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: List photon energies to the ground state:
$n = 2$: $10.2$ eV. $n = 3$: $12.09$ eV. $n = 4$: $12.75$ eV. $n = 5$: $13.06$ eV.

Step 2: Energy needed:
The photon must carry $\phi + K_{max} = 2.65 + 10.1 = 12.75$ eV.

Step 3: Match:
The $n = 4$ to $n = 1$ transition gives exactly 12.75 eV.

Final Answer:
n equals 4, option (C). \[ \boxed{4} \]