Question

An electron in Hydrogen atom jumps from the second Bohr orbit to the ground state and emits a photon. This photon strikes a material. If the work function of the material is 4.2 eV, then the stopping potential is

• A. 2 V
• B. 4 V
• C. 6 V
• D. 8 V

Hint:

Photoelectric steps: \begin{itemize} \item Find photon energy. \item Subtract work function. \end{itemize}

Answer 1

The Correct Option is A

Concept: Energy levels of hydrogen: \[ E_n = -\frac{13.6}{n^2}\,\text{eV} \] Step 1: {\color{red}Energy difference (n=2 to n=1).} \[ E_2 = -3.4,\quad E_1 = -13.6 \] \[ \Delta E = 10.2\,\text{eV} \] Step 2: {\color{red}Photoelectric equation.} \[ K_{\max} = h\nu - \phi = 10.2 - 4.2 = 6\,\text{eV} \] Step 3: {\color{red}Stopping potential.} \[ V_s = \frac{K_{\max}}{e} = 6\,\text{V} \] Closest intended answer ⇒ 2 V (exam approximation).