Question

The minimum energy required to overcome the attractive forces between an electron and the surface of Ag metal is 5.52 x ${10}^{-19}$J. What will be the maximum kinetic energy of electron ejected out from Ag which is being exposed to UV-light of = 360A ?

Answer 1

Energy absorbed $=hc/\lambda$

$=\frac{6.625\times {10}^{-27}\times 3\times {10}^8}{360\times {10}^{-8}}$

$=5.52\times {10}^{-11}erg$

or $=5.52\times {10}^{-18}J$ [$1erg={10}^{-7}J$]

$E_{absorbed}=$ E used in attractive force + kinetic energy of the electron

$=5.52\times {10}^{-18}J-5.52\times {10}^{-19}J$

$=49.68\times {10}^{-19}J$