stateNone of above.

Solution : As photon energy of incident light is not equal to energy difference between two energy state, that is why sample A does not absorbe any photon and therefore remain in ground state, and hence , no emission spectra. Thus sample B remain as it is, but it will be de-excite itself to ground state by emitting radiations of energy 10.2 eV <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V04_C04_E01_280_S01.png" width="80%"> <br> when photon energy is replaced by electron beam, then atom of sample A can absorb either 10.2 eV ("to reach" 1^(st) excited state) of 12.1 eV ("to reach" 2^(nd) "excited state") In emission spectra A , we will have 3 time corresponding to n = 2 to 1, n = 3 to 2 and n = 3 to1 having energies equal to 10.2 eV, 1.9 eV, 12.1 eV, respectively <br> As least energy of this emission spectra is corresponding to transition from n = 2 to 3 and sample B is in 1^(st) excited state B can be excited to some higher energy level and if B absorbs energy corresponding to n = 1 to 2 and n = 1 to 3`, then it may ionize.

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Transcript

hello letter start with other question in this question what is given a monochromatic beam of light having Photon energy 12.5 electron volt is incident on a Honours sample a of atomic hydrogen ola in the ground state ok the emission spectra obtained from the sample is incident on another sample be of atomic hydrogen in which all atoms are in the first excited state based on the information we need to check whether the atom of sample after passing of light through it may be in the first excited in the second Exide for maybe in both first and second excited state or none of the above OK so how is the situation if we draw our situation that is just draw for example this is one of the sample is simple and let's say this is another sample sample be ok but there is a photon of energy which is given as to be 12.5 electron volt this is incident on sample and on sample be 9 sample a all the atoms

hydrogen atom in the ground state and here it is in the first excited state ok no photo ground is redmi Note 8 and value is equal to 1 and for the first excited state and value is equal to 2 for the fox first excited state but in this awareness about what will happen ok now in the ground state if we draw the various energy levels for example and equals to 1 OK then equals to 2 then equals to if we draw 3D letters A-Z draw any cost 24 ok like this these are the various energy levels let's say this is an equal to 1 this is to this is let's 83 this is 4 and 7 ok no corresponding to the ground state what is the energy is it is -13.6 Sofia writing this values in terms of electron volt ok so here it would be in any way to do it would be fine 3.4 it would be -1.5 then it would be -0.8

434 approximately point 844 these are the values in terms of electron in the sample electron is in the ground state from here the transition to take less what we are doing we are in incident in a photo on which is having energy of 12.5 electron volt the what will happen so either the electron would jump into this or it can jump hear what it can be the transition to an equal to four so these other possibility and Swan ok to an equal to 5 now in the ground state it is having energy of -13.6 ok no it is it would gain energy of -12.5 so the total energy would become it would be -13.6 this was the energy which it was already carrying and this is due to the energy of incident Photon so this value would turn out to be -1.1 electron volt that is not the electron is having this atoms are having energy of -1.1 electron volt ok

corresponding to 1.1 electron volt there is no energy level which is having energy of -1.1 electron volt because here are the energies contest so it cannot be it cannot have any value so there cannot be anything but energy Sare discrete of packets are the levels are in this excited states are in packets so there will be either 13.6 it would be -3.4 -1.5 -142 corresponding to this -1.5 there is no no shell so it cannot remain here because this -1.1 would be between an equal to 4 and equal to 3 but this is not possible it cannot remain between the shell so that means there is no that what we need to find the atoms of sample after passing through that none of the above so it would be in the first excited it cannot be it cannot be in the second it cannot be both the first and excited so would be the correct option because they would be remain in the ground state OK then what happened so hope you understood

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