Question:
When photons of energy \(4.2\,\text{eV}\) incident on a photosensitive material of work function \(2.2\,\text{eV}\), photoelectrons are emitted with a maximum linear momentum of \(P\). If photons of energy \(6.2\,\text{eV}\) incident on the same photosensitive material, the maximum linear momentum of the emitted photoelectrons is
When photons of energy \(4.2\,\text{eV}\) incident on a photosensitive material of work function \(2.2\,\text{eV}\), photoelectrons are emitted with a maximum linear momentum of \(P\). If photons of energy \(6.2\,\text{eV}\) incident on the same photosensitive material, the maximum linear momentum of the emitted photoelectrons is
Show Hint
In photoelectric effect,
\[
K_{\max}=E_{\text{photon}}-\phi.
\]
Since
\[
p=\sqrt{2mK_{\max}},
\]
the maximum momentum is proportional to the square root of the maximum kinetic energy:
\[
p\propto \sqrt{K_{\max}}.
\]
Updated On: Jul 9, 2026
- \(4P\)
- \(P\sqrt{2}\)
- \(2P\)
- \(P\sqrt{3}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
According to Einstein's photoelectric equation,
\[
K_{\max}=h\nu-\phi,
\]
where
\[
\phi=\text{work function}.
\]
Also,
\[
K_{\max}=\frac{p^2}{2m}.
\]
Hence,
\[
p\propto \sqrt{K_{\max}}.
\]
Step 1: Calculate the maximum kinetic energy in the first case. Given, \[ E_1=4.2\,\text{eV}, \qquad \phi=2.2\,\text{eV}. \] Therefore, \[ K_1=4.2-2.2. \] \[ K_1=2\,\text{eV}. \] Since the corresponding momentum is \(P\), \[ P^2\propto K_1. \]
Step 2: Calculate the maximum kinetic energy in the second case. Given, \[ E_2=6.2\,\text{eV}. \] Hence, \[ K_2=6.2-2.2. \] \[ K_2=4\,\text{eV}. \]
Step 3: Relate the two momenta. Let the new momentum be \(p_2\). \[ \frac{p_2}{P} = \sqrt{\frac{K_2}{K_1}}. \] \[ \frac{p_2}{P} = \sqrt{\frac{4}{2}}. \] \[ \frac{p_2}{P} = \sqrt{2}. \] Therefore, \[ p_2=P\sqrt{2}. \]
Step 4: Write the final answer. \[ \boxed{p_2=P\sqrt{2}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Calculate the maximum kinetic energy in the first case. Given, \[ E_1=4.2\,\text{eV}, \qquad \phi=2.2\,\text{eV}. \] Therefore, \[ K_1=4.2-2.2. \] \[ K_1=2\,\text{eV}. \] Since the corresponding momentum is \(P\), \[ P^2\propto K_1. \]
Step 2: Calculate the maximum kinetic energy in the second case. Given, \[ E_2=6.2\,\text{eV}. \] Hence, \[ K_2=6.2-2.2. \] \[ K_2=4\,\text{eV}. \]
Step 3: Relate the two momenta. Let the new momentum be \(p_2\). \[ \frac{p_2}{P} = \sqrt{\frac{K_2}{K_1}}. \] \[ \frac{p_2}{P} = \sqrt{\frac{4}{2}}. \] \[ \frac{p_2}{P} = \sqrt{2}. \] Therefore, \[ p_2=P\sqrt{2}. \]
Step 4: Write the final answer. \[ \boxed{p_2=P\sqrt{2}} \] \[ \boxed{\text{Answer = (B)}} \]
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