Photoelectric emission is observed from a metallic surface for frequencies`v_(1) and v_(2)`of the incident light rays `(v_(1)gtv_(2))`.If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ratio `1:k,then the threshold frequency of the metallic surface is

Options

A. `(v_(2)-v_(1))/(k-1)`
B. `(kv_(1)-v_(2))/(k-1)`
C. `(kv_(2)-v_(1))/(k-1)`
D. `(v_(2)-v_(1))/k`

Answer

B

Explanation

The expression for `(K.E) _(max)` of photoelectrons is given as
`(K.E)_(max)=hv-hv_(o)`
For frequency `v_(1)`
`(K.E)_(1max)=h_(1)=hv_(O)`
For Frequency `V_(2)`
`(k.E)_(2max)=hv_(2)-hv_(o)`
Therefore the threshold frequency can be obtained as,
`(K.E)_(1)/((K.E)_ (2))=(hv_(2)-hv_(o))/(hv_(2)-hv_(o)`
`(1)/(k)=(v_1-vo)/(v_2-vo`
(`therefore` the ratio of K.E of Photoelectron is 1:K)
`impliesv_(2)-v_(O)=kV_(1)-Kv_(o)`
`impliesKv_(o)-v_(o)=Kv_(1)-v_(2)`
`implies v_(o)(k-1)=kv_(1)-v_(2)`
`thereforev_(o)=(kv_(1)-v_(2))/(k-1)`