Question

Only a fraction of the electrical energy supplied to a tungsten light bulb is converted to visible light. The rest of the energy shows up as infrared radiation (i.e., heat). A 75-W light bulb converts 15.0% of the energy supplied to it into visible light. Assuming a wavelength of 550 nm, how many photons are emitted by the light bulb per second? (recall 1 W = 1 J/s)

Answer 1

Answer:

$n=3.11\times 10^{19}\ photons\ per\ second$

Explanation:

It is given that,

Power of the light bulb, P = 75 W

Wavelength, $\lambda=550\ nm=550\times 10^{-9}\ m$

It is mentioned that light bulb converts 15.0% of the energy supplied to it into visible light. 15 % of 75 W is,

$E=15\%\ of \ 75\ W=11.25\ J$

Let n is the number of photons emitted by the light bulb per second. It is given by :

$E=\dfrac{nhc}{\lambda}$

$n=\dfrac{E\lambda}{hc}$

$n=\dfrac{11.25\times 550\times 10^{-9}}{6.63\times 10^{-34}\times 3\times 10^8}$

$n=3.11\times 10^{19}\ photons\ per\ second$

Hence, this is the required solution.