Question 11 of 62intermediate🔧 ApplyNumerical2 marks

A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm. Calculate the rate of emission of quanta per second.

Correct Answer

Given Data: Power of the bulb = 2525 watt = 2525 J s−1^{-1} Wavelength of yellow light, λ=0.57μm=0.57×10−6\lambda = 0.57 \mu m = 0.57 \times 10^{-6} m

Step 1: Calculate the energy of one photon The energy (EE) of one photon is given by the expression E=hcλE = \frac{hc}{\lambda} h=6.626×10−34h = 6.626 \times 10^{-34} J s c=3×108c = 3 \times 10^8 m s−1^{-1} E=6.626×10−34 J s×3×108 m s−10.57×10−6 m=3.487×10−19E = \frac{6.626 \times 10^{-34} \text{ J s} \times 3 \times 10^8 \text{ m s}^{-1}}{0.57 \times 10^{-6} \text{ m}} = 3.487 \times 10^{-19} J

Step 2: Calculate the rate of emission of quanta per second Rate of emission of quanta = Total energy emitted per secondEnergy of one photon\frac{\text{Total energy emitted per second}}{\text{Energy of one photon}} Rate of emission = 25 J s−13.487×10−19 J=7.17×1019\frac{25 \text{ J s}^{-1}}{3.487 \times 10^{-19} \text{ J}} = 7.17 \times 10^{19} s−1^{-1}

The rate of emission of quanta per second is 7.17×10197.17 \times 10^{19} s−1^{-1}.

Exercise: EXERCISES | Q: 2.11 | (Page 41)
For More Understanding

Explanation

To find the rate of emission of quanta, we first determine the energy of a single quantum (photon) of the given yellow light using Planck's equation, E=hc/λE = hc/\lambda. Then, we divide the total power of the bulb (which is the total energy emitted per second in Joules) by the energy of a single photon. This gives the number of photons or quanta emitted every second.

Solution Steps

  1. Step 1: Convert the given wavelength to meters: λ=0.57μm=0.57×10−6\lambda = 0.57 \mu m = 0.57 \times 10^{-6} m.

  2. Step 2: Calculate the energy of one photon using E=hcλ=6.626×10−34×3×1080.57×10−6=3.487×10−19E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.57 \times 10^{-6}} = 3.487 \times 10^{-19} J.

  3. Step 3: Calculate the rate of emission of quanta by dividing the bulb's power by the energy of one photon: 253.487×10−19=7.17×1019\frac{25}{3.487 \times 10^{-19}} = 7.17 \times 10^{19} s−1^{-1}.

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