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Question:

A free hydrogen atom absorbs a photon of wavelength λ and gets excited from the state n=1 to the state n=4. Immediately after that, the electron jumps to n = m state by emitting a photon of wavelength β. Let the change in momentum of the atom due to the absorption and the emission be ΔPa and ΔPe respectively. If λ/β = 1, which of the option(s) is/are correct?

Δpa/Δpe = 1

the ratio of kinetic energy of the electron in the state n = m to the state n=1 is 1/4

m=2

β=418nm

Solution:

Correct option is B, m=2
λ/β = (E4 - Em)/(E4 - E1) = 1
(1/m2 - 1)/(1/16 - 1) = 1
(1/m2 - 1)/(-15/16) = 1
1/m2 - 1 = -15/16
1/m2 = 1 - 15/16 = 1/16
m2 = 16
m = 2
β = 1242 × 16/13.6 × 3 ≈ 487nm
K2/K1 = 1/n22 / 1/n12 = n12/n22 = 12/22 = 1/4
as kinetic energy is proportional to 1/n2