Question

A \[100{\rm{ W}}\] bulb emits monochromatic light of wavelength \[400{\rm{ nm}}\]. Calculate number of photons emitted per second by bulb.

Answer 1

Hint:The photon's energy is equal to the product of its frequency and Planck's constant. The ratio of the speed of light and wavelength of the photon gives us the value of the photon's frequency.

Complete step by step answer:
Given:
The power of the bulb is \[P = 100{\rm{ W}}\].
The wavelength of the monochromatic light is \[\lambda = 400{\rm{ nm}} = 400{\rm{ nm}} \times \left( {\dfrac{{{{10}^{ - 9}}{\rm{ m}}}}{{{\rm{nm}}}}} \right) = 400 \times {10^{ - 9}}{\rm{ m}}\].
The expression for the energy of one photon is:
\[E = h\nu \]……(1)
Here h is the Planck's constant and \[\nu \] is the frequency of the photon.
Let us write the expression for the frequency of one photon.
\[\nu = \dfrac{c}{\lambda }\]
Here c is the speed of light.
Substitute \[\dfrac{c}{\lambda }\] for \[\nu \] in equation (1).
\[
E = h\left( {\dfrac{c}{\lambda }} \right)\\
\Rightarrow E = \dfrac{{hc}}{\lambda }
\]……(2)
We know that the value of the speed of light and Planck's constant are given as:
\[
c = 3 \times {10^8}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\\
\Rightarrow h = 6.626 \times {10^{ - 34}}{\rm{Js}}
\]
Substitute \[6.626 \times {10^{ - 34}}{\rm{Js}}\] for h, \[3 \times {10^8}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\] for c and \[400 \times {10^{ - 9}}{\rm{ m}}\] for \[\lambda \] in equation (2).
\[
E = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}{\rm{Js}}} \right)\left( {3 \times {{10}^8}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}} \right)}}{{400 \times {{10}^{ - 9}}{\rm{ m}}}}\\
\Rightarrow E = 4.969 \times {10^{ - 19}}{\rm{ J}}
\]
We also know that the product of the number of photons emitted per second and one photon's energy gives us the value of power.
\[P = n \times E\]
Here n is the number of photons emitted per second.
Substitute \[100{\rm{ W}}\] for P and \[4.969 \times {10^{ - 19}}{\rm{ J}}\] for E in the above expression.
\[
100{\rm{ W}} = n \times 4.969 \times {10^{ - 19}}{\rm{ J}}\\
\Rightarrow 100{\rm{ W}} \times \left( {\dfrac{{{{\rm{J}} {\left/
 {\vphantom {{\rm{J}} {\rm{s}}}} \right.
} {\rm{s}}}}}{{\rm{W}}}} \right) = n \times 4.969 \times {10^{ - 19}}{\rm{ J}}\\
\therefore{\rm{n}} = 2.012 \times {10^{20}}{\rm{ }}{{\rm{1}} {\left/
 {\vphantom {{\rm{1}} {\rm{s}}}} \right.
} {\rm{s}}}
\]

Therefore, the number of photons emitted per second by the bulb is \[2.012 \times {10^{20}}\].

Note: We can remember the value of the speed of light and Plank's constant for a similar type of question. Do not forget to convert the unit of wavelength into a metre. Also, it would be an added advantage to remember the conversions between various units of energy.