Question

Radius of a beam of radiation of wavelength \(5000 \AA\) is $10^{-3} \mathrm{~m}\(. Power of the beam is \)10^{-3} \mathrm{~W}$. This beam is normally incident on a metal of work function \(1.9 \mathrm{eV}\). The charge emitted by the metal per unit area in unit time is \(\ldots \ldots \ldots\) Assume that each incident photon emits one electron. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(1.282 \mathrm{C}\) (B) \(12.82 \mathrm{C}\) (C) \(128.2 \mathrm{C}\) (D) \(1282 \mathrm{C}\)

Short answer

The correct answer is: \(Q = 128.2 \mathrm{C}\) (Option C)

Step-by-step solution

Calculate the energy of each photon

The energy of a photon can be calculated using the formula \(E = \cfrac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. Given that the wavelength of the beam is \(5000 \AA\), we first need to convert it to meters. \(1 \AA = 10^{-10} \mathrm{m}\) Therefore, the wavelength of the beam in meters is: \(\lambda = 5000 \times 10^{-10} \mathrm{m}\) Now, using the values of \(h = 6.625 \times 10^{-34} \mathrm{J \cdot s}\) and \(c = 3 \times 10^8 \mathrm{m/s}\), we can calculate the energy of one photon: \(E = \cfrac{(6.625 \times 10^{-34} \mathrm{J \cdot s})(3 \times 10^8 \mathrm{m/s})}{(5000 \times 10^{-10} \mathrm{m})}\)

Calculate the number of incident photons per second

We are given the power of the beam as \(10^{-3} \mathrm{W}\). Since power is the energy per unit time, we can find the number of photons incident per second by dividing the power by the energy of one photon: \(N = \cfrac{10^{-3} \mathrm{W}}{E}\)

Calculate the charge emitted per unit area in unit time

Since we are given the radius of the beam, we can find the area of the beam using the formula for the area of a circle: \(A = \pi r^2\). The radius of the beam is \(10^{-3} \mathrm{m}\), thus: \(A = \pi (10^{-3} \mathrm{m})^2\) We are told to assume that each incident photon emits one electron. Therefore, the charge emitted by the metal per unit area in unit time can be calculated as: \(Q = \cfrac{Ne}{A}\) Where \(N\) is the number of incident photons per second, \(e\) is the elementary charge (\(1.6 \times 10^{-19} \mathrm{C}\)), and \(A\) is the area of the beam. Plug in the values calculated in the previous steps to find the charge emitted per unit area in unit time: \(Q = \cfrac{Ne}{A}\) Now, compare the calculated value of \(Q\) with the given options to find the correct answer.