The work function of a certain metal is `3.31xx10^-(-19) j`then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength `5000`ÅIS1(GIVEN,H=`6.62XX10^(-34)J_(-s),c=3xx10^(8)ms^(-1),e=1.6xx10^(-19)C)`

Options

A. `2.48 eV`
B. `0,41 ev`
C. `2.07eV`
D. `0.82 eV`

Answer

b

Explanation

According to Einstein equation,
`E=W_0+K.E`
`implies K.E=(hc)/lambda-W_0=(6.62xx10^(-34)xx3xx10^(8))/(5000xx10^10)-3.31xx10^-19`
`=3.972xx10^(-19)-3.31xx10^(-19)`
`=(3.972-3.31)xx10^(-19)`
`=0.662xx10^(-19)`
`therefore E=(K.E)/(e)=(0.662xx10^(-19))/(1.6xx10^(-19))=0,41 eV`