Question:
What is the

(a) momentum,

(b) speed, and

(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

Answer:

Kinetic energy of the electron, E_k = 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: E_k = 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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Question:
What is the

(a) momentum,

(b) speed, and

(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

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Question: What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

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Question:
What is the

(a) momentum,

(b) speed, and

(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.