Question

radiation power of 3.31 watt is falling on a surface .if 20% of light is absorbed and remaining is reflected back,calculate number of photons reflected per second.take the frequency of each photon 10*12 hz

Answer 1

Answer:

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

Explanation:

Given:

• Radiation power, P = 3.31 W.

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

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

$\rm E=h\nu.$

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

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

$\rm E_t=nE=nh\nu.$

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

$\rm P = 80\%\ of \ \dfrac{E_t}{t}\\=\dfrac{80}{100}\times \dfrac{E_t}{t}\\=0.8\times \dfrac{nh\nu}{t}\\\Rightarrow \dfrac nt = \dfrac{P}{0.8\ h\nu}\\=\dfrac{3.31}{0.8\times 6.626\times 10^{-34}\times 10^{12}}\\=6.244\times 10^{18}\ photons\ per\ second.$

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

Answer 2

Answer:

6.244 * 10^21

Explanation:

3.31/0.8*6.244*10^-34*10^12