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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2019 · 12 April · Morning Shift · Physics · Atomic Physics

An excited He ^ + ion emits two photons in succession, with...

An excited He+{He}^{+} ion emits two photons in succession, with wavelengths 108.5nm108.5nm and 30.4nm30.4nm in making a transition to the ground state. The quantum number nn, corresponding to its initial excited state is

(for a photon of wavelength λ\lambda, energy E=1240eVλ(innm)E = \dfrac{1240eV}{\lambda ( innm )})

Options

  1. An=6n = 6
  2. Bn=5n = 5
  3. Cn=7n = 7
  4. Dn=4n = 4

Correct answer

B. n=5n = 5

Step-by-step solution

E1=1240λ1=1240108.5≈11.43eV{E}_{1} = \dfrac{1240}{{\lambda}_{1}} = \dfrac{1240}{108.5} \approx 11.43eV
E2=1240λ2=124030.4≈40.79eV{E}_{2} = \dfrac{1240}{{\lambda}_{2}} = \dfrac{1240}{30.4} \approx 40.79eV
ETotal=E1+E1=52.22eV{E}_{Total} = {E}_{1} + {E}_{1} = 52.22eV
52.22=13.6(2)2[1−1n2]52.22 = 13.6{\left(2\right)}^{2}\left[1 - \dfrac{1}{{n}^{2}}\right]
52.22=54.4(1−1n2)52.22 = 54.4\left(1 - \dfrac{1}{{n}^{2}}\right)
0.96=1−1n20.96 = 1 - \dfrac{1}{{n}^{2}}
1n2=0.04\dfrac{1}{{n}^{2}} = 0.04
n2=10.04=1004=25{n}^{2} = \dfrac{1}{0.04} = \dfrac{100}{4} = 25
n=5n = 5
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