A free hydrogen atom after absorbing a photon of wavelength λ a gets excited from the state n = 1 to…

A free hydrogen atom after absorbing a photon of wavelength λa 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 λe. Let the change in momentum of atom due to the absorption and the emission are Δpa and Δpe, respectively. If λaλe=15. Which of the option(s) is/are correct?
[Use hc=1242 eV nm, 1 nm=10-9 m, h and c are Planck's constant and speed of light, respectively]
  1. λe=418 nm
  2. The ratio of kinetic energy of the electron in the state n=m to the state n=1 is 14
  3. m=2
  4. ΔpaΔpe=12

Solution

Energy for transition of electron from one orbit to other is given by-
E2-E1=13.6Z21n12-1n22=hcλ
⇒1λ=13.6hc.Z21n12-1n22
Now as per question
1λa=13.6hc.Z2112-142
and 1λe=13.6hc.Z21m2-142
⇒λeλa=1-1161m2-116=15m2 16-m2
But λaλe=15given
⇒16-m215 m2=15⇒16-m2=3m2
⇒m2=4
⇒m=2 ⇒C is correct
Now,
1λe=13.6hc.Z21m2-116
=13.61242×114-116 as hc=1242 and z=1
⇒λe=124213.6×163=487 nm ⇒A is incorrect
Now, kinetic energy of electron
K∝1n2
⇒K2K1=1222=14⇒B is correct
Also, ∆P=hλ
⇒∆Pa∆Pe=λeλa=5
⇒D is incorrect.

Asked in: JEE Advanced 2019 (Paper 2)

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