Question

Problem 1.
A bulb emits light of wavelength 4500A.The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second ?​

Answer 1

Answer:

$\displaystyle{Numbers \ of \ photon =2.71\times10^{19} \ sec^{-1}}$

Explanation:

Given ;

Wavelength λ = 4500 × 10⁻¹⁰ m

Energy of photon = h c / λ

We have c and h constant values

Putting in formula to get Energy

$\displaystyle{E=\frac{6.626\times10^{-34}\times3\times10^{8}}{4500\times10^{-10}}}\\\\\\\displaystyle{E=4.42\times10^{-19} \ J}$

Now , energy emitted by bulb ,

$\displaystyle{\implies150\times\dfrac{8}{100} \ J}\\\\\\\displaystyle{\implies12 \ J}$

No. of photon = energy emitted / Energy of one photon

$\displaystyle{Numbers \ of \ photon =\dfrac{12}{4.42\times10^{-19}} \ sec^{-1}}\\\\\\\displaystyle{Numbers \ of \ photon =2.71\times10^{19} \ sec^{-1}}$

Thus we get answer .

Answer 2

ach photon has energy of  hν  

h = planks constant

ν = frequency of light

If there are N photons emitted then the total energy radiated is  Nhν .

Average power is given by  P=Energy/time=E/t=Nhν/t

But  N/t=n  is the number of photons emitted per second.

Therefore,  n=P/hν .

ν=c/λ

So,  n=Pλ/hc

If all the power is not radiated then the effective power = rated power x efficiency.

So the number of photons emitted per second  n=Peffλ/c = P×ηλ/hc  where η is the efficiency.

In the problem power P = 150W, wavelength λ = 4500A and efficiency is 8% = 0.08

So number of photons emitted per second, 

n=(150×0.08*4500*10-10) /(6.6×10−34) *(3*×108) = 27.2*1018 

27.2*1018 number of photons will be emitted per second by the bulb

Hope this will help you