An electron in a hydrogen atom jumps from second Bohr orbit to ground state and the energy difference of the two states is radiated in the form of photons. These are then allowed to fall on a metal surface having a work-function equal to $\text{4.2} \mathrm{eV}$, then the stopping potential is [Energy of electron in ${\mathrm{n}}^{\mathrm{th}}$ orbit $=-\frac{13.6}{{\mathrm{n}}^2} \mathrm{eV}$]

• A

  $2 \mathrm{V}$
• B

  $4 \mathrm{V}$
• Correct answer
  C

  $6 \mathrm{V}$
• D

  $8 \mathrm{V}$

Hence, the correct answer is option: 'C'.