Question 304: Dual Nature of Matter and Radiation

Question

When a photosensitive surface is irradiated by lights of wavelengths ' $\lambda_1$ ' and ' $\lambda_2$ ', kinetic energies of the emitted photoelectrons is ' $E_1$ ' and ' $E_2$ ' respectively. The work function of the photosensitive surface is

Accepted Answer

$$\begin{aligned}
&	ext { From Einstein's photoelectric equation, }\
&\begin{array}{ll} 
& \mathrm{E}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 \
	herefore \quad & \mathrm{E}_1 \lambda_1=\mathrm{hc}-\mathrm{W}_0 \lambda_1 \
& \mathrm{E}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \
	herefore \quad & \mathrm{hc}=\mathrm{E}_1 \lambda_1+\mathrm{W}_0 \lambda_1\quad	ext{.... (i)}
\end{array}
\end{aligned}$$
$$\begin{aligned}
&\begin{array}{ll}
	herefore \quad & \mathrm{E}_2 \lambda_2=\mathrm{hc}-\mathrm{W}_0 \lambda_2 \
& \Rightarrow \mathrm{hc}=\mathrm{E}_2 \lambda_2+\mathrm{W}_0 \lambda_2\quad	ext{.... (ii)}
\end{array}\
&	ext { From equations (i) and (ii), }\
&\begin{array}{ll} 
& E_1 \lambda_1+W_0 \lambda_1=E_2 \lambda_2+W_0 \lambda_2 \
	herefore & E_1 \lambda_1-E_2 \lambda_2=W_0\left(\lambda_2-\lambda_1ight) \
	herefore & W_0=\frac{E_1 \lambda_1-E_2 \lambda_2}{\lambda_2-\lambda_1}
\end{array}
\end{aligned}$$

Answer

$\frac{\left(E_2 \lambda_2-E_1 \lambda_1\right)}{\left(\lambda_2-\lambda_1\right)}$

Answer

$\frac{\left(E_1 \lambda_1+E_2 \lambda_2\right)}{\left(\lambda_2-\lambda_1\right)}$

Answer

$\frac{\left(E_1 \lambda_1-E_2 \lambda_2\right)}{\left(\lambda_2-\lambda_1\right)}$

Answer

$\frac{\left(E_2 \lambda_2+E_1 \lambda_1\right)}{\left(\lambda_1-\lambda_2\right)}$

Dual Nature of Matter and Radiation

When a photosensitive surface is irradiated by lights of wavelengths ' \lambda_1 ' and ' \lambda_2 ', kinetic energies of the emitted photoelectrons is ' E_1 ' and ' E_2 ' respectively. The work function of the photosensitive surface is

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