When a metal surface is irradiated with light of frequency $x \mathrm{~Hz}$, the kinetic energy of emitted photoelectrons is $z \mathrm{~J}$. When the same metal is irradiated with light of frequency $y \mathrm{~Hz}$, the kinetic energy of emitted electrons is $\frac{\mathrm{z}}{3} \mathrm{~J}$. What is the threshold frequency (in Hz ) of metal?

• A

  $\frac{3}{2}(y-x)$
• Correct answer
  B

  $\left(\frac{3 y-x}{2}\right)$
• C

  $\left(\frac{2 y-x}{3}\right)$
• D

  $\frac{2}{3}(y-x)$

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