Light of wavelength \(2000\,\mathring{\mathrm{A}}\) falls on a metal surface which needs minimum energy of \(4.2\) eV to emit the electrons. Kinetic energy of the fastest electron is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2.0\) eV
By Einstein’s photoelectric equation, the fastest electron has kinetic energy equal to the photon energy minus the work function (the minimum energy needed to emit an electron).
\( \displaystyle E = \frac{hc}{\lambda} = \frac{12400\,\mathrm{eV\,\mathring{A}}}{2000\,\mathring{\mathrm{A}}} = 6.2\,\mathrm{eV} \)
\( \displaystyle K_{\max} = E - \phi_0 = 6.2\,\mathrm{eV} - 4.2\,\mathrm{eV} \)
\( \displaystyle K_{\max} = 2.0\,\mathrm{eV} \)