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Structure of Atom question

2023 · 6 Apr · Shift 1 · Q15
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Structure of Atom question

2023 · 6 Apr · Shift 1 · Q15

JEE MainChemistryStructure of AtomNumerical+4 / −1
The wavelength of an electron of kinetic energy 4.50×10−294.50\times10^{-29}4.50×10−29 J is ‾\underline{\hspace{2cm}}​×10−5\times 10^{-5}×10−5 m. (Nearest integer) Given : mass of electron is 9×10−319\times10^{-31}9×10−31 kg, h =6.6×10−34=6.6\times10^{-34}=6.6×10−34 J s
Numerical answer
View written solutionFree

Correct answer: 7

  1. Use de Broglie wavelength relation

For a particle,

λ=hp\lambda = \frac{h}{p}λ=ph​

For a non-relativistic electron,

K=p22mK = \frac{p^2}{2m}K=2mp2​

so

p=2mKp = \sqrt{2mK}p=2mK​

Hence,

λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}λ=2mK​h​
  1. Substitute the given values

Given:

h=6.6×10−34 J s,m=9×10−31 kg,K=4.50×10−29 Jh=6.6\times10^{-34}\ \text{J s},\quad m=9\times10^{-31}\ \text{kg},\quad K=4.50\times10^{-29}\ \text{J}h=6.6×10−34 J s,m=9×10−31 kg,K=4.50×10−29 J

So,

λ=6.6×10−342×9×10−31×4.50×10−29\lambda = \frac{6.6\times10^{-34}}{\sqrt{2\times 9\times10^{-31}\times 4.50\times10^{-29}}}λ=2×9×10−31×4.50×10−29​6.6×10−34​
  1. Simplify inside the square root
2×9×4.50=812\times 9\times 4.50 = 812×9×4.50=81

and

10−31×10−29=10−6010^{-31}\times10^{-29}=10^{-60}10−31×10−29=10−60

Therefore,

2mK=81×10−602mK = 81\times10^{-60}2mK=81×10−60

Thus,

2mK=81×10−60=9×10−30\sqrt{2mK} = \sqrt{81\times10^{-60}} = 9\times10^{-30}2mK​=81×10−60​=9×10−30
  1. Compute the wavelength
λ=6.6×10−349×10−30\lambda = \frac{6.6\times10^{-34}}{9\times10^{-30}}λ=9×10−306.6×10−34​ λ=6.69×10−4\lambda = \frac{6.6}{9}\times10^{-4}λ=96.6​×10−4 λ≈0.733×10−4=7.33×10−5 m\lambda \approx 0.733\times10^{-4} = 7.33\times10^{-5}\ \text{m}λ≈0.733×10−4=7.33×10−5 m
  1. Nearest integer for the blank

The wavelength is of the form

λ≈7.33×10−5 m\lambda \approx 7.33\times10^{-5}\ \text{m}λ≈7.33×10−5 m

So the required nearest integer is

7\boxed{7}7​
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