Radiation of wavelength $\mathrm{\lambda }$ is incident on a photocell. The fastest emitted photoelectron has a speed $\mathrm{v}$. If the wavelength is changed to $\frac{3\mathrm{\lambda }}{4}$, the speed of the fastest emitted photoelectron will be

• A

  $=\mathrm{v}{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$
• B

  $=\mathrm{v}{\left(\frac{3}{4}\right)}^{\frac{1}{2}}$
• Correct answer
  C

  $>\mathrm{v}{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$
• D

  $<\mathrm{v}{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

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