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
• 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
• C. Total number of radial nodes in all the orbitals of $n^{th}$ shell is 14
• D. Total number of angular nodes in all the orbitals of (n - 1) shell is 13

Answer 1

Answer: (C), (D)

Total number of angular nodes in all the orbitals of (n - 1) shell is 13

Number of photons emitted = 6
So,
$\frac{n\times (n-1)}{2}=6$
$n=4$
So excited state is $3^{rd}orn=4$

Photon having highest energy will be $4\to 1$
So its energy will be
= $13.6\times (\frac{1}{1^2}-\frac{1}{4^2})=12.75$

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 - $4\to 2$, $3\to 2$.
C , D are incorrect because number of nodes in $n^{th}$ 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.