Question

How many photons are emitted per second by a 5 mW laser operating at 632.8 nm?
A. $1.6 \times {10^{16}}$
B. $1.6 \times {10^{13}}$
C. $1.6 \times {10^{10}}$
D. $1.6 \times {10^3}$

Answer 1

Hint: To find the required number of photons emitted, we are using the formula
$E = \dfrac{{nhc}}{\lambda }$ Where, E is the energy of laser beam of wavelength $\lambda$, h is the Planck’s constant, c is the speed of light and n is the number of photon emitted.

Complete step-by-step answer:
To calculate the required number of photons emitted.
Calculating the energy of the beam from the given data in the question.
Given:-
$P = 5{\text{ mW}}$
$\Rightarrow P = 5{\text{ mW}}$
$t = 1\sec$

Using the formula: $E = P \times t$, …………………..(i) for energy calculation .We get:
Substituting the given values of P and t in the equation (i)
We have,
$E = 5 \times {10^{ - 3}}$J………………….(ii)
Now we take the formula $E = \dfrac{{nhc}}{\lambda }$, we can also modify it as:
$n = \dfrac{{E\lambda }}{{hc}}$……………………………………………..(iv)
Substituting the given values of E, h, $\lambda$ and c in eqn (iv)[ we take $h = 6.626 \times {10^{ - 34}}{\text{J - s}}$.
We get the required number of photons:-

$$n = \dfrac{{5 \times {{10}^{ - 3}} \times 632.8 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} \\\ \Rightarrow \dfrac{{3196}}{{19.876}} \times {10^{14}} \\\ \Rightarrow 1.591 \times {10^{16}} \\\ \approx 1.6 \times {10^{16}} \\\$$

Hence Option (A) is the correct answer.

Note: To solve such types of questions one should have the ability to implement the short and quick fact based formula. We use the for $E = \dfrac{{nhc}}{\lambda }$ not $\dfrac{{hc}}{\lambda }$ because it is the energy of single electron while on multiplying the n with $\dfrac{{hc}}{\lambda }$ we get the energy of the laser beam. One should also carefully convert the given values into the SI units.