Question

Find the radiation pressure of solar radiation on the surface of earth. Solar constant is $1.4\,kW{m^{ - 2}}$
A) $4.7 \times {10^{ - 5}}\,Pa$
B) $4.7 \times {10^{ - 6}}\,Pa$
C) $2.37 \times {10^{ - 6}}\,Pa$
D) $9.4 \times {10^{ - 6}}\,Pa$

Answer 1

Hint:You can solve this question easily using the formula which gives us, or develops a relation between radiation pressure absorbed, energy flux and the speed of the light in vacuum. Other than that, you may also try and use the formula of basic unit check.

Complete step by step answer:
As told in the hint section of the solution to the question, we will solve the given question using the formula which gives us. Or develops a relation between radiation pressure absorbed, energy flux and speed of light in vacuum. This formula is used very frequently, and is given as:
${P_{ab}} = \dfrac{{{E_f}}}{c}$
Where, ${P_{ab}}$ is the radiation pressure absorbed,
${E_f}$ is the energy flux, or as per given in the above-mentioned question, the solar constant is the energy flux in the given case and,
$c$ is the speed of light in vacuum, which is numerically equal to $3 \times {10^8}\,m/s$
Now, in the question, it is given to us that the solar constant is equal to $1.4\,kW{m^{ - 2}}$
We can say that the energy flux which is needed by us to solve the question is given to us as the solar constant, numerically equal to the value $1.4\,kW{m^{ - 2}}$
We already know the value of speed of light in vacuum, which is $3 \times {10^8}\,m/s$
Substituting the given values in the formula of radiation pressure absorbed as we told above,
${P_{ab}} = \dfrac{{1.4 \times {{10}^3}}}{{3 \times {{10}^8}}}$
Upon further solving, we get:
${P_{ab}} = 4.66 \times {10^{ - 6}}\,Pa$
We can take approximation and car write:
${P_{ab}} \approx 4.7 \times {10^{ - 6}}\,Pa$

Hence, we can see that the correct answer is option (B).

Note:An important point that should be noted is the manipulation of the fact that solar constant is given to us in the question, it was only put in the question so that we can use it as the value of energy flux. Another thing, many students only put the numerical value of energy flux and forget about the fact that the unit given in the question is $kW{m^{ - 2}}$ and not $W{m^{ - 2}}$ and thus, lose marks due to this.