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Dual Nature of Radiation question

2021 · 1 Sep · Shift 2 · Q49
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  5. /2021 · 1 Sep · Shift 2 · Q49

Dual Nature of Radiation question

2021 · 1 Sep · Shift 2 · Q49

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is : [me = mass of electron = 9 ×\times× 10 −-− 31 kg, h = Planck constant = 6.6 ×\times× 6.6 ×\times× 10 −-− 34 Js, kB = Boltzmann constant = 1.38 ×\times× 10 −-− 23 JK −-− 1]
  1. A
    6.26 nm
  2. B
    8.46 nm
  3. C
    2.26 nm
  4. D
    3.25 nm
View written solutionFree

Correct answer: A

  1. Use thermal energy to estimate electron momentum

For a particle in a 3-dimensional ideal gas,

p22m=32kBT\frac{p^2}{2m} = \frac{3}{2}k_B T2mp2​=23​kB​T

So,

p2=3mkBTp^2 = 3mk_B Tp2=3mkB​T

and hence

p=3mkBTp = \sqrt{3mk_B T}p=3mkB​T​
  1. Apply de-Broglie relation

The de-Broglie wavelength is

λ=hp=h3mkBT\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mk_B T}}λ=ph​=3mkB​T​h​
  1. Substitute the given values

Given:

h=6.6×10−34 J s,m=9×10−31 kg,kB=1.38×10−23 J K−1,T=300 Kh = 6.6\times 10^{-34}\ \text{J s}, \quad m = 9\times 10^{-31}\ \text{kg}, \quad k_B = 1.38\times 10^{-23}\ \text{J K}^{-1}, \quad T=300\ \text{K}h=6.6×10−34 J s,m=9×10−31 kg,kB​=1.38×10−23 J K−1,T=300 K

Thus,

λ=6.6×10−343×9×10−31×1.38×10−23×300\lambda = \frac{6.6\times 10^{-34}}{\sqrt{3\times 9\times 10^{-31}\times 1.38\times 10^{-23}\times 300}}λ=3×9×10−31×1.38×10−23×300​6.6×10−34​
  1. Calculate the denominator inside the square root
3×9×1.38×300=111783\times 9\times 1.38\times 300 = 111783×9×1.38×300=11178

So,

3mkBT=11178×10−54=1.1178×10−503mk_B T = 11178\times 10^{-54} = 1.1178\times 10^{-50}3mkB​T=11178×10−54=1.1178×10−50

Therefore,

p=1.1178×10−50=1.1178×10−25≈1.057×10−25p = \sqrt{1.1178\times 10^{-50}} = \sqrt{1.1178}\times 10^{-25} \approx 1.057\times 10^{-25}p=1.1178×10−50​=1.1178​×10−25≈1.057×10−25
  1. Now compute wavelength
λ=6.6×10−341.057×10−25\lambda = \frac{6.6\times 10^{-34}}{1.057\times 10^{-25}}λ=1.057×10−256.6×10−34​ λ≈6.24×10−9 m\lambda \approx 6.24\times 10^{-9}\ \text{m}λ≈6.24×10−9 m λ≈6.24 nm\lambda \approx 6.24\ \text{nm}λ≈6.24 nm

This is approximately

6.26 nm\boxed{6.26\ \text{nm}}6.26 nm​
  1. Check options
  • A: 6.26 nm6.26\,\text{nm}6.26nm ✅
  • B: 8.46 nm8.46\,\text{nm}8.46nm ❌
  • C: 2.26 nm2.26\,\text{nm}2.26nm ❌
  • D: 3.25 nm3.25\,\text{nm}3.25nm ❌

Therefore, the correct option is:

A\boxed{\text{A}}A​
  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They match.

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