Question

A 200 W bulb emits monochromatic light of wavelength 1400 Å and only 10% of the energy is emitted as light. The number of photons emitted by the bulb per second will be
(1) 1.4 x 1018. (2) 1.4 x 1020
(3) 1.4 x 1018 (4) 1.4 x 1024
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Answer 1

Answer:

1.414*10^19 photons emitted.

Explanation:

Energy of each photon is hf.

Where h = planks constant and f = frequency of light.

For N photons emitted the total energy radiated will be  Nhν .

Average power is  P=Energy/time=E/t=Nhν/t.

N/t=n will be the number of photons emitted per second.

So,  n=P/hν.

ν=c/λ.

So,  n=Pλ/hc.

we know that the non radiated will have effective power = rated power x efficiency.

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

Power P = 200W, wavelength λ = 1400A and efficiency is 10% = 0.10.

Accordingly, the number of photons emitted is. 

n=(200×0.10*1400*10^-10) /(6.6×10−34) *(3×10^8) = 1.414*10^19 

Hence, 1.414*10^19  number of photons emitted per second by the bulb.

Answer 2

Answer:

The number of photons emitted by the bulb per second will be $1.4\times 10^{19}$.

Explanation:

Power of monochromatic light bulb = 200 W = 200 Joule/s

10% of the energy is emitted as light

Energy radiated as light per second ,E'= 10% of 200 Joules

$E'=\frac{10}{100}\times 200 Joules= 20 Joules$

Wavelength of the light , λ = 1400 Å $1.4\times 10^{-7} m$

$E=\frac{hc}{\lambda }$

E = energy of photon

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$E=\frac{6.63\times 10^{-34}Js\times 3\times 10^8 m/s}{1.4\tiems 10^{-7} m}$

E = $1.4207\times 10^{-18} Joules$

Let the number of photons be x.

$x\times E=E'$

$x=\frac{E'}{E}=\frac{20 Joules}{1.4207\times 10^{-18} Joules}=1.4\times 10^{19}$

The number of photons emitted by the bulb per second will be $1.4\times 10^{19}$.