Photoelectric Effect JEE 2022: Paper 2 Q16 Solution

What is the photoelectric effect JEE 2022 Paper 2 question?

The intensity change does not enter the stopping-potential calculation in this photoelectric effect JEE 2022 question. The source is JEE Advanced 2022, Paper 2, Physics, Q16, not JEE Main. “Medium difficulty” and “120 seconds” are question-bank labels, not official exam classifications or measured student performance.

A metal illuminated by the first source has a stopping potential of 6.0 V. A replacement source has four times the wavelength and half the intensity; the same metal now gives 0.6 V.

λ_{1} = λ, I_{1} = I; λ_{2} = 4 λ, I_{2} = \frac{I}{2}

Find the first wavelength and the metal’s work function, respectively. Use the supplied constant:

\frac{h c}{e} = 1.24 \times 10^{- 6} J m C^{- 1}

1.72 \times 10^{- 7} m, 1.20 e V

1.72 \times 10^{- 7} m, 5.60 e V

3.78 \times 10^{- 7} m, 5.60 e V

3.78 \times 10^{- 7} m, 1.20 e V

How do you write the energy equation for each source?

The stopping potential measures the maximum electron kinetic energy per unit charge. Combine this relation with Einstein’s photoelectric equation: $K_{max} = e V_{s}$

e V_{s} = \frac{h c}{λ} - Φ

The capital symbol denotes the work-function energy. Dividing by the positive elementary charge gives:

V_{s} = \frac{h c / e}{λ} - \frac{Φ}{e}

Define the work function’s numerical value in electronvolts using the lowercase symbol. The work-function energy divided by the elementary charge has the same numerical value in volts:

Φ = ϕ e V ⟹ \frac{Φ}{e} = ϕ V

The work function is identical because the metal has not changed. With wavelength entered in metres, the two numerical equations are:

6.0 = \frac{1.24 \times 10^{- 6}}{λ} - ϕ

0.6 = \frac{1.24 \times 10^{- 6}}{4 λ} - ϕ

Quadrupling wavelength divides each photon’s energy by four. Intensity measures energy arriving per unit area per unit time, not energy per photon. In the standard photoelectric model, intensity does not set the maximum electron kinetic energy. Do not infer that photocurrent halves here: both wavelength and intensity change.

How do you subtract the equations to find wavelength and work function?

Subtract the second equation from the first to cancel the common work function. Use this official method rather than testing options one at a time: it isolates the wavelength directly.

6.0 - 0.6 = \frac{1.24 \times 10^{- 6}}{λ} - \frac{1.24 \times 10^{- 6}}{4 λ}

Factor out the first-source term. The second term is one-quarter of it, so their difference is three-quarters:

5.4 = \frac{3}{4} [\frac{1.24 \times 10^{- 6}}{λ}]

\frac{1.24 \times 10^{- 6}}{λ} = 5.4 \times \frac{4}{3} = 7.2

This gives the first-source photon energy per unit charge in volts. Rearrange to obtain the first wavelength:

λ = \frac{1.24 \times 10^{- 6}}{7.2} \approx 1.72 \times 10^{- 7} m \approx 172 n m

That determines only the first quantity in the answer pair. Substitute into the first numerical equation to recover the work function: $6.0 = 7.2 - ϕ$ $ϕ = 7.2 - 6.0 = 1.20$

The work function is therefore 1.20 eV. The complete answer is option A:

λ \approx 1.72 \times 10^{- 7} m, Φ = 1.20 e V

Does option A reproduce both stopping potentials?

Yes: the pair gives 6.0 V for the first source and 0.6 V for the second. Use the unrounded first-source photon energy of 7.2 eV rather than recalculating from the rounded wavelength.

First source:

K_{max, 1} = (7.2 - 1.2) e V = 6.0 e V ⟹ V_{s, 1} = 6.0 V

Second source:

E_{γ, 2} = \frac{7.2}{4} e V = 1.8 e V

K_{max, 2} = (1.8 - 1.2) e V = 0.6 e V ⟹ V_{s, 2} = 0.6 V

The second source remains above the emission threshold:

1.8 e V > 1.2 e V

It produces less energetic electrons, not an absence of emission. Both measured stopping potentials must agree, not merely the wavelength printed in an option.

Why can checking only the wavelength lead to option B?

A student can calculate the wavelength correctly, match it in the options and accept B without finding the work function. A and B share the same wavelength, so subtraction alone cannot distinguish them.

λ \approx 1.72 \times 10^{- 7} m

Test B using the photon energies already established. Its claimed work function gives:

K_{max, 1} = (7.2 - 5.6) e V = 1.6 e V

That corresponds to 1.6 V, not the required 6.0 V. The second source rules B out even more directly:

E_{γ, 2} = 1.8 e V < 5.6 e V

Its photons are below the claimed emission threshold, so no photoemission is possible in this model. A negative energy difference is not a physical negative stopping potential. For paired-answer options, solve and verify both quantities.

What three related photoelectric energy-balance questions should you try?

Try changing intensity alone, doubling the wavelength and finding the emission threshold. These original practice variations, not additional verified PYQs, stay within this question bank’s Work, Energy and Power chapter and Photoelectric Effect topic.

Use the original metal and first-source wavelength throughout. Attempt each question before reading its worked answer.

If intensity alone is halved at the original wavelength, what happens to the stopping potential?

It remains 6.0 V because photon energy and work function are unchanged. At fixed wavelength and otherwise unchanged collection conditions, the emission rate decreases. $V_{s} = 6.0 V$

Unlike the original question, this variation changes intensity without changing wavelength. Keep that distinction when predicting electron energy.

What stopping potential would twice the original wavelength produce?

Doubling wavelength halves the photon energy. Subtract the same work function:

E_{γ} = \frac{7.2}{2} e V = 3.6 e V

K_{max} = (3.6 - 1.2) e V = 2.4 e V

V_{s} = 2.4 V

What is the threshold wavelength of this metal?

At threshold, the photon just supplies the work function. The maximum electron kinetic energy is zero:

K_{max} = 0, \frac{h c}{λ_{0}} = Φ

λ_{0} = \frac{h c / e}{Φ / e} = \frac{1.24 \times 10^{- 6}}{1.20} m

λ_{0} \approx 1.03 \times 10^{- 6} m \approx 1033 n m

This is the longest wavelength that meets the emission threshold. Light with a longer wavelength has insufficient photon energy to eject electrons from this metal in the standard model.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Related on JEEnius: JEE Maths Revision Strategy: A 30-Day Main Plan.

Frequently asked questions

Is the photoelectric effect JEE 2022 Paper 2 question from Main or Advanced?

This question is from JEE Advanced 2022, Paper 2, Physics, Q16—not JEE Main. It gives stopping potentials of 6.0 V and 0.6 V for the same metal when the replacement source has four times the wavelength and half the intensity.

What is the answer to JEE Advanced 2022 Paper 2 photoelectric effect Q16?

Option A is correct: the first wavelength is approximately 1.72 × 10⁻⁷ m and the work function is 1.20 eV. Subtracting the two photoelectric equations cancels the common work function and gives a first-source photon energy of 7.2 eV. Substituting back gives the work function as 7.2 − 6.0 = 1.20 eV.

Does halving intensity change the stopping potential?

At fixed wavelength, halving intensity does not change the stopping potential in the standard photoelectric model. Photon energy and the metal's work function determine the maximum electron kinetic energy. In this question, the stopping potential changes because the wavelength increases fourfold, not because intensity halves.

Why is option B wrong if its wavelength is correct?

Option B has the correct wavelength but an incorrect work function of 5.60 eV. It predicts a first stopping potential of 1.6 V rather than 6.0 V. Its claimed work function also exceeds the second source's photon energy of 1.8 eV, so it would prevent photoemission from that source.

What is the threshold wavelength of the metal in this question?

The threshold wavelength is approximately 1.03 × 10⁻⁶ m, or 1033 nm. It follows from dividing the supplied hc/e value, 1.24 × 10⁻⁶ V m, by the work function per unit charge, 1.20 V. Longer wavelengths cannot eject electrons from this metal in the standard photoelectric model.