A 200, W sodium street lamp emits yellow ligh | vedclass

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

Question

(A) The total electrical power input is $P_{in} = 200\, W$.Given the efficiency is $25\%$, the effective power output as light is $P_{out} = 0.25 	im

Vedclass · Accepted Answer

$1.5 	imes 10^{20}$

Vedclass · Answer

(A) The total electrical power input is $P_{in} = 200\, W$.Given the efficiency is $25\%$, the effective power output as light is $P_{out} = 0.25 	imes 200\, W = 50\, W$.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 $P_{out} = nE = n \frac{hc}{\lambda}$.Rearranging for $n$, we get $n = \frac{P_{out} \lambda}{hc}$.Substituting the values: $n = \frac{50 	imes 0.6 	imes 10^{-6}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$.$n = \frac{30 	imes 10^{-6}}{19.89 	imes 10^{-26}} \approx 1.5 	imes 10^{20}$ photons per second.

$A$ $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu m$. Assuming it to be $25\%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is:

Similar Questions

Which of the following is not a property of photons?

Let $n_r$ and $n_b$ be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.

The fact that photons carry energy was established by

An electron-positron pair is created by $\gamma$-rays. What is the minimum energy of the $\gamma$-rays required for this process in $MeV$?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module