Question

A 600 W mercury lamp emits a monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second?
$h=6.626\times {10}^{-34}\text{Js};\text{velocity of light =}3\times {10}^8{\text{ms}}^{-1}$

• A. ${10}^{21}$
• B. ${10}^{22}$
• C. ${10}^{20}$
• D. ${10}^{19}$

Answer 1

The correct option is A ${10}^{21}$

According to Planck's quantum theory, $E=\frac{nhc}{\lambda }$
Where, E = energy of the radiation $=600W=600J/s$
n = number of photons emitted per second
h = Planck's constant $=6.626\times {10}^{-34}Js$
c = speed of light $=3\times {10}^{8}m/s$
$\lambda$ = wavelength of the radiation $=331.3nm=331.3\times {10}^{-9}m$

On subtituting the values in the equation and solving for n,
$n=\frac{E\lambda }{h.c}=\frac{600\times 331.3\times {10}^{-9}}{6.626\times {10}^{-34}\times 3\times {10}^8}$

On solving $n=1\times {10}^{21}photons/second$