Question

A. An electron is moving with a velocity of 2 times 10^6 cm/s. Determine the electron energy in eV, the momentum, and the de Broglie wavelength (in angstroms).

b. The de Broglie wavelength of an electron is 125 angstroms. Determine the electron energy (in eV), the momentum and the velocity.

Answer 1

part a)

p = 1.8222*10^-26 kgm/s

K.E(eV) =1.14*10^-3 eV

λ =0.0364 Angstroms

part b)

p = 5.3*10^-26 kgm/s

V_e = 5.82412*10^4 m/s

K.E(eV) = 9.634*10^-3 eV

Step-by-step explanation:

- Planck's constant h = 6.625*10^-34

- Velocity of an electron V_e = 2*10^6 cm/s

- Mass of an electron m_e = 9.1 * 10^-31 kg

part a)

Find:

Determine the electron energy in eV, the momentum, and the de Broglie wavelength (in angstroms).

Solution:

- The momentum o-f the electron is p:

p = m_e*V_e = (9.1 * 10^-31 )*(2*10^4) = 1.8222*10^-26 kgm/s

- The kinetic energy of the electron is:

K.E = p^2 / 2*m

K.E = (1.82*10^-26)^2 / 2*9.1 * 10^-31 = 1.82*10^-22 J

K.E(eV) = 1.82*10^-22 J * 6.242*10^18 = 1.14*10^-3 eV

- The De-Broglie wavelength is:

λ = h / p

λ = 6.625*10^-34 / 1.822*10^-26

λ = 3.64*10^-12 m = 0.0364 Angstroms

part b)

Find:

The de Broglie wavelength of an electron is 125 angstroms. Determine the electron energy (in eV), the momentum and the velocity.

Solution:

- The De-Broglie wavelength is:

p = h / λ

p = 6.625*10^-34 / 1.25*10^-8

p = 5.3*10^-26 kgm/s

- The momentum p is given by:

V_e = p /m_e

V_e =5.3*10^-26 /9.1 * 10^-31 = 5.82412*10^4 m/s

- - The kinetic energy of the electron is:

K.E = p^2 / 2*m

K.E = (5.3*10^-26)^2 / 2*9.1 * 10^-31 = 1.543406593*10^-21 J

K.E(eV) = 1.543406593*10^-21 J * 6.242*10^18 = 9.634*10^-3 eV