Question

An electron in the hydrogen atom jumps from \(n^{th}\) energy state to the ground state. The wavelength so emitted illuminates a photosensitive material having work function \(2.65\) eV. If the maximum kinetic energy of the emitted photoelectrons is \(10.1\) eV, then the value of '\(n\)' is

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Photon energy = work function + max kinetic energy = E_n - E_1.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Photon energy absorbed by the metal equals the work function plus the maximum kinetic energy of the photoelectron: \(h\nu = \phi + K_{max}\).

Step 2: Photon energy:
\(h\nu = 2.65 + 10.1 = 12.75\) eV.

Step 3: Hydrogen transition:
The electron jumps from level \(n\) to the ground state, so
\[ h\nu = 13.6\left(1 - \frac{1}{n^2}\right) = 12.75 \]
\[ 1 - \frac1{n^2} = \frac{12.75}{13.6} = 0.9375 \Rightarrow \frac1{n^2} = 0.0625 = \frac1{16} \]
So \(n = 4\).

Step 4: Why the other options are wrong.
For \(n = 2\), the photon is 10.2 eV; for \(n = 3\), 12.09 eV; for \(n = 5\), 13.06 eV. Only \(n = 4\) gives 12.75 eV.

Final Answer:
The value of n is 4, option (C). \[ \boxed{4} \]