Question

What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV. Tries 0/20 In this experiment, what is the minimal potential needed to fully stop the electrons if the wavelength of the incident light is 321 nm?

Answer 1

4.24nm

0.385eV

Step-by-step explanation:

Maximum wavelength (λmax) :

λmax = ( hc) /Φ

h = plancks constant = 6.63 * 10^-34

c = speed of light = 3*10^8

1ev = 1.6 * 10^-19

Φ = 2.93eV = 2.93* (1.6*10^-19) = 4.688*10^-19

λmax = [(6.63 * 10^-34) * (3 * 10^8)] / 4.688*10^-19

λmax = 19.89 * 10^-26 / 4.688*10^-19

λmax = 4.242 * 10^-7 m

λmax= 4.24nm

B.)

E = hc / eλ eV

λ = 3.75nm = 3.75 * 10^-7m = 375 *10^-9

E = (6.63 * 10^-34) * (3 * 10^8) / (1.6 * 10^-19) * (375 * 10^-9)

E = 19.89 * 10^-26 / 600 * 10^-28

E = 0.03315 * 10^-26 + 28

E = 0.03315 * 10^2

E = 3.315 eV

Stopping potential : (3.315 eV - 2.93eV) = 0.385eV