Question

A laser gun of power 3 mW and mass 50 gm emits photons of wavelength 500 nm. It is a gravity frequency space and emits for one hour. Find the distance covered by the gun due to recoil in this one hour (approx.)

Answer 1

Hint: While the laser gun emits photons, it applies a force on the photons to move which is equal to the rate of change of momentum of the photons emitted. Due to this force applied by the gun, the shooter experiences an equal backward reaction force that causes the gun to recoil.

Complete step by step solution:

Consider a laser gun of power 3 mW and mass 50 gm emits a photon of wavelength 500 nm. When the gun emits a photon, the gun experiences a backward reaction force due to Newton’s third law of motion, also known as the recoil of the gun. This force is equivalent to the force with which the photons are emitted. Therefore, to calculate the distance covered by the gun, we must know the force with which the photons are emitted.
First, we must know the momentum of each photon for which we will use the de Broglie’s wave equation. Therefore, the momentum (p) of each photon is
$\Rightarrow p=\dfrac{h}{\lambda }$ ………. (1)
We know that force is the rate of change of momentum then, force applied by the gun to emit a photon is
$\Rightarrow {{F}_{photon}}=\dfrac{h}{\lambda t}$ ………. (2)
Since the gun emits photons for one hour, so next we have to find out the number of photons emitted during this period. That can be calculated by dividing the total energy of photons emitted by the gun with the energy of one photon. Therefore, the total energy of photons emitted is
$\Rightarrow {{E}_{total}}=3\times {{10}^{-3}}\times t$ ………. (3)
Next, the energy of one photon is
$\Rightarrow {{E}_{photon}}=\dfrac{hc}{\lambda }$ ………. (4)
Therefore, the total number of photons emitted (N) will be
$\begin{align}
  & \Rightarrow N=\dfrac{{{E}_{total}}}{{{E}_{photon}}} \\
 & \Rightarrow N=\dfrac{3\times {{10}^{-3}}\times t}{\dfrac{hc}{\lambda }} \\
 & \Rightarrow N=\dfrac{3\times {{10}^{-3}}\times \lambda t}{hc} \\
\end{align}$
Therefore, the net force applied by the gun to emit photons for one hour is
\[\begin{align}
  & \Rightarrow {{F}_{net}}=N\times {{F}_{photon}} \\
 & \Rightarrow {{F}_{net}}=\dfrac{3\times {{10}^{-3}}\times \lambda t}{hc}\times \dfrac{h}{\lambda t} \\
 & \Rightarrow {{F}_{net}}={{10}^{-11}}\,N \\
\end{align}\]
According to Newton’s third law of motion, this net force is equivalent to the recoil force acting on the gun.
$\therefore \,{{F}_{net}}={{F}_{recoil}}={{10}^{-11}}\,N$
Due to this recoil force, the gun moves backward with an acceleration
\[\begin{align}
  & \Rightarrow a=\dfrac{{{F}_{recoil}}}{m} \\
 & \Rightarrow a=\dfrac{{{10}^{-11}}}{50\times {{10}^{-3}}} \\
 & \Rightarrow a=2\times {{10}^{-10}}\,m/{{s}^{2}} \\
\end{align}\]
Therefore, the distance covered by the gun due to recoil will be
$\begin{align}
  & \Rightarrow S=\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow S=\dfrac{1}{2}\times 2\times {{10}^{-10}}\times {{(3600)}^{2}} \\
 & \Rightarrow S=1.296\times {{10}^{-3}}\,m \\
 & \Rightarrow S=1.3\,mm \\
\end{align}$

$\therefore$ The distance covered by the laser gun due to recoil is 1.3 mm.

Note:
This question can be solved by using the law of conservation of linear momentum. According to this law, the net linear momentum of the gun due to recoil will be equal to the total momentum of the photons emitted in one hour. From here, we get the velocity with which the gun recoils. Then by putting the value of velocity in the equation $S=\dfrac{1}{2}vt$, we can get the distance covered by the gun during recoil.