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Question: A \(200W\) power source emits monochromatic light of wavelength \(4950\) Angstrom. How many photons ...

A 200W200W power source emits monochromatic light of wavelength 49504950 Angstrom. How many photons per minute are emitted by the source? [Given: h=6.6×10−34Jsh = 6.6 \times {10^{ - 34}}Js]

Explanation

Solution

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 6060, 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== 200W200W
Wavelength of the light emitted== 49504950 Angstrom
Now, we need to convert the above value into SI unit. So,
Wavelength of the light emitted=4950×10−10=4.95×10−7 = 4950 \times {10^{ - 10}} = 4.95 \times {10^{ - 7}}
Planck’s constant is h=6.6×10−34Jsh = 6.6 \times {10^{ - 34}}Js
Speed of light in vacuum c=3×108msc = 3 \times {10^8}\dfrac{m}{s}

We know that the energy of one photon is given by the expression,
E=hcλE = \dfrac{{hc}}{\lambda }
⇒E=6.6×10−34×3×1084.95×10−7\Rightarrow E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4.95 \times {{10}^{ - 7}}}}
⇒E=6.6×3×10−34+7+84.95\Rightarrow E = \dfrac{{6.6 \times 3 \times {{10}^{ - 34 + 7 + 8}}}}{{4.95}}
On further solving the above equation, we get,
E=6.6×3×10−194.95E = \dfrac{{6.6 \times 3 \times {{10}^{ - 19}}}}{{4.95}}
⇒E=4×10−19J.......(1)\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×4×10−19200 = n \times 4 \times {10^{ - 19}}
This can be rearranged and written as,
n=2004×10−19n = \dfrac{{200}}{{4 \times {{10}^{ - 19}}}}
⇒n=50×1019\Rightarrow n = 50 \times {10^{19}}
⇒n=5×1020.....(2)\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 60s60s. So, we will multiply the above expression by 6060.
Number of electrons emitted in one minute =5×1020×60=300×1020 = 5 \times {10^{20}} \times 60 = 300 \times {10^{20}}
The above expression can be simplified and written as 3×10223 \times {10^{22}}.

Therefore, the number of photons per minute is emitted by the source =3×1022 = 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 νo{\nu _o}. The value of hνoh{\nu _o} is known as the work function. So, in order to eject electrons the minimum frequency of light should be νo{\nu _o} and the minimum energy should be hνoh{\nu _o}.