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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Physics
Chapter
Atomic Physics

JEE Advanced 2019 · Paper 2 · Physics · Atomic Physics

A free hydrogen atom after absorbing a photon of wavelength λ _ a...

A free hydrogen atom after absorbing a photon of wavelength λa{\lambda}_{a} gets excited from the state n=1n = 1 to the state n=4.n = 4 . Immediately after that the electron jumps to n=mn = m state by emitting a photon of wavelength λe.{\lambda}_{e} . Let the change in momentum of atom due to the absorption and the emission are Δpa\Delta{p}_{a} and Δpe,\Delta{p}_{e}, respectively. If λaλe=15.\dfrac{{\lambda}_{a}}{{\lambda}_{e}} = \dfrac{1}{5} . Which of the option(s) is/are correct?
[Use hc=1242eVnm,1nm=10−9m,hc = 1242eVnm, 1nm = {10}^{- 9}m, hh and cc are Planck's constant and speed of light, respectively]

Options

  1. Aλe=418nm{\lambda}_{e} = 418nm
  2. BThe ratio of kinetic energy of the electron in the state n=mn = m to the state n=1n = 1 is 14\dfrac{1}{4}
  3. Cm=2m = 2
  4. DΔpaΔpe=12\dfrac{\Delta{p}_{a}}{\Delta{p}_{e}} = \dfrac{1}{2}

Correct answer

B. The ratio of kinetic energy of the electron in the state n=mn = m to the state n=1n = 1 is 14\dfrac{1}{4}

Step-by-step solution

Energy for transition of electron from one orbit to other is given by-
E2−E1=(13.6)Z2(1n12−1n22)=hcλ{E}_{2} - {E}_{1} = \left(13.6\right){Z}^{2}\left(\dfrac{1}{{n}_{1}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right) = \dfrac{hc}{\lambda}
⇒1λ=13.6hc.Z2(1n12−1n22)⇒ \dfrac{1}{\lambda} = \dfrac{13.6}{hc} . {Z}^{2}\left(\dfrac{1}{{n}_{1}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right)
Now as per question
1λa=13.6hc.Z2(112−142)\dfrac{1}{{\lambda}_{a}} = \dfrac{13.6}{hc} . {Z}^{2}\left(\dfrac{1}{{1}^{2}} - \dfrac{1}{{4}^{2}}\right)
and 1λe=13.6hc.Z2(1m2−142)\dfrac{1}{{\lambda}_{e}} = \dfrac{13.6}{hc} . {Z}^{2}\left(\dfrac{1}{{m}^{2}} - \dfrac{1}{{4}^{2}}\right)
⇒λeλa=(1−116)1m2−116=(15)m216−m2⇒ \dfrac{{\lambda}_{e}}{{\lambda}_{a}} = \dfrac{\left(1 - \dfrac{1}{16}\right)}{\dfrac{1}{{m}^{2}} - \dfrac{1}{16}} = \dfrac{\left(15\right){m}^{2}}{16 - {m}^{2}}
But λaλe=15(given)\dfrac{{\lambda}_{a}}{{\lambda}_{e}} = \dfrac{1}{5}\left(given\right)
⇒16−m215m2=15⇒16−m2=3m2⇒ \dfrac{16 - {m}^{2}}{15{m}^{2}} = \dfrac{1}{5} ⇒ 16 - {m}^{2} = 3{m}^{2}
⇒m2=4⇒ {m}^{2} = 4
⇒m=2⇒(C)⇒ m = 2 ⇒ \left(C\right) is correct
Now,
1λe=13.6hc.Z2(1m2−116)\dfrac{1}{{\lambda}_{e}} = \dfrac{13.6}{hc} . {Z}^{2}\left(\dfrac{1}{{m}^{2}} - \dfrac{1}{16}\right)
=13.61242×1(14−116)(ashc=1242andz=1)= \dfrac{13.6}{1242} \times 1\left(\dfrac{1}{4} - \dfrac{1}{16}\right)\left(ashc = 1242andz = 1\right)
⇒λe=124213.6×163=487nm⇒(A)⇒ {\lambda}_{e} = \dfrac{1242}{13.6} \times \dfrac{16}{3} = 487nm ⇒ \left(A\right) is incorrect
Now, kinetic energy of electron
K∝1n2K ∝ \dfrac{1}{{n}^{2}}
⇒K2K1=1222=14⇒(B)⇒ \dfrac{{K}_{2}}{{K}_{1}} = \dfrac{{1}^{2}}{{2}^{2}} = \dfrac{1}{4} ⇒ \left(B\right) is correct
Also, ∆P=hλ∆ P = \dfrac{h}{\lambda}
⇒∆Pa∆Pe=λeλa=5⇒ \dfrac{∆ {P}_{a}}{∆ {P}_{e}} = \dfrac{{\lambda}_{e}}{{\lambda}_{a}} = 5
⇒(D)⇒ \left(D\right) is incorrect.
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