Question

A 100 Watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

• A. 4 $\times$ ${10}^{20}$ $s^{-1}$
• B. 2 $\times$ ${10}^{20}$ $s^{-1}$
• C. 8 $\times$ ${10}^{20}$ $s^{-1}$
• D. 1$\times$ ${10}^{20}$ $s^{-1}$

Answer 1

The correct option is B

2 $\times$ ${10}^{20}$ $s^{-1}$

We know that the wave velocity (in this case - the speed of light in vacuum)

c = $\nu \lambda$ $\Rightarrow$ $\nu =\frac{c}{\lambda }$ and, by Planck's Theory.

E= $h\nu$ $\Rightarrow$ E = $\frac{hc}{\lambda }$

Power of the bulb = 100 watt.

= 100 J $s^{-1}$

Energy of one photon is calculated as:

E = $h\nu$ = $\frac{hc}{\lambda }$

= $\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{400\times {10}^{-9}m}$

= 4.969 $\times$ ${10}^{-19}$J

Number of photons emitted (per second):

= $\frac{100J{s}^{-1}}{4.969\times {10}^{-19}J}$= 2.012 $\times$ ${10}^{20}$ $s^{-1}$