
How many photons are produced in a laser pulse of 0.338 J at 505 nm?
Answer
562.8k+ views
Hint: In order to find how many photons are produced in a laser pulse, we must first know what a photon is. Photon is said to be an elementary particle which moves with the speed of light in vacuum. It is a massless quantity.
Complete step by step answer:
- Let us first understand about the photon. Photon is said to be an elementary particle which will move with the speed of light in vacuum. Photons are massless quantities. Photons will not have any electric charge and it is a very stable particle. Photons will usually interact with other particles like electrons. This type of effect is called the Compton effect. Photons will be created or destroyed, when the radiation is emitted or absorbed.
- Now let us move onto the given problem.
the wavelength of photon is given as 505 nm
\[\lambda = 505 nm = 505 \times {10^{ - 9}}m\]
We can use the following equation to find the energy of the photon
\[E = \dfrac{{h \times c}}{\lambda }\]…… (1)
- We know that
The Planck’s constant is \[h = 6.626 \times {10^{ - 34}}Js\]
The speed of light is \[c = 3 \times {10^8}m{s^{ - 1}}\]
Wavelength of the light is \[\lambda = 505 nm = 505 \times {10^{ - 9}}m\]
Substituting the values of h, c and \[\lambda \] in the equation (1).
\[E = \dfrac{{6.626 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{505 \times {{10}^{ - 9}}m}}\]
\[E = 3.936 \times {10^{ - 19}}J\]
The energy of the photon is found to be \[E = 3.936 \times {10^{ - 19}}J\].
The total number of photons producing a laser pulse of 0.338 J is given by
\[0.338 \times \dfrac{{1photon}}{{3.936 \times {{10}^{ - 19}}J}} = 8.59 \times {10^{17}}photons\]
The number of photons produced in a laser pulse of 0.338 J is \[8.59 \times {10^{17}}photons\].
Note: Let us see some of the applications of the photons.
- photons are usually used in the optical devices.
- It is used in the Geiger-counter.
- The quantum energies of photons are usually used in cameras.
Complete step by step answer:
- Let us first understand about the photon. Photon is said to be an elementary particle which will move with the speed of light in vacuum. Photons are massless quantities. Photons will not have any electric charge and it is a very stable particle. Photons will usually interact with other particles like electrons. This type of effect is called the Compton effect. Photons will be created or destroyed, when the radiation is emitted or absorbed.
- Now let us move onto the given problem.
the wavelength of photon is given as 505 nm
\[\lambda = 505 nm = 505 \times {10^{ - 9}}m\]
We can use the following equation to find the energy of the photon
\[E = \dfrac{{h \times c}}{\lambda }\]…… (1)
- We know that
The Planck’s constant is \[h = 6.626 \times {10^{ - 34}}Js\]
The speed of light is \[c = 3 \times {10^8}m{s^{ - 1}}\]
Wavelength of the light is \[\lambda = 505 nm = 505 \times {10^{ - 9}}m\]
Substituting the values of h, c and \[\lambda \] in the equation (1).
\[E = \dfrac{{6.626 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{505 \times {{10}^{ - 9}}m}}\]
\[E = 3.936 \times {10^{ - 19}}J\]
The energy of the photon is found to be \[E = 3.936 \times {10^{ - 19}}J\].
The total number of photons producing a laser pulse of 0.338 J is given by
\[0.338 \times \dfrac{{1photon}}{{3.936 \times {{10}^{ - 19}}J}} = 8.59 \times {10^{17}}photons\]
The number of photons produced in a laser pulse of 0.338 J is \[8.59 \times {10^{17}}photons\].
Note: Let us see some of the applications of the photons.
- photons are usually used in the optical devices.
- It is used in the Geiger-counter.
- The quantum energies of photons are usually used in cameras.
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