Question

A bank of overhead arc lamps can produce a light intensity of $2500\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}}$ in the $25\,{\text{ft}}$ space simulator facility at NASA. Find the average momentum density of a total absorbing surface.
A.$8.33 \times {10^{ - 6}}{\text{ kg}} \cdot {{\text{m}}^{ - 2}} \cdot {{\text{s}}^{{\text{ - 1}}}}$
B.$8.33 \times {10^{ - 14}}{\text{ kg}} \cdot {{\text{m}}^{ - 2}} \cdot {{\text{s}}^{{\text{ - 1}}}}$
C.$2.78{\text{ kg}} \cdot {{\text{m}}^{ - 2}} \cdot {{\text{s}}^{{\text{ - 1}}}}$
D.$2.78 \times {10^{ - 14}}{\text{ kg}} \cdot {{\text{m}}^{ - 2}} \cdot {{\text{s}}^{{\text{ - 1}}}}$

Answer 1

Use the equation for the average momentum density of a totally absorbing surface. This equation gives the relation between the average momentum density of a totally absorbing surface, intensity of the light and the speed of the light.
Formula used:
The average momentum density of a totally absorbing surface is given by
${p_{avg}} = \dfrac{I}{{{c^2}}}$ …… (1)
Here, ${p_{avg}}$ is the average momentum density of the totally absorbing surface, $I$ is the intensity of the light and $c$ is the speed of the light.

Complete step by step answer:
A bank of overhead arc lamps can produce a light intensity of $2500\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}}$ in the $25\,{\text{ft}}$ space simulator facility at NASA.
Calculate the average momentum density of the totally absorbing surface.
Substitute $2500\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}}$ for $I$ and $3 \times {10^8}\,{\text{m/s}}$ for $c$ in equation (1).
${p_{avg}} = \dfrac{{2500\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}}}}{{{{\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}^2}}}$
$\Rightarrow {p_{avg}} = \dfrac{{2500\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}}}}{{9 \times {{10}^{16}}\,{{\text{m}}^2}{\text{/}}{{\text{s}}^2}}}$
$\Rightarrow {p_{avg}} = 277.7 \times {10^{ - 16}}\,{{\text{m}}^2}{\text{/}}{{\text{s}}^2}$
$\Rightarrow {p_{avg}} = 2.78 \times {10^{ - 14}}\,{\text{kg}} \cdot {{\text{m}}^{ - 2}} \cdot {{\text{s}}^{ - 1}}$
Therefore, the average momentum density of a totally absorbing surface is $2.78 \times {10^{ - 14}}\,{\text{kg}} \cdot {{\text{m}}^{ - 2}} \cdot {{\text{s}}^{ - 1}}$.

So, the correct answer is “Option D”.

Additional Information:
The average momentum density is the momentum per unit volume.
The average momentum density of a totally reflecting surface is given by
${p_{avg}} = \dfrac{{2I}}{{{c^2}}}$
Here, ${p_{avg}}$ is the average momentum density of the totally reflecting surface, $I$ is the intensity of the light and $c$ is the speed of the light.
The momentum density is also denoted by $\dfrac{{dp}}{{dV}}$. Here, $dp$ is the infinitesimal change in the momentum.

Note:
The unit of power in simplified form is ${\text{kg}} \cdot {{\text{m}}^2} \cdot {{\text{s}}^{{\text{ - 3}}}}$. Hence, the unit of the average momentum density becomes ${\text{kg}} \cdot {{\text{m}}^{ - 2}} \cdot {{\text{s}}^{ - 1}}$.
The phenomenon of radiation pressure is caused due to the momentum which is the property of the file and not the moving particles.