Question

Hydrogen atoms in a particular excited state 'n', when all returned to ground state, 6 different photons are emitted. Which of the following is/are incorrect?

• A. Out of 6 different photons, only 2 photons have speed equal to that of visible light.
• B. If highest energy photon 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 in $(n-1{)}^{th}$ shell is 13.

Answer 1

Answer: (A), (C), (D)

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

Number of photons emitted = $6$
So, $\frac{n(n-1)}{2}=6;n=4$
Excited state is $3^{rd}$
Photon having highest energy will get $4\to 1$
So, its energy will be $=13.6(\frac{1}{1^2}-\frac{1}{4^2})=13.6\times \frac{15}{16}=12.75eV$
When it is incident on plate having work function $8eV$ then,
$KE=12.75-8=4.75eV$
$KE$ will be equal to this value or may be less if electron is inner electron.
So option (b) is correct.
Option (a) is incorrect because all photon have equal velocity which is $3\times {10}^{8}m/s$ .

Radial node = n-l-1
For 4th shell,
Radial node of 4s = 3
Radial node of 4p = 6
Radial node of 4d = 5
Totally 4th shell has 14 radial node.

Angular node = l
Angular node of 3s = 0
Angular node of 3p = 3
Angular node of 3d = 10
Totally 3rd shell has 13 angular node.
Options (c) and (d) are also correct statements.