Question

Q. 1: (a) Write down de Broglie’s formula in terms of Kinetic Energy Ek of the particle involved.

(b) Calculate the electron’s and proton’s de Broglie wavelengths when each of them has kinetic energy 25x10-3eV.

(c) The de-Broglie wavelength of electron-beam used in the electron-microscope is 40 pico-meters. The electron-beam is accelerated from rest using potential difference Va, determine the applied voltage.

Answer 1

1: (a)

De-Broglie wavelength, λ = h/mv =h/p where, m is the mass and v is the velocity.

The momentum p and kinetic energy E of a particle can be related as p =

de Broglie’s formula in terms of Kinetic Energy Ek of the particle is λ = h/

(b)

Mass of electron, me = 9.11 x 10^-31 kg

Mass of proton, mp = 1.67 x 10^-27 kg

Kinetic energy of both electron and proton, E = 25 x 10^-3 eV = 25 x10^-3 x 1.6 x 10^-19 J = 40 x 10^-22 J

De-Broglie wavelength of electron,

λ_e = h/ = 6.625 x 10^-34/ 40 x 10^-22 = 0.775 x 10^-8 m

De-Broglie wavelength of proton,

λ_p = h/ = 6.625 x 10^-34/ 40 x 10^-22 = 1.81 x 10^-10m

( c)

De-Broglie wavelength of electron, λ_e = 40 pico meter = 40 x 10^-12 m

Charge of electron, q = 1.6 x 10^-19 C

If Va is the accelerating voltage, the de-Broglie wavelength of electron is λ_e = h/

40 x 10^-12 = 6.625 x 10^-34/

Va = 940 V

the accelerating voltage, Va = 940 V