Light of wavelength lambda_ph falls on a cathode plate inside a vacuum tube as shown in the figure.

Q: Why do we use $K_{\max}$ instead of just any kinetic energy?

The question specifically asks for the *minimum* de Broglie wavelength. Since $\lambda = \frac{h}{\sqrt{2mK}}$, the wavelength is inversely proportional to the square root of kinetic energy. To minimize the wavelength, we must maximize the kinetic energy, which is why we use $K_{\max}$ from the photoelectric equation.

Q: Why doesn't the distance $d$ affect the final kinetic energy?

The work done by an electric field on a charge $q$ moving through a potential difference $V$ is simply $W = qV$. It is completely independent of the path taken or the distance $d$ between the plates. A larger distance just means a weaker electric field, but the total work done over that longer distance remains exactly the same.

Q: What happens if $V$ is negative?

A negative $V$ (stopping potential) would decelerate the electrons. The work done would be $-eV$. If the magnitude of $eV$ is greater than $K_{\max}$, the electrons would be stopped completely before reaching the anode, and there would be no electrons passing through the mesh to even have a de Broglie wavelength!

Q: Why is option (a) incorrect? Doesn't $\lambda_e$ increase when $\lambda_{\text{ph}}$ increases?

Yes, $\lambda_e$ does increase when $\lambda_{\text{ph}}$ increases. However, option (a) claims they increase at the *same rate*. Because $\lambda_{\text{ph}}$ is inside a reciprocal, which is then inside a square root, which is inside another reciprocal, the relationship is highly non-linear. The rates of change (derivatives) are not equal.

Q: What does $V \gg \frac{\phi}{e}$ physically mean?

It means the energy supplied by the external battery ($eV$) is so massive that the initial kinetic energy the electron got from the photon, and the energy it lost escaping the metal ($\phi$), become negligible rounding errors in comparison. The electron's final speed is almost entirely dictated by the battery.

Question

Light of wavelength $\lambda_{	ext{ph}}$ 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 statements(s) is (are) true?

Options:

(A) $\lambda_e$ increases at the same rate as $\lambda_{	ext{ph}}$ for $\lambda_{	ext{ph}} < \frac{hc}{\phi}$

(B) $\lambda_e$ is approximately halved, if $d$ is doubled

(C) $\lambda_e$ decreases with increase in $\phi$ and $\lambda_{	ext{ph}}$

(D) For large potential difference ($V >> \frac{\phi}{e}$), $\lambda_e$ is approximately halved if $V$ is made four times.

Sigma Percentile · Accepted Answer

### The Quantum Journey: From Photons to de Broglie Waves

This problem is a beautiful synthesis of two foundational pillars of modern physics: Einstein's Photoelectric Effect and de Broglie's Matter Waves. Let's break down the journey of an electron from the moment light strikes the cathode to its arrival at the anode.

### The Photoelectric Kickoff

When monochromatic light of wavelength $\lambda_{	ext{ph}}$ illuminates the cathode, photons transfer their energy to the electrons. According to Einstein's photoelectric equation, the maximum kinetic energy ($K_{\max}$) of an ejected electron is the energy of the incident photon minus the work function ($\phi$) of the metal:

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

This is the initial kinetic energy of the fastest electrons just as they leave the cathode surface.

### The Electric Boost

Once free, these electrons find themselves in an electric field created by the potential difference $V$ between the cathode and the anode. The electric field does work on the electrons, accelerating them towards the anode. The work done is simply $eV$.

By the work-energy theorem, the final kinetic energy $K$ of the electrons as they pass through the wire mesh anode is the sum of their initial kinetic energy and the work done by the electric field:

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

### The de Broglie Finale

Now, we switch gears from particle mechanics to wave mechanics. Louis de Broglie postulated that every moving particle has an associated wavelength, given by $\lambda = \frac{h}{p}$, where $p$ is the momentum. Since momentum $p = \sqrt{2mK}$, the de Broglie wavelength $\lambda_e$ of the electrons at the anode is:

$$\lambda_e = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2m\left(\frac{hc}{\lambda_{	ext{ph}}} - \phi + eVight)}}$$

### Analyzing the Extremes

Let's evaluate the given options based on our master equation.

*   **Option (a) & (c):** If we increase $\phi$ or $\lambda_{	ext{ph}}$, the term $\left(\frac{hc}{\lambda_{	ext{ph}}} - \phiight)$ decreases. This makes the entire denominator smaller, which means $\lambda_e$ must **increase**. Furthermore, the relationship is highly non-linear due to the square root and the inverse proportionality, so it certainly doesn't increase at the "same rate".
*   **Option (b):** The distance $d$ between the electrodes does not appear anywhere in our final expression. The work done by an electric field depends only on the potential difference $V$, not the distance over which it is applied. 
*   **Option (d):** Consider a scenario where the accelerating potential $V$ is extremely large, such that $eV \gg \frac{hc}{\lambda_{	ext{ph}}} - \phi$. In this limit, the initial kinetic energy becomes negligible, and the equation simplifies to:

$$\lambda_e \approx \frac{h}{\sqrt{2meV}}$$

Here, we clearly see that $\lambda_e \propto \frac{1}{\sqrt{V}}$. If we increase the potential difference to $4V$, the new wavelength becomes:

$$\lambda_e' \approx \frac{h}{\sqrt{2me(4V)}} = \frac{1}{2} \frac{h}{\sqrt{2meV}} = \frac{\lambda_e}{2}$$

The de Broglie wavelength is indeed halved. Thus, **Option (d)** is the correct statement.

Select Answer:

Visualized Solution

$K_{m a x} = \frac{h c}{λ _{ph}} - ϕ$

$K = K_{m a x} + e V$

$λ_{e} = \frac{h}{2 m K}$

$e V ≫ \frac{h c}{λ _{ph}} - ϕ$

$λ_{e} \propto \frac{1}{V}$

The Sigma Insight: Matter Waves and de Broglie Relation

The Quantum Journey

The Photoelectric Kickoff

The Electric Boost

The de Broglie Finale

Analyzing the Extremes

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kmax​=λph​hc​−ϕ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=Kmax​+eV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λe​=2mK​h​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } eV≫λph​hc​−ϕ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λe​∝V​1​