
How many photons are there in each cubic meter near the earth's surface at any instant?
Answer
588.3k+ views
Hint:The above problem can be resolved using the mathematical formula for the number of photons present in a unit volume near the earth's surface. Various variables are used in this formula consisting of the numerical constant along with some other constant value like the Planck's constant and the speed of light, which completely provides the complete solution.
Complete step by step answer:
The standard formula for the number of photons per unit volume is given as,
\[{n_1} = \dfrac{{P\lambda }}{{4\pi {r^2}hc}}\]
Here, \[\dfrac{P}{{4\pi {r^2}}}\] is a numerical constant and its value is \[1.4 \times {10^3}\], \[\lambda \] is the wavelength of the photon and its standard value can be taken as \[5 \times {10^{ - 7}}\;{\rm{m}}\]. and h is the Planck’s constant and its value is \[6.63 \times {10^{ - 34}}\;{\rm{J/s}}\].
Solve by substituting the values in the above equation as,
\[\begin{array}{l}
{n_1} = \dfrac{{P\lambda }}{{4\pi {r^2}hc}}\\
{n_1} = 1.4 \times {10^3} \times \left( {\dfrac{{5 \times {{10}^{ - 7}}\;{\rm{m}}}}{{6.63 \times {{10}^{ - 34}}\;{\rm{J/s}} \times 3 \times {{10}^8}\;{\rm{m/s}}}}} \right)\\
{n_1} = 3.51 \times {10^{21}}
\end{array}\]
Therefore, the number of photons in each cubic metre near the surface of earth is \[3.51 \times {10^{21}}\].
Note:To resolve the given problem, one must try to understand the concept of photons' ejection from any surface. To achieve the desired result, one must be sure about the mathematical relation of the number of photons removed per unit of the volume. Moreover, the photons are the small packets of energy that travels in the space almost with the speed of light, and they are the basic sub-particles that require some energy to get ejected from any surface, thereby causing the photoelectric emission. In addition, the photon finds its application in the photoelectric effect and in many other fields of quantum mechanics.
Complete step by step answer:
The standard formula for the number of photons per unit volume is given as,
\[{n_1} = \dfrac{{P\lambda }}{{4\pi {r^2}hc}}\]
Here, \[\dfrac{P}{{4\pi {r^2}}}\] is a numerical constant and its value is \[1.4 \times {10^3}\], \[\lambda \] is the wavelength of the photon and its standard value can be taken as \[5 \times {10^{ - 7}}\;{\rm{m}}\]. and h is the Planck’s constant and its value is \[6.63 \times {10^{ - 34}}\;{\rm{J/s}}\].
Solve by substituting the values in the above equation as,
\[\begin{array}{l}
{n_1} = \dfrac{{P\lambda }}{{4\pi {r^2}hc}}\\
{n_1} = 1.4 \times {10^3} \times \left( {\dfrac{{5 \times {{10}^{ - 7}}\;{\rm{m}}}}{{6.63 \times {{10}^{ - 34}}\;{\rm{J/s}} \times 3 \times {{10}^8}\;{\rm{m/s}}}}} \right)\\
{n_1} = 3.51 \times {10^{21}}
\end{array}\]
Therefore, the number of photons in each cubic metre near the surface of earth is \[3.51 \times {10^{21}}\].
Note:To resolve the given problem, one must try to understand the concept of photons' ejection from any surface. To achieve the desired result, one must be sure about the mathematical relation of the number of photons removed per unit of the volume. Moreover, the photons are the small packets of energy that travels in the space almost with the speed of light, and they are the basic sub-particles that require some energy to get ejected from any surface, thereby causing the photoelectric emission. In addition, the photon finds its application in the photoelectric effect and in many other fields of quantum mechanics.
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