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Question: A \(2mW\) laser operates at a wavelength of \(500nm\) . The number of photons that will be emitted p...

A 2mW2mW laser operates at a wavelength of 500nm500nm . The number of photons that will be emitted per second is:
[Given Planck's constant h=6.6×1034Jsh = 6.6 \times {10^{ - 34}}Js ,
speed of light c=3.0×108m/sc = 3.0 \times {10^8}m/s ]
(A) 2×10162 \times {10^{16}}
(B) 1.5×10161.5 \times {10^{16}}
(C) 5×10155 \times {10^{15}}
(D) 1×10161 \times {10^{16}}

Explanation

Solution

Hint In lasers there are mainly three ways of interaction of atoms - spontaneous emission, stimulated emission and absorption of radiation. Power is defined as the energy per unit time and we also know the formula for energy of a photon. So, we can use them and substitute the given values to get the number of photons.
Formula Used:
E=nhcλE = \dfrac{{nhc}}{\lambda }

Complete step by step answer
When a photon has a frequency ν\nu then the energy of the photon will be hνh\nu where h is the Planck’s constant. We can write frequency as a ratio of speed and wavelength, so energy of photon expression for ‘n’ photons will become,
E=nhcλE = \dfrac{{nhc}}{\lambda } ------ (1)
Power is the energy per unit time and in this question, it is given as 2mW=2×103W2mW = 2 \times {10^{ - 3}}W and wavelength is 500nm=500×109m500nm = 500 \times {10^{ - 9}}m
and P=EtP = \dfrac{E}{t} where E is the energy of photon and t is the time
Putting equation (1) in the above expression we have,
P=Et=hcλtP = \dfrac{E}{t} = \dfrac{{hc}}{{\lambda t}} and on rearranging we get,
n=Pλthc\Rightarrow n = \dfrac{{P\lambda t}}{{hc}}
Now let’s put the known values, keep in mind that the number of points emitted per second is asked so we can take t=1sect = 1\sec
Then n=2×103×500×109×16.6×1034×3×108n = \dfrac{{2 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 9}} \times 1}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}
n=1.5×1016\Rightarrow n = 1.5 \times {10^{16}}

Hence the correct option is B.

Note
If the energy of the photon is given in electron-volt(eV), then to change it in joules we can use
1eV=1.6×1019J1eV = 1.6 \times {10^{ - 19}}J
In 1900, Planck proposed a theory that the emission of radiation is not continuous. It takes place in the form of small packets of definite energy called ‘quanta’. Later these were termed as ‘photons’. This theory was further explained by Einstein who said that light travels in a packet of energy called as photons and each photon has hνh\nu energy where h is Planck’s constant and ν\nu is intensity.