91Ó°ÊÓ

The expression for frequency of photon is given by,

$f=\frac{c}{\mathrm{λ}}$ ….. (1)

Here, c is the speed of light and $\mathrm{λ}$is the wavelength.

The energyof a photon is given by,

$E=hf$ ….. (2)

Here, h is Planck’s constant and is the frequency.

Combine the above two equations.

$E=\frac{hc}{\mathrm{λ}}$ ….. (3)

Consider the given data as below.

The wavelength, $\mathrm{λ}=630nm$

Plank’s constant, $h=6.626×{10}^{-34}\mathrm{J}·\mathrm{s}$

The speed of light, $c=3×{10}^{8}m/s$

Substitute the above values in eq 3.

$$\begin{gathered} E=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3×{10}^{8}m/s\right)}{\left(630\mathrm{nm}\right)\left({\displaystyle \frac{1m}{{10}^{9}nm}}\right)} \\ =3.15×{10}^{-19}\mathrm{J} \end{gathered}$$

Rate at which electrical energy is supplied is 60W. But the bulb is only 93% efficient in converting this energy into light.

The power delivered from the light is,

$\begin{array}{c}P=\frac{93}{100}×60\mathrm{W}\\ =55.8\mathrm{W}\end{array}$

$$\begin{gathered} \mathrm{Numberofphotons}=\frac{\mathrm{power}\mathrm{delivered}\mathrm{out}}{\mathrm{energy}\mathrm{of}\mathrm{each}\mathrm{photon}} \\ =\frac{55.8\mathrm{W}}{3.15×{10}^{-19}\mathrm{J}} \\ =1.77×{10}^{20}\mathrm{photons}/\mathrm{s}. \\ =\left(1.77×{10}^{20}\mathrm{photons}/\mathrm{s}\right)\left(\frac{60\mathrm{s}}{\min }\right)\left(\frac{60\min }{\mathrm{h}}\right) \end{gathered}$$

Simplify further.

$$\begin{gathered} \mathrm{Numberofphotons}=6372×{10}^{20}photons/h \\ =6372×{10}^{20}photons/h×730h \\ =4.6×{10}^{26}\mathrm{photons} \end{gathered}$$

Hence, the number of photons arelocalid="1663135942932">4.6×1026photons.