Question

(a) The work function for the surface of aluminium is 4.2eV. Find potential difference will be required to stop the emission of maximum kinetic energy photoelectrons emitted by light of 1800Å wavelenght?
(b) Determine the wavelength of that incident light for which stopping potential will be zero? `(h=6.6xx10^(-34))`

Answer 1

(a) Here, `phi_0=4.2eV=4.2xx1.6xx10^(-19)J`,
`lambda=1800Å=1800xx10^(-10)m=18xx10^-8m`,
`V_0=?`
Maximum KE of the emitted photoelectron
`K_(max)=(hc)/lambda-phi_0=((6.6xx10^(-34))xx(3xx10^8))/(18xx10^-8)`
`-4.2xx1.6xx10^(-19)`
`=11xx10^(-19)-6.72xx110^(-19)=4.28xx10^(-19)`
Stopping potential,
`V_0=(K_(max))/l=(4.28xx10^(-19))/(1.6xx10^(-19))=2.675V`