Question

Light of intensity 2 μW.cm-2 is incident on an iron sample of area 4 cm 2 . Assume that the iron reflects 96% of the incident light and only 3% of the absorbed radiation lie in the violet region, which is above the critical/threshold frequency.

(a) What intensity is available for the photoelectric effect ?

(b) Assuming that all the photons in the violet region have an effective wavelength of 250nm, what is the number of electrons emitted per second?

(c) Calculate the current in the phototubes in amperes ?

(d) If the critical frequency is 1.1×10-15 Hz, what is the iron sample’s work function?

(e) If the photoelectrons are produced by light of λ = 250 nm, calculate their kinetic energies and the stopping potential for the iron sample?

Answer 1

(a) What intensity is available for the photoelectric effect?

I = (3%) (100% - 96%) I₀

I = (0.03) (0.04) I₀

where, I₀ = total intensity of light = 2 x 10^-6 W/cm²

then, we get

I = [(0.03) (0.04) (2 x 10^-6 W/cm²)]

I = 2.4 x 10^-9 W/cm²

I = 2.4 nW/cm²

(b) Assuming that all the photons in violet region have an effective wavelength of 250 nm.

What is the number of electrons emitted per second?

N = I A / h f

N = I A / h c

N = [(250 x 10^-9 m) (2.4 x 10^-9 W/cm²) (4 cm²)] / [(6.62 x 10^-34 J.s) (3 x 10⁸ m/s)]

N = [(2.4 x 10^-15 J.m/s) / (1.986 x 10^-25 J.m)]

N = 1.20 x 10¹⁰ electrons/sec

(c) Calculate the current in phototubes in Amperes?

I = N (1.6 x 10^-19 C/electron)

I = [(1.20 x 10¹⁰ electrons/s) (1.6 x 10^-19 C)]

I = 1.92 x 10^-9 A

(d) If the critical frequency is 1.1 x 10^-15 Hz, what is the iron sample’s work function?

= h f₀

= [(4.14 x 10^-15 eV.s) (1.1 x 10^-15 Hz)]

= 4.5 eV

(e) If the photoelectrons are produced by light, calculate their kinetic energies & the stopping potential for the iron sample?

e V_S = h f -

e V_S = (h c / ) -

e V_S = {[(4.14 x 10^-15 eV.s) (3 x 10⁸ m/s)] / (250 x 10^-9 m)} - (4.5 eV)

e V_S = [(4.96 eV) - (4.5 eV)]

e V_S = 0.46 eV

V_S = 0.46 V