(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`