Question
Ultraviolet radiation of $$6.2\,eV$$ falls on an aluminium surface. $$KE$$ of fastest electron emitted is (work function $$= 4.2\,eV$$ )
A.
$$3.2 \times {10^{ - 21}}J$$
B.
$$3.2 \times {10^{ - 19}}J$$
C.
$$7 \times {10^{ - 25}}J$$
D.
$$9 \times {10^{ - 32}}J$$
Answer :
$$3.2 \times {10^{ - 19}}J$$
Solution :
According to Einstein photoelectric equation
$$\eqalign{
& KE = E = {W_0}\,\,\left[ {_{{W_0} = \,{\text{work}}\,{\text{function}}}^{E\, = \,\,{\text{energy incidented}}}} \right] \cr
& {\text{Here,}}\,\,E = 6.2\,eV \cr
& {W_0} = 4.2\,eV \cr
& \therefore KE = 6.2 - 4.2 = 2.0\,eV \cr
& = 2 \times 1.6 \times {10^{ - 19}} \cr
& = 3.2 \times {10^{ - 19}}\,J \cr} $$