Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is
- 2.8 10-9 m
- 2.8 10-12 m
- < 2.8 10-10 m
- < 2.8 10-9 m
Answer
(1) 2.8 10-9 m
Solution
Energy of the photon: E = hc = 1240 eV nm500 nm = 2.48 eV
Maximum kinetic energy: K = E - 0 = 2.48 - 2.28 = 0.20 eV
= 0.20 1.6 10-19 = 3.2 10-20 J
Minimum de Broglie wavelength: = h 2mK
= 6.63 10-34 2 9.1 10-31 3.2 10-20
= 6.63 10-342.41 10-25 2.8 10-9 m
The emitted electrons have kinetic energies from 0 up to K , their wavelengths are
2.8 10-9 m
Answer: (1)