Question

When a light of wavelength 570nm falls on the surface of potassium metal, electrons are ejected with a velocity of 6.4x104 m/s. What is the minimum energy required to remove an electron from potassium metal?

Answer 1

The minimum energy required to remove an electron from potassium metal is 3.465 x 10⁻¹⁹ J

Step-by-step explanation:

Given;

light wavelength, λ = 570nm

velocity of ejected electron, v = 6.4 x 10⁴ m/s

Minimum energy required to remove an electron from a metal surface is given as;

W₀ = hf₀

where;

W₀ is work function also known as minimum energy required

h is Planck's constant

f₀ is threshold frequency of light

Also,

E = W₀ + K.E

where;

E is the energy of the incident light

W₀ is work function

K.E is the kinetic energy of the electron

`h(c)/(\lambda) = W_o + (1)/(2) mv^2\\\\W_o = h(c)/(\lambda) - (1)/(2) mv^2\\\\`

where;

m is mass of electron

c is speed of light

`W_o = (6.62 *10^(-34)*3*10^8)/(570*10^(-9))\ - (1)/(2) *9.1*10^(-31)*(6.4*10^4)^2\\\\W_o = 3.484*10^(-19) J \ - \ 0.01864*10^(-19) J\\\\W_o = 3.465 *10^(-19) J`

Therefore, the minimum energy required to remove an electron from potassium metal is 3.465 x 10⁻¹⁹ J