Question

A $250\,W$electric bulb of $80\% $ efficiency emits a light of $6626\,\mathop {\text{A}}\limits^ \circ $ wavelength. The number of photons emitted per second by the lamp is: $\left[ {h = 6.636 \times {{10}^{ - 34}}\,Js} \right]$
(A) $1.42 \times {10^{17}}$
(B) $2.18 \times {10^{16}}$
(C) $6.66 \times {10^{20}}$
(D) $2.83 \times {10^{16}}$
(E) $4.25 \times {10^{16}}$

Answer 1

Hint:First we have to find the actual energy of the bulb and then by using the energy of the photon formula, the energy of one photon can be determined. Then by using the number of photons emitted formula, the number of photons emitted can be determined.

Formulae Used:
The energy of photons,
${E_p} = \dfrac{{hc}}{\lambda }$
Where, ${E_p}$ is the energy of photons, $h$ is the Planck’s constant, $c$ is the velocity of light in free space and $\lambda $ is the wavelength of the light.
The number of photons emitted per second,
\[n = \dfrac{{{E_s}}}{{{E_p}}}\]
Where, $n$ is the number of photons, ${E_s}$ is the energy from the source and ${E_p}$ is the energy of the photons.

 Complete step-by-step solution:
The energy of the bulb, $250\,W$
The efficiency of the bulb, $80\% $
The wavelength of the light, $\lambda = 6626\,\mathop {\text{A}}\limits^ \circ $
The value of Planck’s constant, $h = 6.626 \times {10^{ - 34}}\,Js$
Now, the actual energy of the light is which is the source, so
${E_s} = 250 \times 80\% $
Then,
${E_s} = 250 \times \dfrac{{80}}{{100}}$
By simplifying, then
${E_s} = 200\,W$
The energy of photons,
${E_p} = \dfrac{{hc}}{\lambda }\,...............\left( 1 \right)$
By substituting the Planck’s constant, speed of light in free space and he wavelength in the above equation (1),
${E_p} = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6626 \times {{10}^{ - 10}}}}$
On simplifying, then the above equation is written as,
${E_p} = 3 \times {10^{ - 19}}\,W$
The number of photons emitted per second,
\[n = \dfrac{{{E_s}}}{{{E_p}}}\,............\left( 2 \right)\]
By substituting the energy from source and the energy of photons, then
The number of photons emitted per second,
\[n = \dfrac{{200}}{{3 \times {{10}^{ - 19}}}}\]
On dividing the above equation, then the above equation is written as,
$n = 66.66 \times {10^{19}}$
The above equation is also written as,
$n = 6.66 \times {10^{20}}$
Thus, the above equation shows the number of photons emitted per second.
Hence, option (C) is the correct answer.

Note:- The efficiency of the light is only $80\% $ of its original energy, so the energy from the source is taken as $80\% $ of the total energy of the light. The number of photons emitted per second has no unit because the energy of the source is divided by the energy of the photon. So, the units cancel each other out.