Correct Answer - B
For minimum frequency in Lyman series of H -atom, transition n=2to n=1 takes place.
Now , hv=`DeltaE=E_(2)-E_(1)`
or hv=`[(-(13.6)/(2^(2)))-(-(13.6)/(1^(2)))]eV`
`v=(10.2times1.6times10^(-19))/(6.63times10^(-34))`
` therefore v=2.466times 10^(15)Hz`
If the threshold frequency of metal is v or less, photoelectron will come out.