Answer to Question #88637 in Atomic and Nuclear Physics for rejoice

Question

the stopping potential of a photoelestric chamber is 3v the work function is 2.2ev calculate the incident photon energy and frequency

Answer 1

Einstein’s Photoelectric Equation

$$hν=W+K_{max} (1)$$

where hν is the photon energy, W is the work function, Kmax is the maximum kinetic energy of the emitted electrons

The maximum kinetic energy of the emitted electrons is equal to

$$K_{max}=eV (2)$$

where e=-1.6×10^-19 C

In our case, V=3 v, W=2.2 ev

Using (1) and (2) we got:

hν=7 ev or 11.2×10^-19 J

The photon energy is equal to

$$hν (3)$$

where h is Planck's constant

Using (3) we got: ν=1.7×10¹⁵ Hz

Answer:

7 ev or 11.2×10^-19 J

1.7×10¹⁵ Hz