A point source of light of power ' \(P^{\prime}\) and wavelength ' \(\lambda\) ' is emitting light in all directions. The number of photons present in a spherical region of radius ' \(r\) ' to radius \(\mathrm{r}+\mathrm{x}\) with centre at the source is (1) \(\frac{\mathrm{P} \lambda}{4 \pi \mathrm{r}^{2} \mathrm{hc}}\) (2) \(\frac{\mathrm{P} \lambda \mathrm{x}}{\mathrm{hc}^{2}}\) (3) \(\frac{\mathrm{P} \lambda \mathrm{x}}{4 \pi \mathrm{r}^{2} \mathrm{hc}}\) (4) \(\frac{3 P \lambda x}{4 \pi r^{2} h c}\) (5) None of these

Understanding the energy of a single photon is crucial to solving many problems in physics, especially in the context of photon energy calculations. A photon, which is a quantum of light, carries energy that can be calculated using the formula: \( E = \frac{hc}{\text{λ}} \). Here, \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \) joule-seconds, and \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \) meters per second.
For example, if the wavelength \( \text{λ} \) of the photon is known, we can plug it into the formula and calculate the energy. This energy calculation helps in determining the number of photons emitted given a certain power. If the total power of the light source, denoted as \( P \), is known, the number of photons emitted per second (denoted as \( N \)) can be found using the relationship \( P = N \times E \). Rearranging, we get \( N = \frac{P \times \text{λ}}{hc} \). This formula is useful when dealing with point sources of light and understanding their energy output.

To sum up, first calculate the energy of individual photons, and then use the power equation to find out how many photons are generated per second.

The concept of volume is pivotal in many physics problems, especially when dealing with spherical shells. A spherical shell, in essence, is a hollow sphere defined by two radii. In this problem, it's important to define the region between radius \( r \) and \( r + x \). The volume \( V \) of the spherical shell is given by the formula: \( V = 4 \text{π} r^2 x \).
This formula assumes that \( x \) is small compared to \( r \), such that the volume of the shell is essentially the product of the outer surface area of the sphere at radius \( r \) and the shell thickness \( x \). This assumption simplifies calculations and is commonly used in exercises involving thin shells.

For better grasp, imagine blowing up a balloon (spherical shell). If you measure the layer between two very close radius points, the additional volume is thin enough to approximate to the surface area multiplied by the thickness. This simplification helps us in distributing the photons in the next step of our calculation.

Photon distribution is about how photons are spread within a given volume. In problems like this one, understanding how photons are distributed in a spherical shell is critical. Given that the point source emits photons uniformly in all directions, we recognize that after time \( t = \frac{r}{c} \), photons will uniformly spread within a spherical volume.

The number of photons, \( N \), within the spherical shell depends on the power of the light source and the thickness of the shell. Given the total number of photons emitted per second, and using the previously calculated spherical shell volume, we can find the photons in a specific region. The density of photons simplifies to \( \frac{N}{4 \text{π} r^2 c t} \), and with time it becomes \( \frac{N}{4 \text{π} r^2} \frac{1}{c} \times \frac{r}{c} \).
This relationship helps us derive the formula \( \frac{P \times \text{λ} x}{4 \text{π} r^2 h c} \) for the number of photons in the spherical shell of thickness x.

Understanding photon distribution not only solves theoretical problems but also lays the foundation for more advanced topics in physics like radiation transmission and quantum electrodynamics.