Question

(a) The work function for the surface of aluminium is 4.2eV. How much potential difference will be required to stop the emission of maximum energy electrons emitted by light of 2000Å wavelenght?
(b) What will be the wavelength of that incident light for which stopping potential will be zero ? `h=6.6xx10^(-34) Js, c=3xx10^8ms^-1`.

Answer 1

Here, `phi_0=4.2eV`
`=4.2xx1.6xx10^(-19)J,`
`V=?, lambda=2000Å=2000xx10^(-10)m`
Max. K.E. of the emitted photoelectron,
`K_(max)=(hc)/lambda-phi_0`
`(6.6xx10^(-34)xx3xx10^8)/(2000xx10^(-10))-4.2xx1.6xx10^(-19)J`
`=3.18xx10^(-19)J`
stopping potential, `V_0=(K_(max))/e=(3.18xx10^(-19))/(1.6xx10^(-19))`
`=1.9875V`
(b) For threshold wavelength `lambda_0`, the stopping
potential is zero.
`:. lambda_0=(hc)/(phi_0)=(6.6xx10^(-34)xx3xx10^8)/(4.2xx1.6xx10^(-19))`
`=2.946xx10^-7m`
`=2946Å`