Question

Work function of metal is \(2 \mathrm{eV}\). Light of intensity $10^{-5} \mathrm{Wm}^{-2}\( is incident on \)2 \mathrm{~cm}^{2}\( area of it. If \)10^{17}$ electrons of these metals absorb the light, in how much time does the photo electric effectc start? Consider the waveform of incident light. (A) \(1.4 \times 10^{7} \mathrm{sec}\) (B) \(1.5 \times 10^{7} \mathrm{sec}\) (C) \(1.6 \times 10^{7} \mathrm{sec}\) (D) \(1.7 \times 10^{7} \mathrm{sec}\)

Short answer

The time it takes for the photoelectric effect to start is \(1.6 \times 10^{7}\) sec. The correct answer is (C) \(1.6 \times 10^{7} \mathrm{sec}\).

Step-by-step solution

Calculate the total energy incident on the metal surface

The intensity (I) of the light is given as \(10^{-5}\) W/m². The area (A) of the metal surface is given as \(2 cm^{2}\) or \(2 \times 10^{-4} m^{2}\). We can find the power (P) incident on the surface using the formula: \[P = IA\] Plug in the given values: \[P = (10^{-5} \mathrm{Wm}^{-2}) (2 \times 10^{-4} \mathrm{m}^{2})\] \[P = 2 \times 10^{-9} \mathrm{W}\]

Calculate the energy required to eject one electron

The work function of the metal is given as 2 eV. This is the minimum energy required to eject one electron from the metal surface. To convert the energy from eV to Joules, we can use the conversion factor \(1.6 \times 10^{-19} \mathrm{J/eV}\): \[E_{electron} = (2 \mathrm{eV}) (1.6 \times 10^{-19} \mathrm{J/eV})\] \[E_{electron} = 3.2 \times 10^{-19} \mathrm{J}\]

Calculate the total energy required for the given number of electrons

We're given that \(10^{17}\) electrons are involved in the photoelectric effect. To find the total energy required for all these electrons, we can multiply the energy required for one electron by the total number of electrons: \[E_{total} = E_{electron} \times N_{electrons}\] \[E_{total} = (3.2 \times 10^{-19} \mathrm{J}) (10^{17})\] \[E_{total} = 3.2 \times 10^{-2} \mathrm{J}\]

Calculate the time it takes for the electrons to absorb the energy

Now that we have the power incident on the metal surface (from Step 1) and the total energy required for the given number of electrons (from Step 3), we can find the time (t) it takes for the photoelectric effect to start using the formula: \[P = \frac{E_{total}}{t}\] Rearrange the formula to find t: \[t = \frac{E_{total}}{P}\] Plug in the values: \[t = \frac{3.2 \times 10^{-2} \mathrm{J}}{2 \times 10^{-9} \mathrm{W}}\] \[t = 1.6 \times 10^{7} \mathrm{sec}\] Hence, the time it takes for the photoelectric effect to start is \(1.6 \times 10^{7} \mathrm{s}\). The correct answer is (C) \(1.6 \times 10^{7} \mathrm{sec}\).