what is the kinetic energy of an electron?

Question

Experiments with calcium show that it will emit electrons if light above 6.55*10^14 Hz is used. What is the kinetic energy of an electron when light with frequency of 7.26*10^14 Hz shines on the surface?

Thanks

Answer 1

Well, the photon energy is determined by its frequency ν with the following formula: E = hν, where h is the Planck constant: h = 6.626x10^(-34) J·s.
Now, experiments with calcium show that it will emit electrons if light above 6.55*10^14 Hz is used. This means the energy corresponding to 6.55*10^14 Hz is used for the electron to move away from the nuclear. When light with frequency of 7.26*10^14 Hz shines on the surface, the energy corresponding to 6.55*10^14 Hz is STILL needed for the electron to move away from the nuclear. The difference is converted to the kinetic energy of the electron:
KE = h*(7.26*10^14 - 6.55*10^14)
= 6.626x10^(-34)*0.71x10^14 J
= 4.70x10^(-20) J