Question

Indicate which of the following photons can cause emission of photoelectrons from the surface of gallium ( 6.7 × 10–19 J)? A) 300 nm B) 400 nm C)

Answer 1

The energy of a photon is given by:

`E=hf=(hc)/(lambda)`

For (a):

`E = (6.63xx10^(-34)xx3xx10^(8))/(300xx10^(-9))"J"`

`E=6.63xx10^(-19)"J"`

I assume that `6.7xx10^(-19)"J"` refers to the work function of the metal in which case the photon will not have enough energy to eject an electron.

Neither will photon (b) as, without bothering with the calculation, it has a longer wavelength hence a lower frequency and so a lower energy.