Question

A 25 watt bulb emits monochromatic yellow light of wavelength of $0.57\mu m$. Calculate the rate of emission of quanta per second.

Answer 1

Monochromatic light, or one-color light, is an electromagnetic radiation derived from photon emissions from atoms. Photons travel, as energy wave fronts of different lengths and levels of energy. Energy levels determine the frequency of light, and the length of a wave determines its color.

Complete step by step answer:
It is given in the problem that,
Power of bulb$= 25watt$ ($\therefore$ $watt = J{s^{ - 1}}$ )
Wavelength of light=$0.57\mu m = 0.57 \times {10^{ - 6}}m$
To calculate:
We have to calculate Rate of emission of quanta per second which is ,$R = \dfrac{{Power}}{{Energy}} = ?$ where,
${\text{R}}$ is the rate of emission of quanta per second
${\text{P}}$ is the power of bulb
And ${\text{E}}$ is the energy .
The power is provided to us so we have to calculate energy only.
And we know that,
Energy,$E = \dfrac{{hc}}{\lambda }$ ($\lambda = wavelength$ ) ------------(1)
Where,
$h$ is planck's constant and its numerical value is $6.626 \times {10^{ - 34}}$ .
$c$ is the speed of light, $c = 3 \times {10^8}$
Put all these values in eq.(1)
$E = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.57 \times {{10}^{ - 6}}}}$
$E = 34.87 \times {10^{ - 20}}J$
Hence the obtained energy is , $E = 34.87 \times {10^{ - 20}}J$.
Now, to calculate Rate of emission of quanta per second just apply the formula
$R = \dfrac{P}{E}$ ----------------------(2)
Where,
${\text{R}}$ is the rate of emission of quanta per second
${\text{P}}$ is the power of bulb
And ${\text{E}}$ is the energy which we have calculated.
Put all the values in eq.(2)
$R = \dfrac{{25}}{{34.87 \times {{10}^{ - 20}}}}$
$R = 7.17 \times {10^{19}}{s^{ - 1}}$
Hence the rate of emission of quanta per second is $7.17 \times {10^{19}}$ ${s^{ - 1}}$ .

Note:
The unit conversion of units plays an important role in numerical, we have to bring all units in the same order and we should know the conversions like$watt = J{s^{ - 1}}$ (used in our problem). Unit conversion is a multi-step process that involves multiplication or division by a numerical factor, selection of the correct number o.