An electric bulb of 100 W power emits photon | vedclass

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

Question

(C) The power of the bulb is $P = 100 \ W$,which means it emits $100 \ J$ of energy per second.The energy of a single photon is given by the formula $

Vedclass · Accepted Answer

$2.06 	imes 10^{20}$

Vedclass · Answer

(C) The power of the bulb is $P = 100 \ W$,which means it emits $100 \ J$ of energy per second.The energy of a single photon is given by the formula $E = \frac{hc}{\lambda}$.Given values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 410 	imes 10^{-9} \ m$.Substituting these values,the energy of one photon is $E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{410 	imes 10^{-9}} \approx 4.85 	imes 10^{-19} \ J$.The number of photons emitted per second $(n)$ is calculated as $n = \frac{P}{E}$.$n = \frac{100}{4.85 	imes 10^{-19}} \approx 2.06 	imes 10^{20}$ photons per second.

An electric bulb of $100 \ W$ power emits photons of wavelength $410 \ nm$ per second. The number of photons emitted per second is: $(h = 6.63 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s)$

Similar Questions

The wavelength of an electron and a photon are equal. If the energy of the photon is $E$ and the momentum of the electron is $p$,what is the value of $p/E$?

The energy flux of sunlight reaching the surface of the earth is $1.388 \times 10^{3} \; W/m^{2}$. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of $550 \; nm$.

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

The rest energy of an electron is

Positron is the antiparticle of

Vedclass Test Series

Exam Paper Generator

Online Exam Module