Photoelectric Effect JEE 2021: Step-by-Step Solution

What does the photoelectric effect JEE 2021 question ask?

The answer to this photoelectric effect JEE 2021 question is 6 eV, not the intermediate 5 eV for the shared light source. It is from JEE Advanced 2021, Paper 2, Physics, and is a numerical-answer question. The question bank rates it medium and gives a target solving time of 90 seconds; these are bank benchmarks, not official exam classifications.

The metals have these work functions:

P: 4.0 eV.
Q: 4.5 eV.
R: 5.5 eV.

P and Q receive light from the same monochromatic source. R receives light from a different monochromatic source. Their maximum photoelectron kinetic energies satisfy: $E_{P} = 2 E_{Q} = 2 E_{R}$

Find the incident photon energy for R, in eV. No apparatus geometry is specified or needed for this energy-comparison problem.

How should you assign the photon energies in step 1?

Use one photon-energy variable for P and Q, but a separate variable for R. Their shared monochromatic source gives P and Q equal incident photon energies. Their emitted electrons have different maximum kinetic energies because the metals have different work functions.

Einstein’s photoelectric equation is: $K_{max} = h ν - ϕ$

The terms are: $K_{max} : maximum photoelectron kinetic energy$ $h ν : incident photon energy$ $ϕ : work function$

Let the common photon energy for P and Q be: $h ν_{P} = h ν_{Q} = x$

All energy values and variables in the following algebra are expressed in eV. For P and Q, this gives: $E_{P} = x - 4.0$ $E_{Q} = x - 4.5$

R receives light from a different source. Do not assign the same photon-energy variable to all three metals.

How do you find the shared P–Q photon energy in step 2?

The P–Q source supplies photons of energy 5.0 eV. Substitute the two kinetic-energy expressions into the given relation. The ratio applies to electron kinetic energies after subtracting each metal’s work function, not to incident photon energies.

Start with: $E_{P} = 2 E_{Q}$

Substitute the expressions from step 1: $x - 4.0 = 2 (x - 4.5)$

Expand the bracket, including the work-function term: $x - 4.0 = 2 x - 9.0$

Move the constants to one side and the photon-energy terms to the other: $9.0 - 4.0 = 2 x - x$ $x = 5.0$

Now calculate the maximum kinetic energy for P:

E_{P} = 5.0 - 4.0 = 1.0 e V

The 5.0 eV value belongs to the P–Q source, not R’s source. Use P’s electron kinetic energy next to determine R’s electron kinetic energy.

How do you calculate and verify R’s photon energy in step 3?

R’s incident photon energy is 6.0 eV: 5.5 eV supplies its work function and 0.5 eV remains as maximum electron kinetic energy. Use the given relation between P and R, then add R’s work function. $E_{P} = 2 E_{R}$ $1.0 = 2 E_{R}$ $E_{R} = 0.5 e V$

Let R’s incident photon energy, expressed in eV, be: $h ν_{R} = y$

Apply Einstein’s equation and substitute R’s kinetic energy: $E_{R} = y - 5.5$ $0.5 = y - 5.5$

y = 0.5 + 5.5 = 6.0 e V

The final answer is:

6 e V

Enter 6, because the question already specifies eV. Check Q’s kinetic energy and then the complete chained equality:

E_{Q} = 5.0 - 4.5 = 0.5 e V

E_{P} = 1.0 e V

2 E_{Q} = 2 E_{R} = 2 (0.5) = 1.0 e V

Each source also clears the relevant emission threshold:

P receives 5.0 eV, above its 4.0 eV work function.
Q receives 5.0 eV, above its 4.5 eV work function.
R receives 6.0 eV, above its 5.5 eV work function.

No conversion to joules is needed. Every energy is already in eV, and the calculation uses only addition, subtraction and dimensionless ratios.

Why is it wrong to give P and R equal kinetic energies?

That drops the factor of two in the given relation. The mistake is in reading the condition, not in Einstein’s equation. This is a numerical-answer question, so there are no supplied wrong options to analyse.

For illustration, the incorrect assignment is:

E_{R} = E_{P} = 1.0 e V

It leads to:

y = 5.5 + 1.0 = 6.5 e V

The 6.5 eV result is an illustrative error, not an official distractor. Avoid dropping the factor of two by rewriting the chained equality as two separate relations:

E_{Q} = \frac{E_{P}}{2}

E_{R} = \frac{E_{P}}{2}

Both Q and R therefore produce electrons with maximum kinetic energy 0.5 eV. The correct 6 eV photon energy for R restores the required ratio instead of making P and R equal.

How can you practise the same energy-balance method?

Change the photon energy and check both the kinetic-energy ratio and the emission threshold. These are original follow-up exercises using the supplied question’s metals, not additional verified JEE PYQs. Both use photoelectric energy balance within the supplied bank’s chapter grouping, Work, Energy and Power.

What are P’s and Q’s maximum kinetic energies and their ratio under 6.0 eV photons?

Their maximum kinetic energies are 2.0 eV and 1.5 eV, respectively. Subtract the unchanged work functions of 4.0 eV for P and 4.5 eV for Q:

E_{P} = 6.0 - 4.0 = 2.0 e V

E_{Q} = 6.0 - 4.5 = 1.5 e V

E_{P} : E_{Q} = 2.0 : 1.5 = 4 : 3

The original ratio does not persist when the photon energy changes. The new ratio is: $4 : 3 \neq 2 : 1$

Will R emit electrons under the original 5.0 eV P–Q source?

No photoemission occurs in the standard single-photon model. R’s work function is 5.5 eV, so each incident photon falls short by:

5.5 - 5.0 = 0.5 e V

No electron is emitted. Do not report the following as a physical electron kinetic energy:

K_{max} = - 0.5 e V

Check the photon energy against the work function first. Calculate a maximum photoelectron kinetic energy only when the emission threshold is met.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

For a worked example of the same idea, see Three Dimensional Geometry JEE 2009: Matrix-Match Solution.

Frequently asked questions

What is the answer to the JEE Advanced 2021 photoelectric effect question?

The incident photon energy for metal R is 6 eV. Its work function is 5.5 eV and its maximum photoelectron kinetic energy is 0.5 eV, so Einstein's equation gives 5.5 + 0.5 = 6 eV. Enter 6 because the question already specifies eV.

Why can't we use the same photon energy for P, Q and R?

P and Q receive light from the same monochromatic source, so their incident photon energies are equal. R receives light from a different source and needs a separate photon-energy variable. The shared P–Q photon energy is 5 eV, while R's photon energy is 6 eV.

How do you use the relation E_P = 2E_Q = 2E_R?

The relation applies to maximum photoelectron kinetic energies, not incident photon energies. Rewrite it as E_Q = E_P/2 and E_R = E_P/2. With E_P = 1 eV, both E_Q and E_R are 0.5 eV.

Will metal R emit photoelectrons under 5 eV light?

No photoemission occurs in the standard single-photon model because R's work function is 5.5 eV. A 5 eV photon falls short of the emission threshold by 0.5 eV. Do not report a negative photoelectron kinetic energy; no electron is emitted.

Photoelectric Effect JEE 2021: Why the Answer Is 6 eV