1. Given data

• Radius of silver sphere: $R = 1\text{ cm} = 10^{-2}\text{ m}$
• Work function: $\phi = 4.7\text{ eV}$
• Incident wavelength: $\lambda = 200\text{ nm}$

We need the maximum number of photoelectrons emitted before emission stops.

2. Maximum kinetic energy of emitted photoelectrons

Photon energy: $E = \frac{hc}{\lambda} = \frac{1240\text{ eV nm}}{200\text{ nm}} = 6.2\text{ eV}$

So the maximum kinetic energy is $K_{\max} = E - \phi = 6.2 - 4.7 = 1.5\text{ eV}$

3. Stopping condition

As electrons are emitted, the sphere becomes positively charged. Emission stops when the electric potential of the sphere is such that even the most energetic photoelectron cannot escape.

Thus, $eV = K_{\max} = 1.5\text{ eV}$ So, $V = 1.5\text{ V}$

4. Charge required to raise the sphere to this potential

Potential of an isolated conducting sphere: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$ Hence, $Q = 4\pi\varepsilon_0 R V$

Using $4\pi\varepsilon_0 = \frac{1}{9\times 10^9}$ we get $Q = \frac{RV}{9\times 10^9}$

Substitute $R=10^{-2}\text{ m}$ and $V=1.5\text{ V}$: $Q = \frac{10^{-2}\times 1.5}{9\times 10^9} = \frac{1.5\times 10^{-2}}{9\times 10^9}$ $Q \approx 1.67\times 10^{-12}\text{ C}$

5. Number of electrons emitted

If $n$ electrons are emitted, $Q = ne$ So, $n = \frac{Q}{e} = \frac{1.67\times 10^{-12}}{1.6\times 10^{-19}}$ $n \approx 1.04\times 10^7$

Thus the maximum number of emitted photoelectrons is of the form $A\times 10^Z$ with $A\approx 1.04$ and $Z=7$

6. Final answer

$\boxed{7}$

7. Comparison with stored answer

Stored correct answer = $7$.

Our derived answer matches the stored answer.