Question

Radiation power of 3.31 watt is falling on a surface. If 20% of light is observed and remaining is reflected back, calculate no of photon reflected per second. take the frequency of each Photon
${10}^{12}$
Hz
and
$h = 6.62 \times {10}^{ - 34} j - s$

​

Answer 1

Answer:

Answer:

\rm 6.244\times 10^{18}\ photons\ per\ second.6.244×1018 photons per second.

Explanation:

Given:

Radiation power, P = 3.31 W.

Frequency of each photon, \rm \nu =10^{12}\ Hz.ν=1012 Hz.

The radiation power which is falling on the surface is equal to the energy transmitting to the surface per unit time.

The energy of one photon is given by

\rm E=h\nu.E=hν.

h is the Planck's constant, having value \rm 6.626\times 10^{-34}\ Js.6.626×10−34 Js.

If there are n number of photons falling on the surface, then the total energy transmitting to the surface is given by

\rm E_t=nE=nh\nu.Et=nE=nhν.

Since, the radiation power is the energy transmitting to the surface per unit time and only 20% of light is absorbed and remaining is reflected back, therefore, 80% of the light is reflected.

The power of the reflected light is given as

\begin{lgathered}\rm P = 80\%\ of \ \dfrac{E_t}{t}\\=\dfrac{80}{100}\times \dfrac{E_t}{t}\\=0.8\times \dfrac{nh\nu}{t}\\\Rightarrow \dfrac nt = \dfrac{P}{0.8\ h\nu}\\=\dfrac{3.31}{0.8\times 6.626\times 10^{-34}\times 10^{12}}\\=6.244\times 10^{18}\ photons\ per\ second.\end{lgathered}P=80% of tEt=10080×tEt=0.8×tnhν⇒tn=0.8 hνP=0.8×6.626×10−34×10123.31=6.244×1018 photons per second.

It is the number of photons reflected per second from the surface.