Question

What is the de Broglie wavelength of an election, whose kinetic energy is 120 eV ?

(Given: m = 9.11 x 10^-31 kg, h = 6.63 x 10^-34 Js, 1eV = 1.6 x 10^-19 J)

1. 100 nm
2. 0.112 nm
3. 0.8 nm
4. 0.5 nm

Answer 1

Correct Answer - Option 2 : 0.112 nm

CONCEPT:

• De-Broglie wavelength: The wavelength of an electron due to its motion is called the de-Broglie wavelength of the electron.

The de-Broglie wavelength of the electron (λe) is given by:

λe = h/p = h/(mv)

\({λ _e} = \;\frac{h}{{\sqrt {2m\;\left( {KE} \right)} }}\)

Where, h is Planck's constant, P is momentum, v is velocity, m is the mass of the particle, KE is the kinetic energy of the particle.

CALCULATION:

Given that:

m = 9.11 x 10-31 kg, h = 6.63 x 10-34 Js, 1eV = 1.6 x 10-19 J

Kinetic energy (KE) = 120 eV = 120 × 1.6 × 10^-19 J

KE = 192 × 10^-19 J

The de-Broglie wavelength of the electron (λe) is given by:

\({λ _e} = \;\frac{h}{{\sqrt {2m\;\left( {KE} \right)} }}\)

\({λ _e} = \;\frac{6.63 × 10^{-34}}{{\sqrt {2× 9.11 × 10^{-31}\;\left( {192× 10^{-19}} \right)} }}\)

De-Broglie wavelength of the electron (λ_e) = 0.112 × 10^-9 m = 0.112 nm

Hence option 2 is correct.