Question

An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of electron.

Answer 1

(a). Using the concept of the De-broglie wavelength,

λ = h/p

⇒ p = h/λ

⇒ p = 6.62 × 10⁻³⁴/10⁻⁹

∴ p = 6.62 × 10⁻²⁵ N-s.

(b). Energy of the Photon = hc/λ

= 6.62 × 10⁻³⁴ × 3 × 10⁸/10⁻⁹

= 19.86 × 10⁻¹⁷ J.

(c). Kinetic Energy of the Electron = p²/2m

= (6.62 × 10⁻²⁵)²/(2 × 9.1 × 10⁻³¹)

= 43.82 × 10⁻⁵⁰/(18.4 × 10⁻³¹)

= 2.38 × 10⁻¹⁹ J.

Hope it helps.

Answer 2

Given :

Wave length of electron and photon =λ=1nm=10⁻⁹m

a) Momentum of electron: Pe= h/λ=6.63 x 10⁻³⁴/10⁻⁹

where h is planks constant

=6.63 x 10⁻²⁵ kg- m/s

Momentum of photon:

Pph=h/λ=6.63 x 10⁻³⁴/10⁻⁹=6.63 x 10⁻²⁵kg-m/s

b) energy of Photon :

E=hc/λ=6.63 x 10⁻³⁴x 3 x 10⁸/10⁻⁹x 1.6 x 10⁻¹⁹

=19.86x 10⁻¹⁷/1.6 x 10⁻¹⁹

=1243eV or 1.24keV

c) energy of electron : E= p²/2me

=(6.63 x 10⁻²⁵)²/2 x 9.1 x 10⁻³¹ x1.6 x 10⁻¹⁹

=1.51 ev