Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass $m$ are $V_1$ and $V_2$ respectively,then ($h=$ Planck's constant):

  • A
    $V_1+V_2=\left[\frac{2h}{m}(n_1+n_2)\right]^{1/2}$
  • B
    $V_1-V_2=\left[\frac{2h}{m}(n_1-n_2)\right]^{1/2}$
  • C
    $V_1^2+V_2^2=\frac{2h}{m}(n_1+n_2)$
  • D
    $V_1^2-V_2^2=\frac{2h}{m}(n_1-n_2)$

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Similar Questions

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength $(\lambda)$ of incident light and the corresponding stopping potential $(V_0)$ are given below:
$\lambda (\mu m)$$V_0$ (Volt)
$0.3$$2.0$
$0.4$$1.0$
$0.5$$0.4$

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