In a photoelectric experiment, the stopping potential is 0.45 V. The maximum speed of the emitted photoelectrons is (Take mass of electron \(m=9.0\times10^{-31}\ \mathrm{kg}\) and \(e=1.6\times10^{-19}\ \mathrm{C}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(4.0\times10^{5}\ \mathrm{m\,s^{-1}}\)
The stopping potential is just enough to stop the fastest electrons. So the kinetic energy of these electrons equals \( eV_0 \), and the speed follows from \( \frac{1}{2}mv_{\max}^{2}=eV_0 \).
\( \begin{aligned} v_{\max} &= \sqrt{\frac{2eV_0}{m}} \\ &= \sqrt{\frac{2\times1.6\times10^{-19}\times0.45}{9.0\times10^{-31}}} \\ &= \sqrt{1.6\times10^{11}} \\ &= 4.0\times10^{5}\ \mathrm{m\,s^{-1}} \end{aligned} \)
Hence, the fastest photoelectrons leave the metal with a speed of \( 4.0\times10^{5}\ \mathrm{m\,s^{-1}} \).
NCERT (2026-27):
Summary – “4. … The stopping potential is directly related to the maximum kinetic energy of electrons emitted: e V0 = (1/2) m v2max = Kmax.”