Question

n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How fast does the electron move away from the metal

Answer 1

The speed of the electron is 1.371 x 10⁶ m/s.

Step-by-step explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

`E = (hc)/(\lambda) \\\\E = (6.626*10^(-34) *3*10^(8))/(130*10^(-9)) \\\\E = 15.291*10^(-19) \ J`

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

`v = \sqrt{(2K.E)/(m) } \\\\v = \sqrt{(2*8.563*10^(-19))/(9.11*10^(-31))}\\\\v = 1.371 *10^(6) \ m/s`

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.