91Ó°ÊÓ

The energy of a photon having frequency \({\rm{f}}\) is –

\({\bf{E = hf}}\) …… (1)

Here, \(h = 6.626 \times {10^{ - 34}}\;{\rm{Js}}\), is Planck's constant, and \(f\) is the frequency of the incident photon.

The power is defined as the energy per second is obtained as:

\({\bf{P = }}\frac{{\bf{E}}}{{\bf{t}}}\) ...... (2)

Therefore, the number of emitted photons per second is obtained as:

\({\bf{n = }}\frac{{\bf{P}}}{{\bf{E}}}\) …… (3)

Determine the average energy of the photon as:

\(\begin{align}{}E &= 75.0\;{\rm{keV}}\\ &= 75.0 \times {10^3}\;{\rm{eV}}\\ &= 75.0 \times {10^3} \times 1.602 \times {10^{ - 19}}\;{\rm{CV}}\\ &= 1.20 \times {10^{ - 14}}\;{\rm{J}}\end{align}\)

From equation\({\rm{(3)}}\), the number of emitted photons per second is calculated as:

\(\begin{align}{}{\rm{n}} &= \frac{{{\rm{1}}{\rm{.00 W}}}}{{{\rm{1}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - {\rm{14}}}}{\rm{\;J}}}}\\ &= \frac{{{\rm{1}}{\rm{.00\;J\;}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}{{{\rm{1}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - {\rm{14}}}}{\rm{\;J}}}}\\ &= 8.32 \times {10^{13}}\;{\rm{photons}} \cdot {{\rm{s}}^{ - {\rm{1}}}}\end{align}\)

Therefore, the value for number of photons is obtained as \(8.32 \times {10^{13}}\;{\rm{photons}} \cdot {{\rm{s}}^{ - {\rm{1}}}}\).