Maximum Photoelectrons Emitted from Isolated Silver Sphere

To find the maximum number of photoelectrons emitted from the isolated silver sphere, we analyze the photoelectric effect and the electrostatic potential acquired by the sphere.

1. Energy of the Incident Photons:
   The energy of a photon of wavelength $\lambda = 200\text{ nm}$ is given by:
   $E = \frac{hc}{\lambda}$
   Using standard values:
   $hc \approx 1240\text{ eV}\cdot\text{nm} \quad (\text{or } 1242\text{ eV}\cdot\text{nm})$
   $E = \frac{1240\text{ eV}\cdot\text{nm}}{200\text{ nm}} = 6.2\text{ eV}$

2. Maximum Kinetic Energy of Emitted Photoelectrons:
   The work function of the silver sphere is $\Phi = 4.7\text{ eV}$.
   According to Einstein's photoelectric equation, the maximum initial kinetic energy of the emitted photoelectrons is:
   $K_{\max} = E - \Phi = 6.2\text{ eV} - 4.7\text{ eV} = 1.5\text{ eV}$

3. Stopping Condition (Limiting Potential):
   As photoelectrons are emitted, the isolated sphere loses negative charge and acquires a positive charge $Q$. Consequently, an electric potential $V$ develops on the sphere:
   $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$
   where:

• $R = 1\text{ cm} = 10^{-2}\text{ m}$ is the radius of the sphere,
• $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$.

Photoemission ceases when the positive potential of the sphere reaches a value such that even the fastest emitted electrons (with kinetic energy $K_{\max}$) are pulled back to the sphere. This occurs when the potential $V$ equals the stopping potential $V_s$:
$e V_{\max} = K_{\max} = 1.5\text{ eV} \implies V_{\max} = 1.5\text{ V}$

1. Maximum Charge and Number of Emitted Photoelectrons:
   The maximum positive charge accumulated on the sphere is:
   $Q_{\max} = 4\pi\varepsilon_0 R V_{\max}$
   Substituting the given values:
   $Q_{\max} = \frac{10^{-2}\text{ m} \times 1.5\text{ V}}{9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2} = \frac{1.5}{9} \times 10^{-11}\text{ C} = \frac{1}{6} \times 10^{-11}\text{ C} \approx 1.67 \times 10^{-12}\text{ C}$

The total number of photoelectrons emitted, $n$, corresponds to this charge:
$n = \frac{Q_{\max}}{e}$
where $e = 1.6 \times 10^{-19}\text{ C}$.

$n = \frac{\frac{1}{6} \times 10^{-11}}{1.6 \times 10^{-19}} = \frac{10^8}{9.6} \approx 1.04 \times 10^7$

1. Determining the Value of $Z$:
   The question specifies that the maximum number of photoelectrons is written in the form:
   $n = \text{A} \times 10^Z \quad \text{with } 1 < \text{A} < 10$

Since $n \approx 1.04 \times 10^7$:
$\text{A} \approx 1.04 \quad \text{and} \quad Z = 7$

Thus, the value of $Z$ is $7$.