Solution: Isolated Zinc Bead Potential
Key Concept: The bead starts at a negative potential $-V_0$. As light hits it, it emits electrons.
1. Electrons are emitted with $K_{max} = \frac{hc}{\lambda} – \phi$.
2. Since the bead is negative, it repels the emitted electrons, accelerating them as they move to infinity.
3. The bead loses negative charge, so its potential rises (becomes less negative, then positive).
4. Emission stops when the bead becomes positive enough to pull back the max energetic electrons.
Statement (a): Maximum speed immediately after emission
At the surface, before field interaction: $$ \frac{1}{2}mv_{surface}^2 = \frac{hc}{\lambda} – \frac{hc}{\lambda_0} $$ $$ v_{surface} = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right)} $$ Statement (a) is Correct.
At the surface, before field interaction: $$ \frac{1}{2}mv_{surface}^2 = \frac{hc}{\lambda} – \frac{hc}{\lambda_0} $$ $$ v_{surface} = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right)} $$ Statement (a) is Correct.
Statement (b): Maximum speed at great distance
The electron moves from potential $-V_0$ to $0$ (infinity). Work done by field $W = q\Delta V = (-e)(0 – (-V_0)) = -eV_0$. Potential energy $U = qV$. $K_i + U_i = K_f + U_f$ $K_{surface} + (-e)(-V_0) = K_{\infty} + (-e)(0)$ $K_{\infty} = K_{surface} + eV_0$ $$ \frac{1}{2} m v_{\infty}^2 = hc \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right) + eV_0 $$ $$ v_{\infty} = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right) + \frac{2eV_0}{m}} $$ Statement (b) is Correct.
The electron moves from potential $-V_0$ to $0$ (infinity). Work done by field $W = q\Delta V = (-e)(0 – (-V_0)) = -eV_0$. Potential energy $U = qV$. $K_i + U_i = K_f + U_f$ $K_{surface} + (-e)(-V_0) = K_{\infty} + (-e)(0)$ $K_{\infty} = K_{surface} + eV_0$ $$ \frac{1}{2} m v_{\infty}^2 = hc \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right) + eV_0 $$ $$ v_{\infty} = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right) + \frac{2eV_0}{m}} $$ Statement (b) is Correct.
Statement (d): Total number of photoelectrons
Final potential $V_f$ is reached when emission stops: $V_f = \frac{hc}{e}(\frac{1}{\lambda} – \frac{1}{\lambda_0})$. Initial potential $V_i = -V_0$. Capacitance of sphere $C = 4\pi\epsilon_0 r$. Charge lost $\Delta Q = C(V_f – V_i) = 4\pi\epsilon_0 r (V_f – (-V_0))$. Number of electrons $N = \frac{\Delta Q}{e} = \frac{4\pi\epsilon_0 r}{e} (V_f + V_0)$. Substituting $V_f$: $$ N = \frac{4\pi\epsilon_0 r}{e} \left[ \frac{hc}{e} \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right) + V_0 \right] $$ Statement (d) is Correct.
Final potential $V_f$ is reached when emission stops: $V_f = \frac{hc}{e}(\frac{1}{\lambda} – \frac{1}{\lambda_0})$. Initial potential $V_i = -V_0$. Capacitance of sphere $C = 4\pi\epsilon_0 r$. Charge lost $\Delta Q = C(V_f – V_i) = 4\pi\epsilon_0 r (V_f – (-V_0))$. Number of electrons $N = \frac{\Delta Q}{e} = \frac{4\pi\epsilon_0 r}{e} (V_f + V_0)$. Substituting $V_f$: $$ N = \frac{4\pi\epsilon_0 r}{e} \left[ \frac{hc}{e} \left( \frac{1}{\lambda} – \frac{1}{\lambda_0} \right) + V_0 \right] $$ Statement (d) is Correct.
Correct Options: (a), (b), and (d)