Question 30: Dual Nature of Matter and Radiation

Question

A monochromatic source of light operating at 15 kW emits $2.5 	imes 10^{22}$ photons $/ \mathrm{s}$. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to $\_\_\_\_$。
(Take $h=6.6 	imes 10^{-34} \mathrm{~J} . \mathrm{s}$ and $c=3 	imes 10^8 \mathrm{~m} / \mathrm{s}$ ).

Accepted Answer

The power $(\mathrm{P})$ of a light source is the total energy emitted per unit time. If the source emits n photons per second, and each photon has an energy E , then the power is:
$$ \mathrm{P}=\mathrm{n} 	imes \mathrm{E} $$
The given power of the source is $\mathrm{P}=15 \mathrm{~kW}=15 	imes 10^3 \mathrm{~W}$ and the number of photons per second is $\mathrm{n}= 2.5 	imes 10^{22} \mathrm{~s}^{-1}$
$$ E=\frac{P}{n} $$
$$\Rightarrow $$ $\mathrm{E}=\frac{15 	imes 10^3}{2.5 	imes 10^{22}}$
$$\Rightarrow $$ $$ \mathrm{E}=6 	imes 10^{-19} \mathrm{~J} $$
The energy of a photon is related to its wavelength by the formula:
$$ E=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E} $$
Substituting the constants and energy,
The Plank's constant is $\mathrm{h}=6.6 	imes 10^{-34} \mathrm{~J} \cdot \mathrm{~s}$,
The speed of light is $\mathrm{c}=3 	imes 10^8 \mathrm{~m} / \mathrm{s}$
$$ \lambda=\frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{6 	imes 10^{-19}} \mathrm{~m} $$
$$\Rightarrow $$ $\lambda=\frac{19.8 	imes 10^{-26}}{6 	imes 10^{-19}} \mathrm{~m}$
$$\Rightarrow $$ $\lambda=3.3 	imes 10^{-7} \mathrm{~m}$
$$\Rightarrow $$ $$ \lambda=3.3 	imes 10^{-7} 	imes 10^9 \mathrm{~nm}=330 \mathrm{~nm} $$
Visible light typically ranges from approximately 400 nm to 700 nm . And the ultraviolet (UV) radiation ranges from approximately 10 nm to 400 nm . Since 330 nm is less than 400 nm , the radiation belongs to the ultraviolet region.
Therefore, the emitted radiation belongs to the Ultraviolet region.
Hence, the correct option is (D).

Answer

Microwave

Answer

Infrared

Answer

Visible

Answer

Ultraviolet

Dual Nature of Matter and Radiation

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