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In a photoelectric experiment three different lights are incident on a metal of work function $1.5 \mathrm{eV}$. Light $A$ is $200 \mathrm{~nm}$ wavelength with intensity $1.8 \mathrm{~W} / \mathrm{m}^2$, light $B$ is $400 \mathrm{~nm}$ with $1 \mathrm{~W} / \mathrm{m}^2$ and light $C$ is $600 \mathrm{~nm}$ with $0.5 \mathrm{~W} / \mathrm{m}^2$. The current versus voltage is measured. Which graphs correspond to which light?
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2156 Upvotes
Verified Answer
The correct answer is:
III II IV
Given that, work-function, $\phi_0=1.5 \mathrm{eV}$
Let $\lambda_0=$ threshold wavelength.
Then, $\phi_0=\frac{h c}{\lambda_0}$
$$
\begin{aligned}
\Rightarrow \quad 1.5 \mathrm{eV} & =\frac{1242}{\lambda_0}(\mathrm{eV}-\mathrm{nm}) \\
\lambda_0 & =828 \mathrm{~nm}
\end{aligned}
$$
Now, photocurrent is directly proportional to intensity of incident light.
$V_1, V_2, V_3$ and $V_4$ are cut off voltage, which is required to stop the electron having maximum kinetic energy.
Given that, $\lambda_A=200 \mathrm{~nm}$, its corresponding cut off voltage is $V_2$.
$\lambda_B=400 \mathrm{~nm}$, its corresponding cut off voltage is $V_3$.
$\lambda_C=600 \mathrm{~nm}$, its corresponding cut off voltage is $V_4$.
and corresponding photocurrent,
$$
\begin{aligned}
I_A & =1.8 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{III} \\
I_B & =1 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{II} \\
I_C & =0.5 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{IV}
\end{aligned}
$$
Hence, consequently option (d) is correct.
Let $\lambda_0=$ threshold wavelength.
Then, $\phi_0=\frac{h c}{\lambda_0}$
$$
\begin{aligned}
\Rightarrow \quad 1.5 \mathrm{eV} & =\frac{1242}{\lambda_0}(\mathrm{eV}-\mathrm{nm}) \\
\lambda_0 & =828 \mathrm{~nm}
\end{aligned}
$$
Now, photocurrent is directly proportional to intensity of incident light.
$V_1, V_2, V_3$ and $V_4$ are cut off voltage, which is required to stop the electron having maximum kinetic energy.
Given that, $\lambda_A=200 \mathrm{~nm}$, its corresponding cut off voltage is $V_2$.
$\lambda_B=400 \mathrm{~nm}$, its corresponding cut off voltage is $V_3$.
$\lambda_C=600 \mathrm{~nm}$, its corresponding cut off voltage is $V_4$.
and corresponding photocurrent,
$$
\begin{aligned}
I_A & =1.8 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{III} \\
I_B & =1 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{II} \\
I_C & =0.5 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{IV}
\end{aligned}
$$
Hence, consequently option (d) is correct.
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