Correct Answer - 3
The threshold frequency `(v_(0))` corresponding to the wavelength `6500 Å` is `c//lambda_(0)`
Therefore, the threshold energy `=hv_(0)hc//lambda_(0)`
Substituting for `h, c` and `lambda_(0)` we gwt threshold energy `=3.056xx10^(-12)` ergs.
The energy of the incident photons is given by `E=hc//lambda_(0)`, since incident wavelength `lambda=360 Å`. Therefore, incident energy `=55.175xx10^(-12)` erge
The kinetic energy of the photoelectrons will be the difference of incident energy and threshold energy, `:. KE=hv-hv_(0)=(55.175xx10^(-12))-(3.056xx10^(-12))` ergs. `=52.119xx10^(-12)` ergs