A 200 W sodium street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

Options

A. `55.175xx10^(-12)`
B. `3.056xx10^(-12)`
C. `52.119xx10^(-12)`
D. `48.66xx10^(-10)`

Answer

3

Explanation

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