Modern Physics 4 Question 5

5. If the de-Broglie wavelength of an electron is equal to 103 times, the wavelength of a photon of frequency 6×1014Hz, then the speed of electron is equal to

(Take, speed of light =3×108m/s,

Planck’s constant =6.63×1034J-s and

mass of electron =9.1×1031kg )

(Main 2019, 11 Jan I)

(a) 1.45×106m/s

(b) 1.8×106m/s

(c) 1.1×106m/s

(d) 1.7×106m/s

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Solution:

  1. Wavelength of the given photon is given as,

λp=cνp=3×1086×1014m=5×107m

As, it is given that, de-Broglie wavelength of the electron is

λe=103×λp=5×1010m

[ using Eq. (i) ]

Also, the de-Broglie wavelength of an electron is given as,

λe=hp=hmveve=hλeme

Substituting the given values, we get

=6.63×10345×1010×9.1×1031m/s=1.45×106m/s



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