The maximum velocity of an electron emitted by light of wavelength \(\lambda\) incident on the surface of a…

The maximum velocity of an electron emitted by light of wavelength \(\lambda\) incident on the surface of a metal of work function \(\phi\) is [ \(h=\) Planck's constant, \(m=\) mass of electron and \(c=\) speed of light]
  1. \(\sqrt{\frac{2(h c+\lambda \phi)}{m \lambda}}\)
  2. \(\frac{2(h c-\lambda \phi)}{m}\)
  3. \(\sqrt{\frac{2(h c-\lambda \phi)}{m \lambda}}\)
  4. \(\frac{2(h \lambda-\phi)}{m}\)

Solution

According to Einstein's photoelectric equation, maximum kinetic energy of emitted electron is \(\begin{aligned} & K_{\max }=\frac{h c}{\lambda}-\phi \quad \Rightarrow \frac{1}{2} m v_{\max }^2=\frac{h c-\lambda \phi}{\lambda} \\ & \Rightarrow \quad v_{\max }^2=\frac{2(h c-\lambda \phi)}{m \lambda} \Rightarrow v_{\max }=\sqrt{\frac{2(h c-\lambda \phi)}{m \lambda}} \end{aligned}\)

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

Practice more Dual Nature of Matter and Radiation questions on Aicharya