Atomic Structure JEE 2023: Helium-Ion Wavelength Solution

What is the helium-ion wavelength question in JEE Advanced 2023?

The emitted wavelength is 30 nm in this JEE Advanced 2023 Atomic Structure problem. It is from JEE Advanced 2023, Paper 2, Chemistry, Atomic Structure, and requires a numerical answer, not an MCQ choice.

For singly ionised helium, an electron moves from an orbit of radius 105.8 pm to an orbit of radius 26.45 pm. Find the wavelength, in nanometres, of the photon emitted during this transition, using the four supplied constants: $a = 52.9 p m$

R_{H} = 2.2 \times 10^{- 18} J

h = 6.6 \times 10^{- 34} J s

c = 3 \times 10^{8} m s^{- 1}

The question bank classifies this as medium difficulty, with a 120-second benchmark. Neither is an official exam label. Follow the official route: recover the quantum numbers from the radii, calculate the photon energy, then convert that energy into wavelength.

How do the two radii give the quantum numbers?

The initial quantum number is two; the final quantum number is one. Singly ionised helium has one electron, so it is hydrogen-like and follows the Bohr-model radius formula. Its nucleus still contains two protons: the atomic number remains two despite the net ionic charge being plus one.

r_{n} = \frac{a n^{2}}{Z}

Z = 2 r_{n} = \frac{52.9 n^{2}}{2} p m

For the initial orbit, substitute the larger radius. Both the orbit radius and the supplied radius constant are in picometres, so their units cancel without conversion to metres.

105.8 = \frac{52.9 n_{i}^{2}}{2}

n_{i}^{2} = \frac{2 \times 105.8}{52.9} = 4 n_{i} = 2

For the final orbit, substitute the smaller radius into the same formula. The atomic number stays unchanged.

26.45 = \frac{52.9 n_{f}^{2}}{2}

n_{f}^{2} = \frac{2 \times 26.45}{52.9} = 1 n_{f} = 1

Choose the positive roots because principal quantum numbers are positive integers. The transition is: $2 \to 1$

The radius ratio is four, but radius depends on the square of the quantum number. The quantum-number ratio is only two:

\frac{r_{i}}{r_{f}} = 4 \frac{n_{i}}{n_{f}} = \sqrt{4} = 2

How much energy does the emitted photon carry?

The photon carries the following positive energy:

Δ E = 6.6 \times 10^{- 18} J

Here, the energy difference means the positive emitted-photon energy, not the signed change in the electron’s energy. The electron moves to a lower energy level and loses energy; the photon carries that released energy away.

For a hydrogen-like species, use the official expression below. Substitute the atomic number and both quantum numbers before simplifying to keep the nuclear-charge factor visible.

Δ E = R_{H} Z^{2} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})

Δ E = 2.2 \times 10^{- 18} \times 2^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) J

= 2.2 \times 10^{- 18} \times 4 (1 - \frac{1}{4}) J

= 2.2 \times 10^{- 18} \times 4 \times \frac{3}{4} J

= 6.6 \times 10^{- 18} J

The supplied Rydberg constant is an energy constant in joules, not the wavenumber form measured in inverse metres. Its units determine how to use it: insert it directly into the energy formula.

Do not multiply this constant by Planck’s constant and the speed of light again. Do not insert a negative photon energy into the wavelength formula.

How does the photon energy give a wavelength of 30 nm?

Use the photon-energy relation and rearrange for wavelength:

Δ E = \frac{h c}{λ} λ = \frac{h c}{Δ E}

The supplied rounded constants give exactly 30 nm. Keep those values throughout rather than replacing them with more precise constants.

Substitute in SI units so the first wavelength result is in metres. Keep the powers of ten visible until the cancellation is complete.

λ = \frac{(6.6 \times 10^{- 34}) (3 \times 10^{8})}{6.6 \times 10^{- 18}} m

The factors of 6.6 cancel. Dividing by the denominator adds eighteen to the exponent: $- 34 + 8 + 18 = - 8$

λ = 3 \times 10^{- 8} m

Convert metres into nanometres by dividing by the length of one nanometre. Enter only the numerical value, since the question specifies the unit.

1 n m = 10^{- 9} m

λ = \frac{3 \times 10^{- 8}}{10^{- 9}} n m = 30 n m

Wavelength = 30 n m Numerical entry = 30

Why does dropping the nuclear-charge factor give 120 nm?

Dropping the squared atomic-number factor makes the calculated wavelength four times too large. This numerical-answer question has no supplied options: 120 nm is a calculated wrong result, not an official distractor.

The mistake is to identify the transition correctly, then calculate its energy as though the species were hydrogen. For singly ionised helium, the required factor is four:

2 \to 1 Z^{2} = 4

Omitting that factor gives:

Δ E_{w r o n g} = 2.2 \times 10^{- 18} (1 - \frac{1}{4}) = 1.65 \times 10^{- 18} J

λ_{w r o n g} = \frac{1.98 \times 10^{- 25}}{1.65 \times 10^{- 18}} m = 1.2 \times 10^{- 7} m = 120 n m

Wavelength is inversely proportional to photon energy. An energy four times too small therefore produces a wavelength four times too large, even when the unit conversion is correct.

Check the two charge dependences separately. The radius formula contains the inverse atomic number, while the transition-energy formula contains its square:

r_{n} \propto \frac{1}{Z} Δ E \propto Z^{2}

Net ionic charge is not the atomic number.

How can I practise the radius-to-wavelength method?

Use the same sequence for both problems: radii first, energy second, wavelength last. These are author-created practice questions based on the worked example, not additional verified JEE PYQs. Use the same four supplied constants and attempt each before reading its answer check.

What wavelength does hydrogen emit from 211.6 pm to 52.9 pm?

A hydrogen atom moves from an orbit of radius 211.6 pm to one of radius 52.9 pm. Find the emitted wavelength in nanometres; the answer check is 120 nm.

Hydrogen has atomic number one. Recover the quantum numbers before using the energy expression:

Z = 1 n_{i}^{2} = \frac{211.6}{52.9} = 4 n_{f}^{2} = \frac{52.9}{52.9} = 1

n_{i} = 2, n_{f} = 1 2 \to 1

Δ E = 2.2 \times 10^{- 18} (1 - \frac{1}{4}) = 1.65 \times 10^{- 18} J

λ = \frac{1.98 \times 10^{- 25}}{1.65 \times 10^{- 18}} m = 120 n m

Here, 120 nm is valid because the atom is hydrogen. It was wrong for the original helium ion because that nucleus has twice the atomic number.

What wavelength does singly ionised helium emit from 238.05 pm to 105.8 pm?

Singly ionised helium moves from radius 238.05 pm to radius 105.8 pm. Find the emitted wavelength in nanometres; the answer check is 162 nm.

Z = 2 n_{i}^{2} = \frac{2 \times 238.05}{52.9} = 9 n_{f}^{2} = \frac{2 \times 105.8}{52.9} = 4

n_{i} = 3, n_{f} = 2 3 \to 2

Δ E = 2.2 \times 10^{- 18} \times 4 (\frac{1}{4} - \frac{1}{9}) = \frac{11}{9} \times 10^{- 18} J

λ = \frac{1.98 \times 10^{- 25}}{(11 / 9) \times 10^{- 18}} m = 1.62 \times 10^{- 7} m = 162 n m

If either answer differs, locate the first incorrect line before repeating the calculation. Then use How to Study Atomic Structure JEE: A Six-Step Plan to plan your next revision session.

Frequently asked questions

What is the answer to the JEE Advanced 2023 helium-ion wavelength question?

The emitted wavelength is 30 nm using the supplied constants. The electron moves from n = 2 to n = 1, releasing a photon with energy 6.6 × 10⁻¹⁸ J. Enter 30 as the numerical answer because the question specifies nanometres.

How do I find quantum numbers from the orbit radii of He+?

Use the Bohr radius formula r = an²/Z, with a = 52.9 pm and Z = 2 for He+. Rearranging gives n² = rZ/a, so 105.8 pm gives n = 2 and 26.45 pm gives n = 1. No conversion to metres is needed when both radius values use picometres.

Why is Z equal to 2 for He+ instead of 1?

Z is the atomic number, which counts protons in the nucleus, not the net ionic charge. Removing one electron from helium leaves two protons and one electron, so He+ is hydrogen-like but still has Z = 2.

Why am I getting 120 nm instead of 30 nm for the helium-ion question?

Omitting the Z² factor from the transition-energy formula gives 120 nm for this problem. For He+, Z² = 4, so leaving it out makes the photon energy four times too small and the wavelength four times too large. Retaining that factor gives 30 nm with the supplied constants.