Light of wavelength \lambda ph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is \phi and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de-Broglie wavelength of the electrons passing through the anode is \lambda e, which of the following statement(s) is (are) true?

For large potential difference (V >> \phi /e), \lambda e is approximately halved if V is made four times

Dual Nature of Radiation PYQ — JEE Advanced 2016 · Shift 2 · Q50 with Solution

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Correct answer: D

Photoelectric emission from the cathode

When light of wavelength $\lambda_{ph}$ falls on the cathode, the maximum kinetic energy of emitted photoelectrons is

$K_{\max}=\frac{hc}{\lambda_{ph}}-\phi$

This is valid only if

$\frac{hc}{\lambda_{ph}} > \phi \quad \Rightarrow \quad \lambda_{ph}<\frac{hc}{\phi}$

Effect of applied potential difference

The anode is a wire mesh, so electrons can pass through it. The minimum de-Broglie wavelength of electrons passing through the anode corresponds to the maximum momentum, i.e. the maximum kinetic energy at the anode.

If the anode is at higher potential by $V$ , electrons gain additional energy $eV$ . Hence

$K=\frac{hc}{\lambda_{ph}}-\phi + eV$

So the momentum is

$p=\sqrt{2mK} = \sqrt{2m\left(\frac{hc}{\lambda_{ph}}-\phi+eV\right)}$

Therefore the minimum de-Broglie wavelength is

=\frac{h}{\sqrt{2m\left(\frac{hc}{\lambda_{ph}}-\phi+eV\right)}}$$ Note that $d$ does not appear in this expression; only the potential difference $V$ matters. --- 3. **Check each option** ### Option A Claims: $\lambda_e$ increases at the same rate as $\lambda_{ph}$ for $\lambda_{ph}<hc/\phi$. But $$\lambda_e \propto \frac{1}{\sqrt{\frac{hc}{\lambda_{ph}}-\phi+eV}}$$ This is not a linear dependence on $\lambda_{ph}$. So $\lambda_e$ does increase as $\lambda_{ph}$ increases (while emission is possible), but **not at the same rate**. So, **A is false**. --- ### Option B Claims: $\lambda_e$ is approximately halved if $d$ is doubled. Since $$\lambda_e=\frac{h}{\sqrt{2m\left(\frac{hc}{\lambda_{ph}}-\phi+eV\right)}}$$ there is **no dependence on $d$**. Doubling $d$ changes the electric field if $V$ is fixed, but not the total energy gained by electron from cathode to anode. So, **B is false**. --- ### Option C Claims: $\lambda_e$ decreases with increase in $\phi$ and $\lambda_{ph}$. From the formula, - increasing $\phi$ decreases kinetic energy, so $\lambda_e$ **increases**, not decreases. - increasing $\lambda_{ph}$ decreases photon energy $hc/\lambda_{ph}$, so kinetic energy decreases and hence $\lambda_e$ **increases**, not decreases. So, **C is false**. --- ### Option D Claims: For large potential difference $(V\gg \phi/e)$, $\lambda_e$ is approximately halved if $V$ is made four times. For large $V$, $$K \approx eV$$ Hence $$\lambda_e \approx \frac{h}{\sqrt{2meV}} \propto \frac{1}{\sqrt{V}}$$ If $V \to 4V$, then $$\lambda_e \to \frac{1}{\sqrt{4}}\lambda_e = \frac{1}{2}\lambda_e$$ So, **D is true**. --- 4. **Final derived answer** The only correct option is: $$\boxed{D}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: D

Photoelectric emission from the cathode

When light of wavelength $\lambda_{ph}$ falls on the cathode, the maximum kinetic energy of emitted photoelectrons is

$K_{\max}=\frac{hc}{\lambda_{ph}}-\phi$

This is valid only if

$\frac{hc}{\lambda_{ph}} > \phi \quad \Rightarrow \quad \lambda_{ph}<\frac{hc}{\phi}$

Effect of applied potential difference

If the anode is at higher potential by $V$ , electrons gain additional energy $eV$ . Hence

$K=\frac{hc}{\lambda_{ph}}-\phi + eV$

So the momentum is

$p=\sqrt{2mK} = \sqrt{2m\left(\frac{hc}{\lambda_{ph}}-\phi+eV\right)}$

Therefore the minimum de-Broglie wavelength is

=\frac{h}{\sqrt{2m\left(\frac{hc}{\lambda_{ph}}-\phi+eV\right)}}$$ Note that $d$ does not appear in this expression; only the potential difference $V$ matters. --- 3. **Check each option** ### Option A Claims: $\lambda_e$ increases at the same rate as $\lambda_{ph}$ for $\lambda_{ph}<hc/\phi$. But $$\lambda_e \propto \frac{1}{\sqrt{\frac{hc}{\lambda_{ph}}-\phi+eV}}$$ This is not a linear dependence on $\lambda_{ph}$. So $\lambda_e$ does increase as $\lambda_{ph}$ increases (while emission is possible), but **not at the same rate**. So, **A is false**. --- ### Option B Claims: $\lambda_e$ is approximately halved if $d$ is doubled. Since $$\lambda_e=\frac{h}{\sqrt{2m\left(\frac{hc}{\lambda_{ph}}-\phi+eV\right)}}$$ there is **no dependence on $d$**. Doubling $d$ changes the electric field if $V$ is fixed, but not the total energy gained by electron from cathode to anode. So, **B is false**. --- ### Option C Claims: $\lambda_e$ decreases with increase in $\phi$ and $\lambda_{ph}$. From the formula, - increasing $\phi$ decreases kinetic energy, so $\lambda_e$ **increases**, not decreases. - increasing $\lambda_{ph}$ decreases photon energy $hc/\lambda_{ph}$, so kinetic energy decreases and hence $\lambda_e$ **increases**, not decreases. So, **C is false**. --- ### Option D Claims: For large potential difference $(V\gg \phi/e)$, $\lambda_e$ is approximately halved if $V$ is made four times. For large $V$, $$K \approx eV$$ Hence $$\lambda_e \approx \frac{h}{\sqrt{2meV}} \propto \frac{1}{\sqrt{V}}$$ If $V \to 4V$, then $$\lambda_e \to \frac{1}{\sqrt{4}}\lambda_e = \frac{1}{2}\lambda_e$$ So, **D is true**. --- 4. **Final derived answer** The only correct option is: $$\boxed{D}$$ This matches the stored correct answer.

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