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Dual Nature of Radiation question

2019 · 11 Jan · Shift 1 · Q62
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Dual Nature of Radiation question

2019 · 11 Jan · Shift 1 · Q62

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
If the de Broglie wavelength of an electron is equal to the 10–3 times the wavelength of a photon of frequency 6 ×\times× 1014 Hz, then the speed of electron is equal to : (Speed of light = 3 ×\times× 108 m/s, Planck's constant = 6.63 ×\times× 10–34 J.s, Mass of electron = 9.1 ×\times× 10–31 kg)
  1. A
    1.7 ×\times× 106 m/s
  2. B
    1.45 ×\times× 106 m/s
  3. C
    1.1 ×\times× 106 m/s
  4. D
    1.8 ×\times× 106 m/s
View written solutionFree

Correct answer: B

  1. Find the wavelength of the photon

For a photon,

λγ=cν\lambda_{\gamma} = \frac{c}{\nu}λγ​=νc​

Given:

c=3×108 m/s,ν=6×1014 Hzc = 3\times 10^8\ \text{m/s}, \qquad \nu = 6\times 10^{14}\ \text{Hz}c=3×108 m/s,ν=6×1014 Hz

So,

λγ=3×1086×1014=0.5×10−6=5×10−7 m\lambda_{\gamma} = \frac{3\times 10^8}{6\times 10^{14}} = 0.5\times 10^{-6} = 5\times 10^{-7}\ \text{m}λγ​=6×10143×108​=0.5×10−6=5×10−7 m
  1. Use the given relation for electron wavelength

The de Broglie wavelength of the electron is 10−310^{-3}10−3 times the photon wavelength:

λe=10−3λγ=10−3×5×10−7=5×10−10 m\lambda_e = 10^{-3}\lambda_{\gamma} = 10^{-3}\times 5\times 10^{-7} = 5\times 10^{-10}\ \text{m}λe​=10−3λγ​=10−3×5×10−7=5×10−10 m
  1. Apply de Broglie relation for electron

For an electron,

λe=hmv\lambda_e = \frac{h}{mv}λe​=mvh​

Thus,

v=hmλev = \frac{h}{m\lambda_e}v=mλe​h​

Substitute the values:

v=6.63×10−34(9.1×10−31)(5×10−10)v = \frac{6.63\times 10^{-34}}{(9.1\times 10^{-31})(5\times 10^{-10})}v=(9.1×10−31)(5×10−10)6.63×10−34​
  1. Calculate the denominator
(9.1×10−31)(5×10−10)=45.5×10−41=4.55×10−40(9.1\times 10^{-31})(5\times 10^{-10}) = 45.5\times 10^{-41} = 4.55\times 10^{-40}(9.1×10−31)(5×10−10)=45.5×10−41=4.55×10−40

So,

v=6.63×10−344.55×10−40=6.634.55×106v = \frac{6.63\times 10^{-34}}{4.55\times 10^{-40}} = \frac{6.63}{4.55}\times 10^6v=4.55×10−406.63×10−34​=4.556.63​×106 v≈1.457×106 m/sv \approx 1.457\times 10^6\ \text{m/s}v≈1.457×106 m/s
  1. Match with the options
v≈1.45×106 m/sv \approx 1.45\times 10^6\ \text{m/s}v≈1.45×106 m/s

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer also gives B, so they agree.

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