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Dual Nature of Radiation question

2020 · 6 Sep · Shift 2 · Q36
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Dual Nature of Radiation question

2020 · 6 Sep · Shift 2 · Q36

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to : (Given : nitrogen molecule weight : 4.64 ×\times× 10–26 kg, Boltzman constant: 1.38 ×\times× 10–23 J/K, Planck constant : 6.63 ×\times× 10–34 J.s)
  1. A
    0.44 Ao\mathop A\limits^oAo​
  2. B
    0.34 Ao\mathop A\limits^oAo​
  3. C
    0.20 Ao\mathop A\limits^oAo​
  4. D
    0.24 Ao\mathop A\limits^oAo​
View written solutionFree

Correct answer: D

  1. Use de-Broglie relation

For a particle,

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}λ=ph​=mvh​

Here the nitrogen molecule is moving with r.m.s. speed at temperature TTT, so

vrms=3kTmv_{\text{rms}} = \sqrt{\frac{3kT}{m}}vrms​=m3kT​​

Therefore,

λ=hm3kTm=h3mkT\lambda = \frac{h}{m\sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3mkT}}λ=mm3kT​​h​=3mkT​h​
  1. Substitute the given values

Given:

m=4.64×10−26 kgm = 4.64\times 10^{-26}\ \text{kg}m=4.64×10−26 kg k=1.38×10−23 J/Kk = 1.38\times 10^{-23}\ \text{J/K}k=1.38×10−23 J/K T=400 KT = 400\ \text{K}T=400 K h=6.63×10−34 J⋅sh = 6.63\times 10^{-34}\ \text{J·s}h=6.63×10−34 J⋅s

So,

λ=6.63×10−343×4.64×10−26×1.38×10−23×400\lambda = \frac{6.63\times 10^{-34}}{\sqrt{3\times 4.64\times 10^{-26}\times 1.38\times 10^{-23}\times 400}}λ=3×4.64×10−26×1.38×10−23×400​6.63×10−34​
  1. Calculate the quantity inside the square root

First,

3×400=12003\times 400 = 12003×400=1200

Now,

4.64×1.38=6.40324.64\times 1.38 = 6.40324.64×1.38=6.4032

Thus,

3mkT=1200×6.4032×10−49=7683.84×10−49=7.68384×10−463mkT = 1200\times 6.4032\times 10^{-49} = 7683.84\times 10^{-49} = 7.68384\times 10^{-46}3mkT=1200×6.4032×10−49=7683.84×10−49=7.68384×10−46

Hence,

3mkT=7.68384×10−46\sqrt{3mkT} = \sqrt{7.68384\times 10^{-46}}3mkT​=7.68384×10−46​ =7.68384×10−23≈2.77×10−23= \sqrt{7.68384}\times 10^{-23} \approx 2.77\times 10^{-23}=7.68384​×10−23≈2.77×10−23
  1. Now compute the wavelength
λ=6.63×10−342.77×10−23\lambda = \frac{6.63\times 10^{-34}}{2.77\times 10^{-23}}λ=2.77×10−236.63×10−34​ λ≈2.39×10−11 m\lambda \approx 2.39\times 10^{-11}\ \text{m}λ≈2.39×10−11 m
  1. Convert into angstrom

Since,

1 A˚=10−10 m1\ \mathring{A} = 10^{-10}\ \text{m}1 A˚=10−10 m

so,

λ=2.39×10−11 m=0.239 A˚\lambda = 2.39\times 10^{-11}\ \text{m} = 0.239\ \mathring{A}λ=2.39×10−11 m=0.239 A˚

This is closest to

0.24 A˚0.24\ \mathring{A}0.24 A˚
  1. Check options
  • A: 0.44 A˚0.44\ \mathring{A}0.44 A˚
  • B: 0.34 A˚0.34\ \mathring{A}0.34 A˚
  • C: 0.20 A˚0.20\ \mathring{A}0.20 A˚
  • D: 0.24 A˚0.24\ \mathring{A}0.24 A˚

So the correct option is D.

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