JEE Main 2022 · 25 July · evening · Single correct · +4 / −1

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ. The value of principal quantum number 'n' of the excited state will be : (R: Rydberg constant)

  1. (a) √((λ R)/(λ-1))
  2. (b) √((λ R)/(λ R-1))
  3. (c) √(λ/(λ R-1))
  4. (d) √((λ R²)/(λ R-1))
Answer and solution

Answer (b) √((λ R)/(λ R-1))

Solution

Physics sir · 4 steps

Write the Rydberg formula for a jump that ends at n = 1, then just solve it for n.

  1. Set up the jump. The electron falls from level n to the ground state, n = 1. The Rydberg formula gives the wavelength of the photon emitted in this jump.

    1/λ = R(1/1² - 1/n²) = R(1 - 1/n²)

  2. Isolate the n term. Divide both sides by R: 1/(λ R) = 1 - 1/n². So 1/n² = 1 - 1/(λ R) = (λ R - 1)/(λ R).
  3. Flip and root. Take the reciprocal to get n² = (λ R)/(λ R - 1), then take the square root.
  4. So. n = √((λ R)/(λ R - 1)), which is option (b).

The idea to keep. Since R has units of inverse length, λ R is a pure number. A quantum number must come out unitless, so any answer where λ and R do not appear together as λ R is wrong.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Atomic Physics

Asked about 110 times in JEE Main since 2020. This question is from 2022.