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. $(h=6.62×10^{-34}Js)$
Answer & explanation
Correct answer: option 1
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)
$N=\frac{3960}{6.626×10^{-19}}$
$≈6×10^{21}$