JEE MainPhysics

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Atoms and Nuclei: JEE Main Physics Question with Solution

The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is

Energy level diagram of hydrogen atom showing levels n=1, 2, 3, 4 with transitions A (4->3), B (4->2), C (3->2), D (3->1).

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\mathrm{C}

\mathrm{D}

\mathrm{B}

\mathrm{A}

Step-by-step solutionView

Correct answer

The transition with the shortest wavelength corresponds to the greatest energy difference in emission, which is transition D from $n = 5$ to $n = 1$ .

Option analysis

Why each option works or fails

A ·

\mathrm{C}

Believing that the transition between the highest numbered adjacent levels gives the shortest wavelength because the energy level index is highest. Recognize that photon energy is inversely proportional to wavelength ( $E = hc/\lambda$ ), so the shortest wavelength requires the maximum energy difference $\Delta E$ , not adjacent high- $n$ transitions where levels are tightly packed.

B ·

\mathrm{D}

Correct option. Transition D is an emission jump from $n = 5$ directly to the ground state $n = 1$ , yielding $\Delta E = -0.54\text{ eV} - (-13.6\text{ eV}) = 13.06\text{ eV}$ , the largest energy release and thus the shortest wavelength. None needed.

C ·

\mathrm{B}

Assuming that a smaller energy jump corresponds directly to a shorter wavelength, confusing proportionality between energy and wavelength. Recall that $\lambda = hc/\Delta E$ . As $\Delta E$ increases, $\lambda$ decreases, meaning the smallest energy gap produces the longest wavelength, not the shortest.

D ·

\mathrm{A}

Thinking transition A ( $n=2$ to $n=1$ ) represents the highest energy emission without checking whether transitions from higher states to $n=1$ release more energy. Compare $\Delta E$ across all downward transitions: $n=5 \to n=1$ (transition D) releases $13.06\text{ eV}$ , which exceeds the $10.2\text{ eV}$ released by $n=2 \to n=1$ (transition A).

Reviewed route

Solution

StepWorking

01concept

For a radiative transition emitting a photon of wavelength $\lambda$ , energy conservation gives $\Delta E = \frac{hc}{\lambda}$ , which implies $\lambda = \frac{hc}{\Delta E}$ . Therefore, the emission transition with the shortest wavelength corresponds to the transition having the maximum energy difference $\Delta E$ directed from a higher energy state to a lower energy state.

02option_verdict

Transition C corresponds to an upward transition from $n = 1$ to $n = 2$ , which represents absorption rather than emission. The energy change is $10.2\text{ eV}$ , which is smaller than $12.09\text{ eV}$ .

03option_verdict

Transition D is a downward emission transition from $n = 3$ ( $-1.51\text{ eV}$ ) to $n = 1$ ( $-13.6\text{ eV}$ ). The energy of the emitted photon is $\Delta E = -1.51 - (-13.6) = 12.09\text{ eV}$ . This is the maximum energy difference among all given emission transitions, yielding the shortest wavelength $\lambda = \frac{hc}{12.09\text{ eV}}$ .

04option_verdict

Transition B is an emission transition from $n = 2$ ( $-3.4\text{ eV}$ ) to $n = 1$ ( $-13.6\text{ eV}$ ) with $\Delta E = 10.2\text{ eV}$ , which is less than the energy difference of transition D ( $12.09\text{ eV}$ ), so it emits a longer wavelength.

05option_verdict

Transition A is an emission transition from $n = 3$ ( $-1.51\text{ eV}$ ) to $n = 2$ ( $-3.4\text{ eV}$ ) with $\Delta E = 1.89\text{ eV}$ , which has the smallest energy gap and hence the longest wavelength.

✓discriminator

Shortest wavelength means maximum photon energy in emission (downward arrow). Transition D drops all the way from $n=3$ to $n=1$ , covering the largest energy gap ( $12.09\text{ eV}$ ), so it must be D.

Hints that build this answer step by step

How is the wavelength $\lambda$ of an emitted photon related to the energy difference $\Delta E$ between the two levels?
$\lambda = \frac{hc}{\Delta E}$ , so the shortest wavelength corresponds to the largest energy gap.
Which downward transition in the hydrogen atom energy diagram results in the largest energy difference $\Delta E$ ?
Transition D ( $n = 5 \to n = 1$ ), with $\Delta E \approx 13.06\text{ eV}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does shortest wavelength mean maximum energy?

Because photon energy is inversely proportional to wavelength: $E = hc/\lambda \implies \lambda \propto 1/E$ .

Answer

The transition with the shortest wavelength corresponds to the greatest energy difference in emission, which is transition D from $n = 5$ to $n = 1$ .

Why each option works or fails

A: $\mathrm{C}$ - Believing that the transition between the highest numbered adjacent levels gives the shortest wavelength because the energy level index is highest. Recognize that photon energy is inversely proportional to wavelength ( $E = hc/\lambda$ ), so the shortest wavelength requires the maximum energy difference $\Delta E$ , not adjacent high- $n$ transitions where levels are tightly packed.
B · correct: $\mathrm{D}$ - Correct option. Transition D is an emission jump from $n = 5$ directly to the ground state $n = 1$ , yielding $\Delta E = -0.54\text{ eV} - (-13.6\text{ eV}) = 13.06\text{ eV}$ , the largest energy release and thus the shortest wavelength. None needed.
C: $\mathrm{B}$ - Assuming that a smaller energy jump corresponds directly to a shorter wavelength, confusing proportionality between energy and wavelength. Recall that $\lambda = hc/\Delta E$ . As $\Delta E$ increases, $\lambda$ decreases, meaning the smallest energy gap produces the longest wavelength, not the shortest.
D: $\mathrm{A}$ - Thinking transition A ( $n=2$ to $n=1$ ) represents the highest energy emission without checking whether transitions from higher states to $n=1$ release more energy. Compare $\Delta E$ across all downward transitions: $n=5 \to n=1$ (transition D) releases $13.06\text{ eV}$ , which exceeds the $10.2\text{ eV}$ released by $n=2 \to n=1$ (transition A).

Step-by-step solution

concept: For a radiative transition emitting a photon of wavelength $\lambda$ , energy conservation gives $\Delta E = \frac{hc}{\lambda}$ , which implies $\lambda = \frac{hc}{\Delta E}$ . Therefore, the emission transition with the shortest wavelength corresponds to the transition having the maximum energy difference $\Delta E$ directed from a higher energy state to a lower energy state.
option_verdict: Transition C corresponds to an upward transition from $n = 1$ to $n = 2$ , which represents absorption rather than emission. The energy change is $10.2\text{ eV}$ , which is smaller than $12.09\text{ eV}$ .
option_verdict: Transition D is a downward emission transition from $n = 3$ ( $-1.51\text{ eV}$ ) to $n = 1$ ( $-13.6\text{ eV}$ ). The energy of the emitted photon is $\Delta E = -1.51 - (-13.6) = 12.09\text{ eV}$ . This is the maximum energy difference among all given emission transitions, yielding the shortest wavelength $\lambda = \frac{hc}{12.09\text{ eV}}$ .
option_verdict: Transition B is an emission transition from $n = 2$ ( $-3.4\text{ eV}$ ) to $n = 1$ ( $-13.6\text{ eV}$ ) with $\Delta E = 10.2\text{ eV}$ , which is less than the energy difference of transition D ( $12.09\text{ eV}$ ), so it emits a longer wavelength.
option_verdict: Transition A is an emission transition from $n = 3$ ( $-1.51\text{ eV}$ ) to $n = 2$ ( $-3.4\text{ eV}$ ) with $\Delta E = 1.89\text{ eV}$ , which has the smallest energy gap and hence the longest wavelength.
discriminator: Shortest wavelength means maximum photon energy in emission (downward arrow). Transition D drops all the way from $n=3$ to $n=1$ , covering the largest energy gap ( $12.09\text{ eV}$ ), so it must be D.

Atoms and Nuclei: JEE Main Physics Question with Solution

What feels right?

The transition with the shortest wavelength corresponds to the greatest energy difference in emission, which is transition D from $n = 5$ to $n = 1$ .

Why each option works or fails

Solution

01concept

02option_verdict

03option_verdict

04option_verdict

05option_verdict

✓discriminator

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