Question

An atom of a rare earth element is illuminated at a wavelength of 7562 Ångstroms. The surface of the metal emits a photoelectron with a kinetic energy of 1.95 x 10-20J. Calculate the threshold frequency for this metal. (1 angstrom = 10^-10m)

Answer 1

Answer:

0.625*10e-20=Energy and frequency=9.45*10e10

Explanation:

1. We know the photon will imapart energy and when the electrons gets the energy it gets released from the atom.

2. If the electron gains only threshold energy then it will get emitted but won't have kinetic energy

3.So, We know Kinetic energy=Photons energy- Threshold energy of metal

4.1.95*10e-20=hc/λ- threshold energy

5. We found energy comes out to tobe 0.625*10e-20J

6. hv=energy

7.0.625*10e-20/6.62*e-34=9.45*10e10=frequency