The number of photons emitted per second by a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The power $P$ of the bulb is the total energy emitted per second. The energy of a single photon is given by $E = \frac{hc}{\lambda}$.If $n$ is the

Vedclass · Accepted Answer

$2 	imes 10^{20}$

Vedclass · Answer

(D) The power $P$ of the bulb is the total energy emitted per second. The energy of a single photon is given by $E = \frac{hc}{\lambda}$.If $n$ is the number of photons emitted per second,then the total power is $P = n 	imes E = n \frac{hc}{\lambda}$.Rearranging for $n$,we get $n = \frac{P \lambda}{hc}$.Given: $P = 66 \ W$,$\lambda = 600 \ nm = 600 	imes 10^{-9} \ m$,$h = 6.6 	imes 10^{-34} \ J \cdot s$,and $c = 3 	imes 10^8 \ m/s$.Substituting the values:$n = \frac{66 	imes 600 	imes 10^{-9}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}$$n = \frac{66 	imes 600 	imes 10^{-9}}{19.8 	imes 10^{-26}}$$n = \frac{39600 	imes 10^{-9}}{19.8 	imes 10^{-26}}$$n = 2000 	imes 10^{17} = 2 	imes 10^{20}$.Thus,the number of photons emitted per second is $2 	imes 10^{20}$.

The number of photons emitted per second by a bulb of $66 \ W$ power emitting waves of wavelength $600 \ nm$ is . . . . . . . $(h = 6.6 \times 10^{-34} \ J \cdot s)$

Similar Questions

What is the charge on a photon?

The momentum of a photon is $2 \times 10^{-16} \text{ g} \cdot \text{cm/s}$. Its energy is:

Energy of a photon whose frequency is $10^{12} \text{ MHz}$ is
[ Planck's constant,$h = 6.63 \times 10^{-34} \text{ Js}, e = 1.6 \times 10^{-19} \text{ C}$ ]

If we express the energy of a photon in $KeV$ and the wavelength in $\mathring{A}$,then the energy of a photon can be calculated from the relation:

The wavelength of a photon with energy $35 \text{ keV}$ is . . . . . . .
$(h = 6.625 \times 10^{-34} \text{ J s}, c = 3 \times 10^{8} \text{ m/s}, 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module