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Question: A small metal plate (work function \(=2 \mathrm{eV}\) ) is placed at a distance of \(2 \mathrm{~m}\)...

A small metal plate (work function =2eV=2 \mathrm{eV} ) is placed at a distance of 2 m2 \mathrm{~m} from a monochromatic light source of wavelength 4.8×10−7 m4.8 \times 10^{-7} \mathrm{~m} and power 1.0 watt. The light falls normally on the plate. The number of photons striking the metal plate per second per unit area will be
A.4.82×10124.82\times {{10}^{12}}
B.4.82×10144.82\times {{10}^{14}}
C.4.82×10164.82\times {{10}^{16}}
D.4.82×10184.82\times {{10}^{18}}

Explanation

Solution

The energy carried by a single photon is photon energy. The amount of energy is directly proportional to the electromagnetic frequency of the photon and is thus inversely proportional to the wavelength, equivalently. The greater the frequency of the photon, the higher its energy is. A photon is an EM-radiation quantum. Its energy is given by E=hf=hcλ(E=h f=\dfrac{h c}{\lambda}( and is related to the radiation frequency f and wavelength λ
Formula used: E=hf=hcλE=hf=\dfrac{hc}{\lambda } (energy of a photon)

Complete step-by-step solution
Photoelectric effect, the phenomenon in which when electromagnetic radiation is absorbed, electrically charged particles are released from or within a material. When light falls on it the effect is often defined as the ejection of electrons from a metal plate. In the form of energy and its provision by
E=hf=hcλE=h f=\dfrac{h c}{\lambda}
Given in the question
Power of source P=1.0\mathrm{P}=1.0 watt
Distance of metal plate from source r=2 m\mathrm{r}=2 \mathrm{~m}
Thus, intensity of incident radiation at metal plate I=P4πr2\mathrm{I}=\dfrac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}
∴I=1.04π(2)2=0.0199Wm−2\therefore \mathrm{I}=\dfrac{1.0}{4 \pi(2)^{2}}=0.0199 \mathrm{Wm}^{-2}
Wavelength of source of light λ=4.8×10−7 m\lambda=4.8 \times 10^{-7} \mathrm{~m}
Thus, energy of incident photon E=hcλ\mathrm{E}=\dfrac{\mathrm{hc}}{\lambda}
∴E=(6.626×10−34)(3×108)4.8×10−7\therefore \text{E}=\dfrac{\left( 6.626\times {{10}^{-34}} \right)\left( 3\times {{10}^{8}} \right)}{4.8\times {{10}^{-7}}}
⇒4.14×10−19 J\Rightarrow 4.14\times {{10}^{-19}}~\text{J}
Number of photons striking per second per unit area N=IE\mathrm{N}=\dfrac{\mathrm{I}}{\mathrm{E}}
N=0.01994.14×10−19\text{N}=\dfrac{0.0199}{4.14\times {{10}^{-19}}}
⇒N=0.01994.14×10−19=4.82×1016\Rightarrow \mathrm{N}=\dfrac{0.0199}{4.14 \times 10^{-19}}=4.82 \times 10^{16} photons per second per unit area.
∴\therefore The number of photons striking the metal plate per second per unit area will be ⇒N=0.01994.14×10−19=4.82×1016\Rightarrow \mathrm{N}=\dfrac{0.0199}{4.14 \times 10^{-19}}=4.82 \times 10^{16}
Hence the correct option is (c).

Note: Photoelectric effect, the phenomenon in which when electromagnetic radiation is absorbed, electrically charged particles are released from or within a material. When light falls on it the effect is often defined as the ejection of electrons from a metal plate. When electromagnetic radiation, such as light, hits a material, the photoelectric effect is the emission of electrons. In this manner, electrons emitted are called photoelectrons.