When a metallic surface is illuminated with radiation of wavelength $\mathrm{\lambda }$, the stopping potential is $\mathrm{V}$. If the same surface is illuminated with radiation of wavelength $2\mathrm{\lambda }$, the stopping potential is $\frac{\mathrm{V}}{4}$. The threshold wavelength for the metallic surface is:

• A

  $4\mathrm{\lambda }$
• B

  $5\mathrm{\lambda }$
• C

  $\frac{5}{2}\mathrm{\lambda }$
• Correct answer
  D

  $3\mathrm{\lambda }$

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