Question

A photocathode has a work function of 2.4 ev. the photocathode is illuminated withmonochromatic radiation whose photon energy is 3.5 ev. the wavelength of the illuminatingradiation is closest to:

Answer 1

The photon energy is equal to

`E=3.5 eV = 5.6 \cdot 10^(-18)J`
We also know that the energy of a photon is equal to

`E=hf`
where h is the Planck constant and f is the photon frequency. By re-arranging this equation, we can find the photon frequency:

`f= (E)/(h)= (5.6 \cdot 10^(-18)J)/(6.6 \cdot 10^(-34) Js)=8.48 \cdot 10^(15)Hz`

And now we can find its wavelength by using the relationship between frequency, wavelength and a speed of a photon (which is the speed of light, c):

`\lambda= (c)/(f)= (3 \cdot 10^8 m/s)/(8.48 \cdot 10^(15)Hz)=3.54 \cdot 10^(-8) m`