Question

A $200W$ power source emits monochromatic light of wavelength $4950$ Angstrom. How many photons per minute are emitted by the source? [Given: $h = 6.6 \times {10^{ - 34}}Js$]

Answer 1

Hint: In order to solve this question, first we will find out the energy of one photon. Then with the help of energy of one photon, we will find out the number of photons emitted in one second. At last, we will multiply the number of photons emitted in one second by $60$, to get the value of the number of photons emitted in one minute.

Complete step by step answer:
Given, the power of the source that emits monochromatic light$ = $$200W$
Wavelength of the light emitted$ = $$4950$ Angstrom
Now, we need to convert the above value into SI unit. So,
Wavelength of the light emitted$ = 4950 \times {10^{ - 10}} = 4.95 \times {10^{ - 7}}$
Planck’s constant is $h = 6.6 \times {10^{ - 34}}Js$
Speed of light in vacuum $c = 3 \times {10^8}\dfrac{m}{s}$

We know that the energy of one photon is given by the expression,
$E = \dfrac{{hc}}{\lambda }$
$\Rightarrow E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4.95 \times {{10}^{ - 7}}}}$
$\Rightarrow E = \dfrac{{6.6 \times 3 \times {{10}^{ - 34 + 7 + 8}}}}{{4.95}}$
On further solving the above equation, we get,
$E = \dfrac{{6.6 \times 3 \times {{10}^{ - 19}}}}{{4.95}}$
$\Rightarrow E = 4 \times {10^{ - 19}}J.......(1)$

We know that power can be defined as the product of the number of photons emitted in one second and the energy of one photon.
Power= number of photons emitted in one second $ \times $ energy of one photon
On putting the required values in the above expression, we get,
$200 = n \times 4 \times {10^{ - 19}}$
This can be rearranged and written as,
$n = \dfrac{{200}}{{4 \times {{10}^{ - 19}}}}$
$\Rightarrow n = 50 \times {10^{19}}$
$\Rightarrow n = 5 \times {10^{20}}.....(2)$

The above value obtained is the number of electrons emitted in one second but we need to find out the number of electrons emitted in one minute, i.e., in $60s$. So, we will multiply the above expression by $60$.
Number of electrons emitted in one minute $ = 5 \times {10^{20}} \times 60 = 300 \times {10^{20}}$
The above expression can be simplified and written as $3 \times {10^{22}}$.

Therefore, the number of photons per minute is emitted by the source $ = 3 \times {10^{22}}$.

Note: When a ray of light having a certain minimum frequency hits a metal surface, then the electrons will be ejected. This minimum frequency required to eject electrons is called the threshold frequency ${\nu _o}$. The value of $h{\nu _o}$ is known as the work function. So, in order to eject electrons the minimum frequency of light should be ${\nu _o}$ and the minimum energy should be $h{\nu _o}$.