Question

when light of wavelength 248 nm falls on a metal of threshold energy 3.0 ev , the de-broglie wavelength of emitted electrons is A°.

Answer 1

The wavelength of the electrons is obtained as 0.087 A°.

The de Broglie wavelength is a concept in quantum mechanics introduced by Louis de Broglie. It suggests that particles, such as electrons, exhibit both particle and wave-like properties.

The threshold energy is 3ev or 4.8 *
`10^{-19` J

The energy of the light;

E = hc/λ

E = 6.6 *
`10^{-34` * 3 *
`10^8`/248 *
`10^{-9`

E = 7.98 *
`10^{-19` J

KE = 7.98 *
`10^{-19` J - 4.8 *
`10^{-19` J

= 3.18 *
`10^{-19` J

The velocity of the electrons= √2KE/m

= √2 * 3.18 *
`10^{-19` /9.11 *
`10^{-31`

= 8.35 *
`10^5` m/s

But momentum = mv

λ= h/mv

λ= 6.6 *
`10^{-34`/9.11 *
`10^{-31` * 8.35 *
`10^5`

= 8.7 *
`10^{-12` m or 0.087 A°