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Question: A Laser light of wavelength \(600nm\) is used to weld retinal detachment. If a Laser pulse of width ...

A Laser light of wavelength 600nm600nm is used to weld retinal detachment. If a Laser pulse of width 60ms60ms and 0.5kW0.5kW is used, the approximate number of photons in the pulse are:
Take planck's constant h=6.62×104Jsh = 6.62 \times {10^{ - 4}}Js
A. 1020{10^{20}}
B. 1018{10^{18}}
C. 1022{10^{22}}
D. 1019{10^{19}}

Explanation

Solution

This question is about the particle part of light which is believed to be a wave, thermionic emission which involves emission of light particles as a result of heat or radiant heat that strikes the surface of a plate.

Complete step by step answer:
We are given a Laser light of wavelength 600nm600nm is used to weld retinal detachment and the width of the Laser pulse is 60ms60ms and power is 0.5kW0.5kW.
We have to find the approximate no. of photos present in the pulse.
Photons are the particles which get emitted when light strikes a metal surface.
We need to first generate the formula that connects energy, number of photons and time, thereafter connect the parameters together.
Energy from the laser = hcλ\dfrac{{hc}}{\lambda }
The ratio of change in number of photon with time or pulse = ΔnΔt\dfrac{{\Delta n}}{{\Delta t}}
The wavelength is in nanometer which is converted to practical unit meter which is 109m{10^{ - 9}}m, the pulse is also converted to practical unit seconds 103s{10^{ - 3}}s.
So the Power generated = ΔnΔt×hcλ\dfrac{{\Delta n}}{{\Delta t}} \times \dfrac{{hc}}{\lambda }
0.5×103=Δn60×103×6.62×1034×3×108660×1090.5 \times {10^3} = \dfrac{{\Delta n}}{{60 \times {{10}^{ - 3}}}} \times \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{660 \times {{10}^{ - 9}}}}
We would then cross multiply and make Δn\Delta n the subject of the relation or formula.
Δn=0.5×103×60×103×660×1096.62×1034×3×108 Δn=996.97×1017  \Delta n = \dfrac{{0.5 \times {{10}^3} \times 60 \times {{10}^{ - 3}} \times 660 \times {{10}^{ - 9}}}}{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \\\ \to \Delta n = 996.97 \times {10^{17}} \\\
We can also move the decimal point three steps to the left hand side, where the answer will now look like
Δn=0.99×1020 Δn1020  \Delta n = 0.99 \times {10^{20}} \\\ \Delta n \simeq {10^{20}} \\\
So the right option for this question is Option A, 1020{10^{20}}
The number of photons in the pulse is approximately equal to ten raised to the power of 20.

Note: This again is applied in other light packed properties where the radiation of light on some surfaces is important, phenomena like photoelectric emission which involves the liberation of photons or quanta of light.