Optics JEE 2017: Photoelectron Wavelength Ratio Solved

What is the correct answer to the optics JEE 2017 photoelectron question?

Option D is correct: in this optics JEE 2017 question, the small-change ratio is proportional to the cube of the electron wavelength divided by the square of the incident wavelength. The question is from JEE Advanced 2017, Paper 2, Physics. It is a single-correct MCQ, rated medium on the JEEnius question-bank scale. The question bank’s expected solving time is 120 seconds, not an official exam allowance or measured student average.

The complete problem is: light falls on a material whose work function and incident wavelength satisfy

Work function = ϕ_{0}, incident wavelength = λ < \frac{h c}{ϕ_{0}} .

The fastest emitted electron has de Broglie wavelength $λ_{d} .$

A change in incident wavelength produces a change in that electron’s wavelength: $Δ λ ⟶ Δ λ_{d} .$

Determine which expression the following ratio is proportional to:

\frac{Δ λ_{d}}{Δ λ} .

A $\frac{λ_{d}}{λ}$
B

\frac{λ_{d}^{2}}{λ}

\frac{λ_{d}^{3}}{λ}

\frac{λ_{d}^{3}}{λ^{2}}

How do you express the maximum kinetic energy in two ways?

Equate the photon energy left after overcoming the work function to the electron’s kinetic energy expressed through its de Broglie wavelength. “Fastest” selects the electron with maximum kinetic energy, not an average emitted electron.

Einstein’s photoelectric equation gives

K_{max} = \frac{h c}{λ} - ϕ_{0} .

The constants are

h = Planck's constant, c = speed of light .

The stated threshold condition makes the photon energy exceed the work function, so the maximum kinetic energy is positive. For electron momentum and mass,

p = momentum, m = electron mass,

the kinetic energy is

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

Use de Broglie’s relation and substitute, keeping the square:

p = \frac{h}{λ_{d}},

K_{max} = \frac{(h / λ_{d})^{2}}{2 m} = \frac{h^{2}}{2 m λ_{d}^{2}} .

Both expressions describe the same fastest electron. Equating them gives the equation to differentiate:

\frac{h^{2}}{2 m λ_{d}^{2}} = \frac{h c}{λ} - ϕ_{0} .

How does differentiation give option D?

Differentiating the electron wavelength’s inverse square produces an inverse cube; differentiating the incident wavelength’s reciprocal produces an inverse square. Rearranging puts the electron wavelength cubed in the numerator, giving option D. The electron wavelength depends on the incident wavelength, so the left side needs the chain rule. Planck’s constant, light speed, electron mass and the material’s work function remain constant.

Differentiate with respect to incident wavelength:

\frac{d}{d λ} (\frac{h^{2}}{2 m} λ_{d}^{- 2}) = \frac{d}{d λ} (h c λ^{- 1} - ϕ_{0}) .

The chain rule on the left gives

\frac{d (λ_{d}^{- 2})}{d λ} = - 2 λ_{d}^{- 3} \frac{d λ_{d}}{d λ} .

On the right, the reciprocal derivative and constant derivative are

\frac{d (λ^{- 1})}{d λ} = - λ^{- 2}, \frac{d ϕ_{0}}{d λ} = 0 .

Substitution gives

- \frac{h^{2}}{m λ_{d}^{3}} \frac{d λ_{d}}{d λ} = - \frac{h c}{λ^{2}} .

Cancel the negative signs:

\frac{h^{2}}{m λ_{d}^{3}} \frac{d λ_{d}}{d λ} = \frac{h c}{λ^{2}} .

Multiply both sides by

\frac{m λ_{d}^{3}}{h^{2}} .

The result is

\frac{d λ_{d}}{d λ} = \frac{m c}{h} \frac{λ_{d}^{3}}{λ^{2}} .

For sufficiently small changes, use the slope at the original wavelength:

\frac{Δ λ_{d}}{Δ λ} \approx \frac{m c}{h} \frac{λ_{d}^{3}}{λ^{2}} \propto \frac{λ_{d}^{3}}{λ^{2}} .

Hence option D. The fixed proportionality factor is $\frac{m c}{h} .$

This factor is independent of both wavelengths. It is retained in the equality, not set equal to unity.

How do you check the sign, units and meaning of the changes?

The derivative is positive and dimensionless, as required. Increasing incident wavelength lowers photon energy and maximum electron kinetic energy. Electron momentum falls, so its de Broglie wavelength increases.

Both wavelengths have units of length. The factors on the right have units

[\frac{m c}{h}] = L^{- 1}, [\frac{λ_{d}^{3}}{λ^{2}}] = L,

so their product is dimensionless.

The official solution interprets the stated changes differentially. The derivative at the original wavelength is not generally an exact ratio for an arbitrary finite change. Both incident wavelengths must remain below the emission threshold:

λ < \frac{h c}{ϕ_{0}}, λ + Δ λ < \frac{h c}{ϕ_{0}} .

Near threshold, the slope changes rapidly. A particularly small change may therefore be needed for a reliable linear approximation.

Why does incorrectly dropping the work function produce option A?

The incorrect route replaces the available kinetic energy with the entire photon energy, without justification:

\frac{h c}{λ} - ϕ_{0} ⟶ \frac{h c}{λ} .

This changes the energy balance and creates a square-root dependence on incident wavelength. Differentiating that reduced relation appears to select A.

The altered energy equation is

\frac{h^{2}}{2 m λ_{d}^{2}} = \frac{h c}{λ} .

It gives

λ_{d}^{- 2} \propto λ^{- 1}, λ_{d} \propto \sqrt{λ},

and therefore

\frac{d λ_{d}}{d λ} = \frac{λ_{d}}{2 λ} .

A constant has zero derivative, but it need not have zero value. The work function disappears during differentiation while still affecting the electron wavelength through the original energy equation.

The given condition guarantees emission:

λ < \frac{h c}{ϕ_{0}} .

It does not establish the stronger condition

\frac{h c}{λ} ≫ ϕ_{0} .

Neglecting the work function can be a high-photon-energy approximation. That approximation is not supplied here and cannot replace the general result.

Can you solve two related photoelectric wavelength questions?

Use the energy balance for the first drill and convert the derivative into a fractional-change relation for the second. These are original practice questions from the same photoelectric-effect topic, not additional verified JEE PYQs. For both, define the threshold wavelength as

λ_{0} = \frac{h c}{ϕ_{0}} .

Question 1. When the incident wavelength is half the threshold wavelength, find the maximum kinetic energy and the fastest electron’s de Broglie wavelength.

λ = \frac{λ_{0}}{2} .

Worked check: the photon energy is

E_{γ} = \frac{h c}{λ_{0} / 2} = 2 ϕ_{0} .

Therefore,

K_{max} = 2 ϕ_{0} - ϕ_{0} = ϕ_{0}, λ_{d} = \frac{h}{\sqrt{2 m ϕ_{0}}} .

Question 2. At an initial wavelength below threshold, express the small fractional electron-wavelength change in terms of the fractional incident-wavelength change. Evaluate the multiplier at half the threshold wavelength, then calculate the response to a small 0.2% increase there.

Worked check: multiply the derived slope by the incident-to-electron wavelength ratio:

\frac{λ}{λ_{d}} \frac{d λ_{d}}{d λ} = \frac{m c}{h} \frac{λ_{d}^{2}}{λ} .

Substitute the energy-balance result:

λ_{d}^{2} = \frac{h^{2}}{2 m (h c / λ - ϕ_{0})} .

Thus,

\frac{m c}{h} \frac{λ_{d}^{2}}{λ} = \frac{h c}{2 (h c - ϕ_{0} λ)} = \frac{λ_{0}}{2 (λ_{0} - λ)} .

Therefore,

\frac{Δ λ_{d}}{λ_{d}} \approx \frac{λ_{0}}{2 (λ_{0} - λ)} \frac{Δ λ}{λ} .

At half threshold, the multiplier is 1, so sufficiently small fractional wavelength changes are approximately equal. The numerical check is

\frac{Δ λ_{d}}{λ_{d}} \approx 1 \times 0.002 = 0.002 = 0.2 % .

The growing multiplier near threshold means greater sensitivity; it does not extend this result to or beyond threshold. Before using the linear approximation, check that the wavelength change is much smaller than the remaining gap: $| Δ λ | ≪ λ_{0} - λ .$

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Frequently asked questions

What is the correct answer to the optics JEE 2017 photoelectron question?

Option D is correct in the JEE Advanced 2017 Paper 2 question. For sufficiently small changes, the ratio of the electron-wavelength change to the incident-wavelength change is proportional to the electron wavelength cubed divided by the incident wavelength squared.

How do you derive the photoelectron wavelength ratio?

Equate the maximum kinetic-energy expressions: h²/(2mλ_d²) = hc/λ − φ₀, where λ_d is the fastest electron's de Broglie wavelength and λ is the incident wavelength. Differentiate with respect to λ using the chain rule to obtain dλ_d/dλ = (mc/h)(λ_d³/λ²). The factor mc/h is constant, so the required proportionality gives option D.

Why can't we ignore the work function in this question?

The threshold condition guarantees photoelectric emission but does not imply that photon energy is much greater than the work function. Although the constant work function has zero derivative, it still affects the electron wavelength through the original energy equation. Dropping it without justification changes the energy balance and leads to the option A expression.

Is the wavelength-change ratio valid for large changes?

The derivative gives a local slope, so it is not generally the exact ratio for an arbitrary finite change. Both incident wavelengths must remain below the emission threshold. Near threshold, the incident-wavelength change should be much smaller than the remaining gap to threshold for a reliable linear approximation.

Optics JEE 2017: Photoelectron Wavelength Ratio Solved

What is the correct answer to the optics JEE 2017 photoelectron question?

How do you express the maximum kinetic energy in two ways?

How does differentiation give option D?

How do you check the sign, units and meaning of the changes?

Why does incorrectly dropping the work function produce option A?

Can you solve two related photoelectric wavelength questions?

Frequently asked questions

What is the correct answer to the optics JEE 2017 photoelectron question?

How do you derive the photoelectron wavelength ratio?

Why can't we ignore the work function in this question?

Is the wavelength-change ratio valid for large changes?

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