An electron in a TV tube has a speed of 6 × 10⁷ m/s. What is the de Broglie wavelength associated with the electron (m = 9.11 × 10^-31 kg)?

Concept:

De Broglie proposed that the wavelength λ associated with a particle of momentum p is given by:

\(\lambda = \frac{h}{p} = \frac{h}{{mv}}\)

where m is the mass of the particle and v its speed. Planck’s Constant, h = 6.623 × 10^-34 Js,

Above equation is known as the de Broglie relation and the wavelength λ of the matter-wave is called de Broglie wavelength.

Thus, the significance of de Broglie equation lies in the fact that it relates the particle character to the wave character of matter.

Important Points

According to Planck’s quantum theory, the energy of a photon is described as E = hν

According to Einstein’s mass energy relation, E = mc²

Frequency ν can be expressed in terms of wavelength λ as, ν = c/λ

hν = mc² ⇒ hν/c = mc ⇒ λ = h/mc, (This equation is applicable for a photon)

Calculation:

Given:

m = 9.11 × 10-31 kg, v = 6 × 107 m/s

\(\lambda = \frac{h}{{mv}} = \frac{{6.623 \times {{10}^{ - 34}}}}{{9.11 \times {{10}^{ - 31}} \times 6 \times {{10}^7}}} = 1.21 \times {10^{ - 11}}m = 0.121 \) Å