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Question: If \[5%\]of the energy supplied to a bulb is irradiated as visible light, how many quanta are emitte...

If 55%of the energy supplied to a bulb is irradiated as visible light, how many quanta are emitted per second by a 100 W100\text{ }Wlamp? Assume wavelength of visible light of5.6×105cm5.6\times {{10}^{-5}}cm.
A. 1.4×1019 B. 3×103 C. 1.4×1019 D. 3×104 \begin{aligned} & \text{A}\text{. 1}\text{.4}\times \text{1}{{\text{0}}^{19}} \\\ & \text{B}\text{. 3}\times \text{1}{{\text{0}}^{3}} \\\ & \text{C}\text{. 1}\text{.4}\times \text{1}{{\text{0}}^{-19}} \\\ & \text{D}\text{. 3}\times \text{1}{{\text{0}}^{4}} \\\ \end{aligned}

Explanation

Solution

Hint: Calculate energy radiated as visible light in joule per second unit. Use the formula of energy which will give the relation between a number of quanta, wavelength, speed of light and energy. We also know that each photon has an energy equal to hνh\nu . Unit of a number of quanta is joule/secjoule/\sec .
Formula used:
E=nhcλE=\dfrac{nhc}{\lambda }
Where,
E=EnergyE=\text{Energy}
h= Plank !!!! s constanth=\text{ Plank }\\!\\!'\\!\\!\text{ s constant}
λ=Wwavelength\lambda =W\text{wavelength}
n= No of quantan=\text{ No of quanta}
c= Velocity of lightc=\text{ Velocity of light}

Complete step by step answer:
In this question, we have given the wavelength of visible light(λ=5.6×105cm)(\lambda =5.6\times {{10}^{-5}}cm) and power.
55%of the energy supplied to a bulb which is irradiated as visible light.
So the energy radiate by visible light is given by
Energy radiated as visible light=100×5100=5joule/sec\text{=100}\times \dfrac{5}{100}=5joule/\sec
We also the formula of energy as,
E=nhcλ n=Eλhc \begin{aligned} & E=\dfrac{nhc}{\lambda } \\\ & \therefore n=\dfrac{E\lambda }{hc} \\\ \end{aligned}
Put the value of wavelength and energy in the above equation
n=5×5.6×107m(6.62×1034)(3×108)=1.4×1019photons/persecn=\dfrac{5\times 5.6\times {{10}^{-7}}m}{(6.62\times {{10}^{-34}})(3\times {{10}^{8}})}=1.4\times {{10}^{19}}photons/per\sec
Quanta are emitted per second by a 100 W100\text{ }Wlamp1.4×1019photons/persec1.4\times {{10}^{19}}photons/per\sec .
Additional information:
It was believed that until 1901, that radiation of energy took the form of continuous electromagnetic waves. However this concept failed to expand the discrete nature of black body radiation in 1901, Max Planck proposed a new theory known as planck's quantum theory. According to this theory radiation of energy does not take place continuously but in the form of discrete packets or quanta. This theory could explain satisfactorily the black body radiation, photoelectric emission, line spectrum of hydrogen Atom and content effect.
Number of quanta depends on wavelength, energy, planck's constant and velocity of light. Greater the value of wavelength, more will be energy and hence the number of quanta will be more.
Answer - (A)

Note: Students should know that energy radiation takes place either when a particle absorbs light or when an electron jumps from higher energy orbit to lower energy orbit. According to planck's quantum theory, the energy is radiated or absorbed in the form of discrete packets called ‘quanta’ or ‘photon’. Each photon has an energy equal tohνh\nu .