Solveeit Logo
Login

Question

Question: A \(25{\text{ watt}}\) bulb emits monochromatic yellow light of wavelength of \(0.57\mu {\text{m}}\)...

A 25 watt25{\text{ watt}} bulb emits monochromatic yellow light of wavelength of 0.57μm0.57\mu {\text{m}}. Calculate the rate of emission of quanta per second.

Explanation

Solution

Monochromatic light is an electromagnetic wave derived from the photon emissions from atoms. In simple terms it is known as a one colour light. Photons travel in forms of energy wave fronts having different levels of energy. The energy levels determine the frequency or wavelength of the radiation which determines its colors.
Formulae used: R=PER = \dfrac{P}{E}
Where, RR is the rate of emission of quanta from the bulb per second
P is the power of the bulb
And E is the energy of the electromagnetic radiation emitted

Complete step by step answer:
We are given in the question that the power of the bulb is 25 watt25{\text{ watt}} or 25 Js−125{\text{ J}}{{\text{s}}^{ - 1}}.
The wavelength of the yellow light emitted is λ=0.57μm\lambda = 0.57\mu m or 5.7×10−7m5.7 \times {10^{ - 7}}m
We have to find the rate of emission of quanta per second which is given by the ratio of power of the bulb to energy of the emitted radiations.
We know the wavelength of the light and can thus find the energy easily using E=hcλE = \dfrac{{hc}}{\lambda }
hh represents the Planck’s constant and its value is 6.626×10−346.626 \times {10^{ - 34}}
And cc represents the speed of light which is 3×108 m s−13 \times {10^8}{\text{ m }}{{\text{s}}^{ - 1}}.
Substituting the values to find the energy, we get,
E=6.626×10−34×3×108 m s−15.7×10−7mE = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}{\text{ m }}{{\text{s}}^{ - 1}}}}{{5.7 \times {{10}^{ - 7}}m}}
⇒E=34.87×10−20 J\Rightarrow E = 34.87 \times {10^{ - 20}}{\text{ J}}
Now to find the rate of emission of quanta, we use the formula R=PER = \dfrac{P}{E}
Substituting the values, we get,
R=2534.87×10−20R = \dfrac{{25}}{{34.87 \times {{10}^{ - 20}}}}

∴R=7.17×1019 s−1\therefore R = 7.17 \times {10^{19}}{\text{ }}{{\text{s}}^{ - 1}}
Note:
The energy of a photon is given by Planck's quantum theory. According to Planck's quantum theory, the energy radiated by any substance is released in the form of fixed packets of energy known as photons. The energy of one photon or quanta is directly proportional to its frequency. Hence, we get the equation,
E α νE{\text{ }}\alpha {\text{ }}\nu or E=hν=hcλE = h\nu = h\dfrac{c}{\lambda }