Light of wavelength 400 nm is incident continuously on a caesium ball (work function 1.9 eV). The maximum potential to which the ball will be charged is (Take \( hc=1240\,\mathrm{eV\,nm} \).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.2 V
As the light ejects photoelectrons, the ball loses electrons and becomes positively charged. Its rising potential pulls the emitted electrons back. Charging stops when this potential can stop even the fastest photoelectrons. The potential then equals the stopping potential \( V_0 \), with \( eV_0=K_{\mathrm{max}} \).
The energy of each photon is:
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{1240\,\mathrm{eV\,nm}}{400\,\mathrm{nm}} \\ &= 3.1\,\mathrm{eV} \end{aligned} \)
By Einstein’s photoelectric equation, the maximum kinetic energy is:
\( \begin{aligned} K_{\mathrm{max}} &= E-\phi_0 \\ &= 3.1\,\mathrm{eV}-1.9\,\mathrm{eV} \\ &= 1.2\,\mathrm{eV} \end{aligned} \)
\( eV_0=1.2\,\mathrm{eV} \)
\( \Rightarrow\; V_0=1.2\,\mathrm{V} \)
Therefore, the ball charges up to a maximum potential of 1.2 V.
NCERT (2026-27):
Effect of Potential on Photoelectric Current – “The interpretation of the observation in terms of photoelectrons is straightforward. All the photoelectrons emitted from the metal do not have the same energy. Photoelectric current is zero when the stopping potential is sufficient to repel even the most energetic photoelectrons, with the maximum kinetic energy (Kmax), so that”
\( K_{\mathrm{max}}=eV_0 \qquad (11.1) \)