A monochromatic source emitting light of wavelength 300 nm has a power output of 33 W. Calculate the number of protons emitted by this source in 2 minutes. $(h=6.62×10^{-34}Js)$

$6×10^{21}$

The correct answer is Option (1) → $6×10^{21}$

The energy E of a single photon is related to its wavelength λ by the equation -

$E=\frac{hc}{λ}$

where,

λ = wavelength of light = $300 nm=3×10^{-9}m$

$E=\frac{(6.626×10^{-34})(3.0×10^{8})}{300×10^{-9}}$

$=\frac{1.9878×10^{-25}}{300×10^{-9}}=6.626×10^{-19}J$

Now,

Total energy $E_{total}$ emitted in t = 2 sec = $P×t$

$=120×33$

$=3960J$

∴ $E_{total}$ = E × No. of Photons (N)

$N=\frac{3960}{6.626×10^{-19}}$

$≈6×10^{21}$