Question

If the stopping potential in a photoelectric experiment is measured to be 1.82 V, the maximum speed of the emitted electrons, in ms\(^{-1}\), is
(mass of the electron = \(9.1\times10^{-31}\) kg)}

• A. \(8.0\times10^5\)
• B. \(2.3\times10^5\)
• C. \(3.0\times10^5\)
• D. \(7.3\times10^6\)
• E. \(5.3\times10^{11}\)

Hint:

For stopping potential problems: \[ eV_s=\frac{1}{2}mv_{\max}^2 \] This directly gives the maximum kinetic energy of the emitted electron.

Answer 1

The Correct Option is A

In photoelectric effect: \[ eV_s=\frac{1}{2}mv_{\max}^2 \] So, \[ v_{\max}=\sqrt{\frac{2eV_s}{m}} \] Given: \[ V_s=1.82\text{ V} \] \[ e=1.6\times10^{-19}\text{ C} \] \[ m=9.1\times10^{-31}\text{ kg} \] Substitute: \[ v_{\max}=\sqrt{\frac{2(1.6\times10^{-19})(1.82)}{9.1\times10^{-31}}} \] \[ =\sqrt{\frac{5.824\times10^{-19}}{9.1\times10^{-31}}} \] \[ =\sqrt{6.4\times10^{11}} \] \[ v_{\max}=8.0\times10^5\text{ ms}^{-1} \]
Hence, the correct answer is: \[ \boxed{(A)\ 8.0\times10^5\text{ ms}^{-1}} \]