Class 12 Physics - Chapter Dual Nature Of Radiation And Matter NCERT Solutions | What is the (a) momentum, (b) spee
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Kinetic energy of the electron, Ek = 120 eV
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
(a) For the electron, we can write the relation for kinetic energy as: Ek = 1/2 mv2
Where, v = Speed of the electron
Momentum of the electron, p = mv = 9.1 × 10−31 × 6.496 × 106 = 5.91 × 10−24 kg ms−1
Therefore, the momentum of the electron is 5.91 × 10−24 kg m s−1 .
(b) Speed of the electron, v = 6.496 × 106 m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as: λ = h/p = 6.6 x 10-34 / 5.91 x 10-24 = 1.116 x 10-10 m = 0.112 nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
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