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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2019 · 10 April · Evening Shift · Physics · Atomic Physics

In L i ^ + + ,electron in first Bohr orbit is excited to a level by...

InLi++L{i}^{+ +},electron in first Bohr orbit is excited to a level by a radiation of wavelengthλ\lambda.When the ion gets de excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ\lambda ?

(Given: h=6.63×10−34Js;c=3×108ms−1h = 6.63 \times {10}^{- 34}Js; c = 3 \times {10}^{8}m{s}^{- 1} )

Options

  1. A10.8nm10.8nm
  2. B9.4nm9.4nm
  3. C11.4nm11.4nm
  4. D12.3nm12.3nm

Correct answer

A. 10.8nm10.8nm

Step-by-step solution

Total Number of possible spectral lines that can be emmited when atom de-excites from state nnto groundstate is C2nC2 n
∴nC2=6⇒n(n−1)2=6⇒n=4∴ {}^{n}{C}_{2} = 6 ⇒ \dfrac{n ( n - 1 )}{2} = 6 ⇒ n = 4
1λ=Rz2(112−142)=1.097×107×9×(1−116)\dfrac{1}{\lambda} = R{z}^{2}\left(\dfrac{1}{{1}^{2}} - \dfrac{1}{{4}^{2}}\right) = 1.097 \times {10}^{7} \times 9 \times \left(1 - \dfrac{1}{16}\right)
⇒λ=10.8nm⇒ \lambda = 10.8nm
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