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Initially a photon of wavelength $\lambda_1$ falls on photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to $\lambda_2$, the maximum energy of the electron emitted becomes $E_2$. Then value of $h c$ ( $h=$ Planck's constant, $c=$ velocity of light) is
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Verified Answer
The correct answer is:
$h c=\frac{E_1-E_2}{\lambda_2-\lambda_1} \cdot\left(\lambda_1 \lambda_2\right)$
From equation of photoelectric effect, we have
$$
\begin{aligned}
& E_1=\frac{h c}{\lambda_1}-W \\
& E_2=\frac{h c}{\lambda_2}-W
\end{aligned}
$$
where, $W$ is work function.
$$
\begin{aligned}
& E_1+W=\frac{h c}{\lambda_1} \\
& E_2+W=\frac{h c}{\lambda_2}
\end{aligned}
$$
From Eq. (iv)
$$
W=\frac{h c}{\lambda_2}-E_2,
$$
$\therefore$ Putting this value in Eq. (iii), we have
$$
\begin{aligned}
& E_1+\frac{h c}{\lambda_1}-E_2-\frac{h c}{\lambda_1} \\
\Rightarrow \quad & E_1-E_2=h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right) \\
\Rightarrow \quad & E_1-E_2=h c\left(\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right) \\
\Rightarrow \quad h c= & \frac{\left(E_1-E_2\right) \lambda_1 \lambda_2}{\left(\lambda_2-\lambda_1\right)}
\end{aligned}
$$
$$
\begin{aligned}
& E_1=\frac{h c}{\lambda_1}-W \\
& E_2=\frac{h c}{\lambda_2}-W
\end{aligned}
$$
where, $W$ is work function.
$$
\begin{aligned}
& E_1+W=\frac{h c}{\lambda_1} \\
& E_2+W=\frac{h c}{\lambda_2}
\end{aligned}
$$
From Eq. (iv)
$$
W=\frac{h c}{\lambda_2}-E_2,
$$
$\therefore$ Putting this value in Eq. (iii), we have
$$
\begin{aligned}
& E_1+\frac{h c}{\lambda_1}-E_2-\frac{h c}{\lambda_1} \\
\Rightarrow \quad & E_1-E_2=h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right) \\
\Rightarrow \quad & E_1-E_2=h c\left(\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right) \\
\Rightarrow \quad h c= & \frac{\left(E_1-E_2\right) \lambda_1 \lambda_2}{\left(\lambda_2-\lambda_1\right)}
\end{aligned}
$$
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