Question

When monochromatic photons incident on a photosensitive material of cut-off wavelength \(496\,\mathrm{nm}\), photoelectrons are emitted with a maximum velocity of \(8\times10^5\,\mathrm{m\,s^{-1}}\). Then the energy of the incident photons is nearly \[ \left(m_e=9\times10^{-31}\,\mathrm{kg},\; e=1.6\times10^{-19}\,\mathrm{C}\right) \]

• A. \(6.1\,\mathrm{eV}\)
• B. \(3.7\,\mathrm{eV}\)
• C. \(4.3\,\mathrm{eV}\)
• D. \(5.2\,\mathrm{eV}\)

Hint:

Einstein's photoelectric equation: \[ \boxed{ h\nu=\phi+K_{\max} } \] where \[ \boxed{ \phi=\frac{hc}{\lambda_0} =\frac{1240}{\lambda_0(\mathrm{nm})}\,\mathrm{eV}. } \]

Answer 1

The Correct Option is C

Step 1: Calculate the work function. The work function is \[ \phi=\frac{hc}{\lambda_0} =\frac{1240}{496} =2.5\,\mathrm{eV}. \]

Step 2: Calculate the maximum kinetic energy. \[ K_{\max} = \frac12 mv^2 = \frac12(9\times10^{-31})(8\times10^5)^2 = 2.88\times10^{-19}\,\mathrm{J}. \] In electron volts, \[ K_{\max} = \frac{2.88\times10^{-19}}{1.6\times10^{-19}} = 1.8\,\mathrm{eV}. \]

Step 3: Find the energy of the incident photon. Using Einstein's photoelectric equation, \[ E=\phi+K_{\max} =2.5+1.8 =4.3\,\mathrm{eV}. \] Hence, \[ \boxed{4.3\,\mathrm{eV}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.