Question

The work function for cesium is 1.96 eV. (a) Find the cutoff wavelength for the metal, (b) what is the maximum kinetic energy for the emitted electrons when 425 nm light is incident on the metal

Answer 1

Answer:

Explanation:

λ = hc/¢

Where

h = the Plank constant 6.63 x 10-34 Is

C = 3.0×10^8

¢= 1.96eV

= (6.63×10^-34Js)×(3×10^8)÷( 1.96eV) × 1eV/1.6×10^-19J

= (1.989×10^-25)÷( 1.96eV)×1eV/1.6×10^-19J

= 6.342×10^-7m

B) maximum kinetic energy

= K=hf−ϕ ........1

ϕ = hc

Where

​ h = constant 6.63 x 10^-34Js

ϕ= 1.96eV

Recall

λ =425×10^-9m

f = frequency in Hz

f = c / λ

C = 3.0×10^8

f = 3.0×10^8 / 425×10^-9m

f = 0.000705Hz

From equation 1

K = (6.63 x 10^-34Js×0.000705Hz )- 6.63 x 10^-34Js×3.0×10^8

= 4.68×10^-37 - 1.989×10^-25

= - 1.98×10^-25J