A 150 W lamp emits light of mean wavelength of `5500Å`. If the efficiency is `12%` , find out the number of photons emitted by the lamp in one second.

Explanation

Given data `:`
Power of light emitted P =150 W
Wavelength of light emitted `lambda = 5500Å`
To find `:`
No. of photons emitted by the lamp in one second n = ?
Each photon has energy E =hv
v frequency of light
If N photons emitted, then the total energy radio Nhv.
Average power `P = ( "Energy")/("time") = ( E )/( t ) = ( Nhv)/( t ) `
But n= `( N )/( t )` is the No. of photons emitted per sec.
`:. n = ( rho)/( hv)` where ` v = ( c )/( lambda)`
So `n =( rho lambda)/( hC )`
If all the power is not radiated then the effective power = rated power `xx` efficiency
`:.` The number of photons emitted per sec.
`n = ( rho_("eff).lambda)/(hC ) = ( rho xx etalambda)/(hC) `[ where `eta`- efficiency ]
`n = (rho eta lambda)/( hC)`
`= ( 150 xx 0.12 xx 550 xx 10^(-10))/( 6.626 xx 10^(-34) xx 3 xx 10^(8)) =( 99000 xx 10^(-10))/( 19.878 xx 10^(-26))`
`= ( 99 xx 10^(-7))/( 19.878 xx 10^(-26))`
`n = 4.98 xx 10^(19)`