Question 268: Dual Nature of Matter and Radiation

Question

When a metal surface is illuminated by light of wavelength $\lambda_1$ and $\lambda_2$, the maximum velocities of photoelectrons ejected are V and 2 V respectively. The work function of the metal is ( $\mathrm{h}=$ Planck's constant, $\mathrm{c}=$ velocity of light, $\lambda_1>\lambda_2$ )

Accepted Answer

Given that: $\quad \mathrm{V}_{	ext {max }_1}=\mathrm{V}$ and $\mathrm{V}_{	ext {max }_2}=2 \mathrm{~V}$ From Einstein's photoelectric equation,
$$ \mathrm{E}=\frac{\mathrm{hc}}{\lambda_1}-\phi_0\,\,\,\,\,\,\, \ldots\left(	ext { where } E=\frac{1}{2} m V_{\max }^2ight)$$
From the given data, we get
$$ \begin{aligned} & \frac{1}{2} m V^2=\frac{h c}{\lambda_1}-\phi_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\ & \frac{1}{2} m(2 V)^2=\frac{4 h c}{\lambda_2}-4 \phi_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $$
Subtracting (i) from (ii),
$$ \begin{aligned} 	herefore \quad & \frac{\mathrm{hc}}{\lambda_1}-\frac{4 \mathrm{hc}}{\lambda_2}=-3 \phi_0 \ & \phi_0=\frac{\mathrm{hc}}{3}\left[\frac{4}{\lambda_2}-\frac{1}{\lambda_1}ight]=\frac{\mathrm{hc}}{3}\left[\frac{4 \lambda_1-\lambda_2}{\lambda_1 \lambda_2}ight] \end{aligned} $$

Answer

$\frac{\mathrm{hc}}{2 \lambda_1 \lambda_2}\left(\lambda_1-\lambda_2\right)$

Answer

$\frac{\mathrm{hc}}{\lambda_1 \lambda_2}\left(\lambda_1-\lambda_2\right)$

Answer

$\frac{\mathrm{hc}}{\lambda_1 \lambda_2}\left(\lambda_1+\lambda_2\right)$

Answer

$\frac{\mathrm{hc}}{3 \lambda_1 \lambda_2}\left(4 \lambda_2-\lambda_1\right)$

Dual Nature of Matter and Radiation

When a metal surface is illuminated by light of wavelength \lambda_1 and \lambda_2, the maximum velocities of photoelectrons ejected are V and 2 V respectively. The work function of the metal is ( \mathrm{h}= Planck's constant, \mathrm{c}= velocity of light, \lambda_1>\lambda_2 )

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