Question

\(2 \mathrm{nW}\) light of wave length \(4400 \AA\) is incident on photo sensitive surface of \(\mathrm{Cs}\). If quantum efficiency is \(0.5 \%\), what will be the value of photoelectric current? (A) \(1.56 \times 10^{-6} \mu \mathrm{A}\) (B) \(2.56 \times 10^{-6} \mu \mathrm{A}\) (C) \(4.56 \times 10^{-6} \mu \mathrm{A}\) (D) \(3.56 \times 10^{-6} \mu \mathrm{A}\)

Short answer

The short answer to the problem is: The photoelectric current is approximately \(3.54 \times 10^{-6} \mu \mathrm{A}\), which is closest to option (D). Therefore, the correct answer is: (D) \(3.56 \times 10^{-6} \mu \mathrm{A}\).

Step-by-step solution

Calculate the energy of a single photon

First, we need to find the energy of a single photon. We can use the relationship between energy, Planck's constant, and wavelength to find the energy of one photon: \[E = \frac{hc}{λ}\] Where \(E\) is the energy of the photon, \(h\) is the Planck constant (\(6.63 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(3 \times 10^{8} \mathrm{\frac{m}{s}})\), and λ is the wavelength in meters. Since the wavelength is given in Angstroms, we need to convert it to meters first: \[4400 \mathrm{\AA} = 4400 \times 10^{-10} \mathrm{m}\] Now, we can substitute the values into the formula: \[E = \frac{ (6.63 \times 10^{-34} \mathrm{Js}) (3 \times 10^{8} \mathrm{\frac{m}{s}}) }{4400 \times 10^{-10} \mathrm{m}} = 4.52 \times 10^{-19} \mathrm{J}\]

Calculate the number of incident photons per second

Next, we need to find the number of incident photons per second. We can do this by dividing the incident light power by the energy of a single photon: \[\mathrm{Number\:of\:photons\:per\:second} = \frac{P}{E}\] Where \(P\) is the incident light power in watts and \(E\) is the energy of a single photon. Given the incident light power of \(2 \mathrm{nW}\), we first convert it to watts: \[2 \mathrm{nW} = 2 \times 10^{-9} \mathrm{W}\] Now, we can substitute the values into the formula: \[\mathrm{Number\:of\:photons\:per\:second} = \frac{2 \times 10^{-9} \mathrm{W}}{4.52 \times 10^{-19} \mathrm{J}} = 4.426 \times 10^{9} \mathrm{photons/s}\]

Calculate the number of emitted photoelectrons per second

We know the quantum efficiency, which is the ratio of the number of emitted photoelectrons to the number of incident photons. We can use this value to find the number of emitted photoelectrons per second: \[\mathrm{Number\:of\:emitted\:photoelectrons\:per\:second} = \mathrm{Quantum\:efficiency} \times \mathrm{Number\:of\:incident\:photons\:per\:second}\] Given the quantum efficiency of \(0.5 \% = 0.005\), we can substitute the values into the formula: \[\mathrm{Number\:of\:emitted\:photoelectrons\:per\:second} = 0.005 \times 4.426 \times 10^{9} \mathrm{photons/s} = 2.213 \times 10^{7} \mathrm{photoelectrons/s}\]

Calculate the photoelectric current

Finally, we can calculate the photoelectric current using the number of emitted photoelectrons per second and the elementary charge (charge of an electron): \[I = e \times \mathrm{Number\:of\:emitted\:photoelectrons\:per\:second}\] where \(e = 1.6 \times 10^{-19} \mathrm{C}\) and \(I\) is the photoelectric current. Now, we can substitute the values into the formula: \[I = (1.6 \times 10^{-19} \mathrm{C}) \times (2.213 \times 10^{7} \mathrm{photoelectrons/s}) = 3.54 \times 10^{-12} \mathrm{A}\] Converting to microAmperes: \[I = 3.54 \times 10^{-12} \mathrm{A} \times \frac{10^6}{1} \mathrm{\frac{\mu A}{A}} = 3.54 \times 10^{-6} \mu \mathrm{A}\] The photoelectric current is approximately \(3.54 \times 10^{-6} \mu \mathrm{A}\), which is closest to option (D). Therefore, the correct answer is: (D) \(3.56 \times 10^{-6} \mu \mathrm{A}\).