Question
A certain metal when irradiated by light $$\left( {\upsilon = 3.2 \times {{10}^{16}}Hz} \right)$$ emits photoelectrons with twice of $$K.E.$$ as did photoelectrons when the same metal is irradiated by light $$\left( {\upsilon = 2.0 \times {{10}^{16}}Hz} \right).$$ The $${\upsilon _0}$$ of the metal is
A.
$$1.2 \times {10^{14}}\,Hz$$
B.
$$8 \times {10^{15}}\,Hz$$
C.
$$1.2 \times {10^{16}}\,Hz$$
D.
$$4 \times {10^{12}}\,Hz$$
Answer :
$$8 \times {10^{15}}\,Hz$$
Solution :
$$\eqalign{
& {\left( {K.E.} \right)_1} = h{\upsilon _1} - h{\upsilon _0} \cr
& {\left( {K.E.} \right)_2} = h{\upsilon _2} - h{\upsilon _0} \cr
& {\text{As}}\,\,\,{\left( {K.E.} \right)_1} = 2 \times {\left( {K.E.} \right)_2} \cr
& \therefore \,\,\left( {h{\upsilon _1} - h{\upsilon _0}} \right) = 2\left( {h{\upsilon _2} - h{\upsilon _0}} \right) \cr} $$
$${\text{or}}$$ $${\upsilon _0} = 2{\upsilon _2} - {\upsilon _1} = 2 \times \left( {2 \times {{10}^{16}}} \right) - \left( {3.2 \times {{10}^{16}}} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.8 \times {10^{16}}\,Hz\,\,\,{\text{or}}\,\,\,8 \times {10^{15}}\,Hz$$