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

Hint: 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.