Answer
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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:
\[E = \dfrac{{nhc}}{\lambda }\]
Complete step by step answer
When a photon has a frequency $\nu $ then the energy of the photon will be $h\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,
$E = \dfrac{{nhc}}{\lambda }$ ------ (1)
Power is the energy per unit time and in this question, it is given as $2mW = 2 \times {10^{ - 3}}W$ and wavelength is $500nm = 500 \times {10^{ - 9}}m$
and $P = \dfrac{E}{t}$ where E is the energy of photon and t is the time
Putting equation (1) in the above expression we have,
$P = \dfrac{E}{t} = \dfrac{{hc}}{{\lambda t}}$ and on rearranging we get,
$ \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 = 1\sec $
Then $n = \dfrac{{2 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 9}} \times 1}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$
$ \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
$1eV = 1.6 \times {10^{ - 19}}J$
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 $h\nu $ energy where h is Planck’s constant and $\nu $ is intensity.
Formula Used:
\[E = \dfrac{{nhc}}{\lambda }\]
Complete step by step answer
When a photon has a frequency $\nu $ then the energy of the photon will be $h\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,
$E = \dfrac{{nhc}}{\lambda }$ ------ (1)
Power is the energy per unit time and in this question, it is given as $2mW = 2 \times {10^{ - 3}}W$ and wavelength is $500nm = 500 \times {10^{ - 9}}m$
and $P = \dfrac{E}{t}$ where E is the energy of photon and t is the time
Putting equation (1) in the above expression we have,
$P = \dfrac{E}{t} = \dfrac{{hc}}{{\lambda t}}$ and on rearranging we get,
$ \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 = 1\sec $
Then $n = \dfrac{{2 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 9}} \times 1}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$
$ \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
$1eV = 1.6 \times {10^{ - 19}}J$
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 $h\nu $ energy where h is Planck’s constant and $\nu $ is intensity.
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