A photoelectric material having work-function phi_0 is illuminated with light of wavelength lambda (

Q: Why do we only consider the *fastest* photoelectron?

The photoelectric equation $K_{\text{max}} = \frac{hc}{\lambda} - \phi_0$ specifically gives the *maximum* kinetic energy, which corresponds to the fastest electron. Electrons emitted from deeper within the metal will have less kinetic energy due to collisions.

Q: Why can we use differentiation for $\Delta\lambda$ and $\Delta\lambda_d$?

When the changes $\Delta\lambda$ and $\Delta\lambda_d$ are very small, they can be approximated by the differentials $d\lambda$ and $d\lambda_d$. This allows us to use calculus to find the relationship between the changes.

Q: What happens to the work function $\phi_0$ when we differentiate?

The work function $\phi_0$ is a constant property of the specific metal being used. Since the derivative of a constant is zero, it disappears when we differentiate the equation with respect to wavelength.

Q: Why is there a negative sign when differentiating $\frac{1}{\lambda}$?

By the power rule of differentiation, $\frac{d}{d\lambda}(\lambda^{-1}) = -1 \cdot \lambda^{-2} = -\frac{1}{\lambda^2}$. The negative signs on both sides of our differentiated equation eventually cancel out.

Q: Does this proportionality hold for large changes in wavelength?

No, this linear proportionality using derivatives is an approximation that is only valid for infinitesimally small changes (or very small $\Delta\lambda$). For large changes, you would need to calculate the exact values using the original equation.

Question

A photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda \left(\lambda < \frac{hc}{\phi_0}\right)$. The fastest photoelectron has a de-Broglie wavelength $\lambda_d$. A change in wavelength of the incident light by $\Delta\lambda$ results in a change $\Delta\lambda_d$ in $\lambda_d$. Then, the ratio $\frac{\Delta\lambda_d}{\Delta\lambda}$ is proportional to

Options:

(A) $\frac{\lambda_d^2}{\lambda^2}$

(B) $\frac{\lambda_d}{\lambda}$

(C) $\frac{\lambda_d^3}{\lambda}$

(D) $\frac{\lambda_d^3}{\lambda^2}$

Sigma Percentile · Accepted Answer

The photoelectric effect is one of the most beautiful phenomena in modern physics, bridging the gap between the wave and particle natures of light. In this problem, we are asked to find how a small change in the wavelength of incident light ($\Delta\lambda$) affects the de-Broglie wavelength ($\Delta\lambda_d$) of the fastest emitted photoelectron. Let's break this down step-by-step.

### Analyzing the Setup

We start with Einstein's photoelectric equation, which is the cornerstone of this phenomenon. When a photon of wavelength $\lambda$ strikes a metal surface with a work function $\phi_0$, it transfers its energy to an electron. The maximum kinetic energy ($K_{	ext{max}}$) of the emitted electron is given by:

$$K_{	ext{max}} = \frac{hc}{\lambda} - \phi_0$$

This equation tells us that the energy of the incident photon ($\frac{hc}{\lambda}$) is used to overcome the work function ($\phi_0$), and the remaining energy becomes the kinetic energy of the fastest electron.

### The Master Equation

To relate this to the de-Broglie wavelength, we need to express the kinetic energy in terms of momentum ($p$). We know from classical mechanics that:

$$K_{	ext{max}} = \frac{p^2}{2m}$$

where $m$ is the mass of the electron. Now, we bring in the de-Broglie hypothesis, which states that the momentum of a particle is related to its wavelength ($\lambda_d$) by:

$$p = \frac{h}{\lambda_d}$$

Substituting this momentum into our kinetic energy expression, we get:

$$K_{	ext{max}} = \frac{1}{2m}\left(\frac{h}{\lambda_d}ight)^2 = \frac{h^2}{2m \lambda_d^2}$$

Now, we can equate this to our original photoelectric equation to form our master equation:

$$\frac{h^2}{2m \lambda_d^2} = \frac{hc}{\lambda} - \phi_0$$

### Differentiating for the Change

The question asks for the ratio of the change in $\lambda_d$ to the change in $\lambda$. Since these changes ($\Delta\lambda$ and $\Delta\lambda_d$) are small, we can use calculus and differentiate our master equation.

Let's differentiate both sides. Remember that $h$, $c$, $m$, and $\phi_0$ are constants. The derivative of $\lambda_d^{-2}$ is $-2\lambda_d^{-3} d\lambda_d$, and the derivative of $\lambda^{-1}$ is $-\lambda^{-2} d\lambda$.

$$\frac{h^2}{2m} \left( -2 \lambda_d^{-3} d\lambda_d ight) = hc \left( - \lambda^{-2} d\lambda ight)$$

Simplifying the negative signs and the constants, we get:

$$\frac{h^2}{m \lambda_d^3} d\lambda_d = \frac{hc}{\lambda^2} d\lambda$$

### Final Calculation

Finally, we rearrange the terms to isolate the ratio $\frac{d\lambda_d}{d\lambda}$:

$$\frac{d\lambda_d}{d\lambda} = \frac{mc}{h} \frac{\lambda_d^3}{\lambda^2}$$

Since $m$, $c$, and $h$ are all constants, we can clearly see the proportionality:

$$\frac{\Delta\lambda_d}{\Delta\lambda} \propto \frac{\lambda_d^3}{\lambda^2}$$

**This elegant result shows how the quantum nature of the electron responds to changes in the incident light, matching option (d).**

Select Answer:

Visualized Solution

$K_{max} = \frac{h c}{λ} - ϕ_{0}$

$K_{max} = \frac{p ^{2}}{2 m}$

$p = \frac{h}{λ _{d}}$

$Master Equation$

$Differentiating$

$Final Ratio$

The Sigma Insight: Photoelectric Effect

Analyzing the Setup

The Master Equation

Differentiating for the Change

Final Calculation

Similar Questions

Radiation of wavelength $λ$ is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed to $\frac{3 λ}{4}$ , the speed of the fastest emitted electron will be

When the wavelength of radiation falling on a metal is changed from $500 nm$ to $200 nm$ , the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to

Two identical photocathodes receive the light of frequencies $f_{1}$ and $f_{2}$ , respectively. If the velocities of the photoelectrons coming out are $v_{1}$ and $v_{2}$ respectively, then

Two identical, photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ . If the velocities of the photoelectrons (of mass $m$ ) coming out are respectively $v_{1}$ and $v_{2}$ , then

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 nm$ to $400 nm$ . The decrease in the stopping potential is close to $(\frac{h c}{e} = 1240 nm V)$