Atomic Structure JEE 2021: Helium Atom Recoil Velocity

How much does a helium atom recoil after absorbing a 330 nm photon in JEE Advanced 2021?

The resulting change in velocity of the He atom is 30 cm s^{-1}. A helium atom at rest absorbs a photon of wavelength exactly 330 nm. Momentum conservation between photon and atom sets photon momentum equal to m Δv, where m is the mass of one helium atom from the molar mass of 4 g mol^{-1} divided by N_A.

The values supplied are h = 6.6 × 10^{-34} J s, N_A = 6 × 10^{23} mol^{-1}, molar mass He = 4 g mol^{-1}. These produce exactly 30 when units stay consistent from the first step.

What is the official step-by-step solution for the JEE Advanced 2021 helium recoil velocity numerical?

The official solution starts with photon momentum $p=\frac{h}{\lambda }$. Momentum conservation gives $\Delta v=\frac{h}{m\lambda }$ where m is mass of one He atom.

Calculate single-atom mass first:

$m=\frac{4\times {10}^{-3}{\text{kg mol}}^{-1}}{6\times {10}^{23}{\text{mol}}^{-1}}=6.67\times {10}^{-27}\text{kg}$.

Next, $\lambda =330\times {10}^{-9}\text{m}=3.30\times {10}^{-7}\text{m}$.

The denominator is $(6.67\times {10}^{-27})\times (3.30\times {10}^{-7})\approx 2.20\times {10}^{-33}$.

Then $\Delta v=\frac{6.6\times {10}^{-34}}{2.20\times {10}^{-33}}=0.30{\text{m s}}^{-1}$ exactly.

Final conversion gives 0.30 m s^{-1} = 30 cm s^{-1}.

Copy these exact lines. The arithmetic is built to finish in under 90 seconds.

What method mistakes produce the wrong numerical in this JEE Advanced 2021 recoil problem?

One concrete reasoning error is using molar mass 4 g mol^{-1} directly in place of single-atom mass. This shifts the answer by a factor of 6 × 10^{23} and gives 5 × 10^{-25} instead of 30.

A second error is treating the problem as energy conservation $\frac{1}{2}m(\Delta v{)}^2=\frac{hc}{\lambda }$ instead of pure momentum conservation. The resulting wrong numerical is off by several orders of magnitude and fails to match the required 30.

Stick to the official momentum route. It alone matches the JEE answer key.

Why does the official solution choose those specific approximations for the JEE Advanced 2021 numerical?

The official approximations are deliberate to yield the clean integer 30 that JEE expects. They begin with 4/6 = 2/3 ≈ 0.6667 leading to 6.67 × 10^{-27} kg.

Next, 6.67 × 3.3 = 22.011 ≈ 22.0 produces denominator exactly 2.20 × 10^{-33}. Division then gives 6.6 / 2.2 = 3 exactly, producing 0.3 m s^{-1} before unit conversion.

More precise h = 6.626 would not give the clean integer 30 expected in the numerical answer. Expect this pattern in every future photon-atom recoil question that JEE sets for clean cancellation.

What are two related practice questions on photon momentum from atomic structure?

Question 1: A hydrogen atom at rest emits a photon of wavelength 121.6 nm; calculate recoil speed of the atom (use m_H ≈ 1.67 × 10^{-27} kg).

Question 2: Calculate de Broglie wavelength of a helium atom moving at 30 cm s^{-1} and compare its magnitude with the absorbed photon wavelength.

Both questions require the same mass conversion from molar mass to single-particle mass in kg and identical unit handling as the JEE Advanced 2021 problem. Solve them with the official momentum template.

How should you prepare for recoil velocity numericals in JEE Advanced atomic structure?

Always convert molar mass to single-particle mass in kg for momentum problems. JEE Advanced numericals in atomic structure test consistency of units more than formula recall.

Practice all photon-electron and photon-atom momentum questions from 2015-2023 past papers. Search every JEE Main paper from 2002 and every Advanced paper from 2007 by chapter in the past-paper archive to find each variant with its worked solution.

Revise photoelectric-effect numericals from the JEE Main archive for related practice. This habit turns the 30 cm s^{-1} template into automatic marks on every similar recoil-velocity numerical.

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).

If that step was the hard part, work through Circle JEE 2018: Deriving E1 and E2 Loci.