Question

What is the maximum number of electrons that can be emitted if a potassium metal surface absorbs 3.50✖️10^-3 J of radiation at a wavelength of 250.0 nm? The work function (binding energy) of potassium metal is 3.845✖️10^-19 J. What is the velocity of the elctron emitted?

Answer 1

#(a):

Given wavelength of absorbed radiation, = 250.0 nm * (10^-9 m / 1 nm) = 2.50*10^-7 m

=> Frequency of the absorbed light, = C / = 3*10⁸ m.s^-1 / 2.50*10^-7 m = 1.2*10¹⁵ Hz or s^-1  

Suppose there are "n" photons in the radiation of 3.50*10^-3 J.

=> energy of a single absorbed photon, E = h* = 6.626*10^-34 J.s * 1.2*10¹⁵ s^-1 = 7.95*10^-19 J

"n" number of photons will also produce "n" number of electrons

=> maximum number of electrons formed by 3.50*10^-3 J energy = Total energy / energy of single photon

= 3.50*10^-3 J / 7.95*10^-19 J = 4.40*10¹⁵ electrons (Answer)

#(b): energy of a single absorbed photon, E = 7.95*10^-19 J

work function, = 3.845*10^-19 J

=> Amount of absorbed energy that converts to kinetic energy of free electron, E_e = E - =  7.95*10^-19 J -  3.845*10^-19 J

= 4.105*10^-19 J

Also E_e = (1/2)*m_e * V²

where me = mass of electron = 9.11*10^-31 Kg

=> V = square root(2E_e / m_e)

=> V = square root(2*4.105*10^-19 J / 9.11*10^-31 Kg )  

=> V = 9.49*10⁵ m/s (Answer)