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Question

A hydrogen atom has an electrons in a particular excited state 'n', when it returns to the ground state, 6 different photons are emitted. Which of the following is/are incorrect:

A
Out of the 6 different photons only 2 photons are in the visible light region
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B
If highest energy proton emitted from the above sample is incident on the metal plate having work function 8 eV, KE of liberated photo-electron may be equal to or less than 4.75 eV
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C
Total number of radial nodes in all the orbitals of nth shell is 14
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D
Total number of angular nodes in all the orbitals of (n - 1) shell is 13
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Solution

The correct option is D Total number of angular nodes in all the orbitals of (n - 1) shell is 13
Number of photons emitted = 6
So,
n × (n1)2=6
n=4
So excited state is 3rd or n=4

Photon having highest energy will be 41
So its energy will be
= 13.6× (112142)=12.75

When it is incident on plate having work function 8 ev
KE = 12.75 - 8 = 4.75

KE will be equal to this value or may be lower if the electron is an inner electron. So, option B is right, but we are looking for incorrect anwers.
Option A is right too as two transitions happen in the Balmer series - 42, 32.
C , D are incorrect because number of nodes in nth and (n - 1) shell are 6 (4s, 4p and 4d; 3+2+1+0 radial nodes) and 3 (3s, 3p, 3d; 2+1+0 Angular nodes) respectively.


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