Question

A small 50 kg vehicle is designed to be moved in free space by a lamp that emits 100 watts of red light of $\lambda = 6630{{\text{A}}^\circ }$. What is its acceleration?
A) $6 \cdot 6 \times {10^{ - 9}}m{s^{ - 2}}$
B) $3 \times {10^{ - 9}}m{s^{ - 2}}$
C) $2 \cdot 2 \times {10^{ - 9}}m{s^{ - 2}}$
D) $4 \cdot 44 \times {10^{ - 9}}m{s^{ - 2}}$

Answer 1

Hint: The light emitted by the photons is used to carry the energy of the light. The linear momentum is always conserved and when the light particles will move with some momentum then the rocket will start moving with the same momentum in the opposite direction as the linear momentum is conserved.

Formula used:
$P = Nhv$ where P is the power, N is the number of the particles and $v$ is the frequency of the wave. Also $p = \dfrac{h}{\lambda }$ where p is the momentum h is the Planck’s constant and $\lambda $ is the wavelength of the wave.

Complete step by step answer:
As it is given that a vehicle of 50 kg has a lamp that emits 100 watts of red light of $\lambda = 6630{{\text{A}}^\circ }$ wavelength.
Let us calculate the number of photons that the red light emits.
$ \Rightarrow P = Nhv$ Where P is the power, N is the number of the particles and $v$ is the frequency of the wave.
$ \Rightarrow N = \dfrac{P}{{hv}}$
It is given that the power of the lamp is 100 watts also $v = \dfrac{c}{\lambda }$where $v$ if the frequency, $c$ is the speed of light and $\lambda $ is the wavelength of the wave.
So the relation becomes,
$ \Rightarrow N = \dfrac{P}{{hv}}$
$ \Rightarrow N = \dfrac{{P \cdot \lambda }}{{h \cdot c}}$
Solving the above relation we get,
$ \Rightarrow N = \dfrac{{\left( {100} \right) \cdot \left( {6630 \times {{10}^{ - 10}}} \right)}}{{\left( {6 \cdot 63 \times {{10}^{ - 34}}} \right) \cdot \left( {3 \times {{10}^8}} \right)}}$
$ \Rightarrow N = 3 \cdot 33 \times {10^{20}}{s^{ - 1}}$
According to de Broglie’s hypothesis,
$p = \dfrac{h}{\lambda }$ Where p is the momentum h is the Planck’s constant and $\lambda $ is the wavelength of the wave.
So the momentum would be,
$ \Rightarrow p = \dfrac{h}{\lambda }$
On substituting the corresponding values, we get
$ \Rightarrow p = \dfrac{{6 \cdot 63 \times {{10}^{ - 34}}}}{{6630 \times {{10}^{ - 10}}}}$
On simplification,
$ \Rightarrow p = \dfrac{{6 \cdot 63 \times {{10}^{ - 34}}}}{{6630 \times {{10}^{ - 10}}}}$
$ \Rightarrow p = {10^{ - 27}}N - s$.
As the force is given by,
$ \Rightarrow F = p \cdot N$ Where p is the momentum of the wave and N is the number of particles per second.
$ \Rightarrow F = \left( {{{10}^{ - 27}}} \right) \cdot \left( {3 \cdot 33 \times {{10}^{20}}} \right)$
$ \Rightarrow F = 3 \cdot 33 \times {10^{ - 7}}N$
As the force is the product of mass and acceleration, therefore
$ \Rightarrow F = m \cdot a$
$ \Rightarrow a = \dfrac{F}{m}$
On substituting the corresponding values, we get
$ \Rightarrow a = \dfrac{{3 \cdot 33 \times {{10}^{ - 7}}}}{{50}}$
On simplification,
$ \Rightarrow a = 6 \cdot 66 \times {10^{ - 9}}\dfrac{m}{{{s^2}}}$

Therefore, the acceleration of the vehicle is equal to $6 \cdot 66 \times {10^{ - 9}}\dfrac{m}{{{s^2}}}$. The correct option is option A.

Note:
The unit of wavelength is very small as compared to the standard unit of distance, it is advisable to convert all physical quantities into a single unit which should be the standard unit of measurement of distance for the calculation of the acceleration in this way there will be no unit related mistakes.