Question

Read the paragraph carefully and select the proper choice from given multiple choices. According to Einstein when a photon of light of frequency for wavelength \(\lambda\) is incident on a photo sensitive metal surface of work function \(\Phi\). Where \(\Phi<\mathrm{hf}\) (here \(\mathrm{h}\) is Plank's constant) then the emission of photo-electrons place takes place. The maximum K.E. of emitted photo electrons is given by $\mathrm{K}_{\max }=\mathrm{hf}-\Phi .\( If the there hold frequency of metal is \)\mathrm{f}_{0}$ then \(\mathrm{hf}_{0}=\Phi\) (i) A metal of work function \(3.3 \mathrm{eV}\) is illuminated by light of wave length \(300 \mathrm{~nm}\). The maximum \(\mathrm{K} . \mathrm{E}>\) of photo- electrons is $=\ldots \ldots \ldots . \mathrm{eV} .\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}\right)$ (A) \(0.825\) (B) \(0.413\) (C) \(1.32\) (D) \(1.65\) (ii) Stopping potential of emitted photo-electron is $=\ldots \ldots . \mathrm{V}$. (A) \(0.413\) (B) \(0.825\) (C) \(1.32\) (D) \(1.65\) (iii) The threshold frequency fo $=\ldots \ldots \ldots \times 10^{14} \mathrm{~Hz}$. (A) \(4.0\) (B) \(4.2\) (C) \(8.0\) (D) \(8.4\)

Short answer

The short answers to the given tasks are as follows: (i) The maximum kinetic energy of photo-electrons is \(0.825 \, eV\). (ii) The stopping potential of emitted photo-electrons is \(0.825 \, V\). (iii) The threshold frequency is \(8 \times 10^{14} \, Hz\).

Step-by-step solution

Find the energy of the incident light

We can find the energy of the incident light by converting the given wavelength to frequency using the formula \(f=c / \lambda\). Given \(\lambda = 300 \, nm = 300 \times 10^{-9} \, m\) and \(c = 3 \times 10^{8} \, m/s\), we get \(f = (3 \times 10^{8}) / (300 \times 10^{-9}) = 1 \times 10^{15} \, Hz\). The energy of the incident light in Joules is then given by \(hf = (6.6 \times 10^{-34}) \times 1 \times 10^{15} = 6.6 \times 10^{-19} \, J\). To convert this from Joules to eV, we use the conversion factor \(1 \, eV = 1.6 \times 10^{-19} \, J\), getting \(hf = (6.6 \times 10^{-19}) / (1.6 \times 10^{-19}) = 4.125 \, eV\).

Find the maximum kinetic energy

We subtract the work function of the metal from the total energy we found in the previous step to find the maximum kinetic energy of the photo-electrons, as given by the equation \(K_{\max} = hf - \Phi\). Here, \(\Phi = 3.3 \, eV\). So, \(K_{\max} = 4.125 - 3.3 = 0.825 \, eV\). Thus, the answer to (i) is (A).

Find the stopping potential of emitted photo-electrons

The maximum kinetic energy in eV is equal to the stopping potential in volts. Therefore, the stopping potential is \(0.825 \, V\). Hence, the answer to (ii) is (B).

Find the threshold frequency

The threshold frequency is the frequency of light that gives exactly enough energy to the electrons to just overcome the work function of the metal. It can be found using the formula \(hf_0 = \Phi\). Rearranging we have \(f_0 = \Phi / h\). Using \(\Phi = 3.3 \, eV = 3.3 \times 1.6 \times 10^{-19} \, J\) and \(h = 6.6 \times 10^{-34} \, J.s\), we get \(f_0 = (3.3 \times 1.6 \times 10^{-19}) / (6.6 \times 10^{-34}) = 8 \times 10^{14} \, Hz\). The answer to (iii) is therefore (C).