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

Question #88637
the stopping potential of a photoelestric chamber is 3v the work function is 2.2ev calculate the incident photon energy and frequency
1
Expert's answer
2019-04-29T09:33:22-0400

Einstein’s Photoelectric Equation


"h\u03bd=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


"h\u03bd (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



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