Question

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

Answer 1

Given condition,

Kinetic energy of the Electron  = 120 eV.

= 120 × 1.6 × 10⁻¹⁹ J.

= 192 × 10⁻¹⁹ J

We know mass of the Electron = 9.1 × 10⁻³¹ kg.

Now,

(a). For Momentum,

We know,

 p² = 2K.E.×  m

p ² = 2 × 192 × 10⁻¹⁹ × 9.1 × 10⁻³¹

⇒ p² = 3494.4 × 10⁻⁵⁰

⇒ p = 59.11 × 10⁻²⁵ N-s.

(b). For Speed,

K.E. = 1/2 × mv²

v² = 2 K.E./m

⇒ v² = 2 × 192 × 10⁻¹⁹/9.1 × 10⁻³¹

⇒ v² = 42.2 × 10¹²

⇒ v = 6.5 × 10⁶ m/s.

(c). For De-Broglie Wavelength,

λ = h/mv

∴ λ = 6.62 × 10⁻³⁴/(9.1 × 10⁻³¹ × 6.5 × 10⁶)

⇒  λ = 0.119 × 10⁻⁹

∴ λ = 1.19 × 10⁻¹⁰ m.

Hope it helps.

Answer 2

Given :

Kinetic energy =KE= 120ev=120x 1.6x 10⁻¹⁹J

a) momentum = p =√2meV=√2mk.e [ since K.E=1/2mv2= ev]

where mass of electron is 9.1 x 10⁻³¹kg

=√2x9.1x10⁻³¹x⁻³¹

=5.91x 10⁻²⁴ kg-m/s

b) Momentum=p=mv

v=p/m

=5.91x 10⁻²⁴ /9.1 x 10⁻³¹

=6.5 x 10⁶ m/s

c) De brogile wave length =λ=12.27/√v A°

=12.27/√120 A°

=0.112 x 10⁻⁹=0.112nm