A silver of radius `1 cm` and work function `4.7 eV` is suspended from an insulating thread in freepace. It is under continuous illumination of `200 nm` wavelength light. As photoelectron are emitted the sphere gas charged and acquired a potential . The maximum number of photoelectron emitted from the sphere is `A xx 10^(e) (where 1lt A lt 10)` The value of `z` is

Options

A. 6
B. 7
C. 8
D. 9

Explanation

To solve the problem step by step, we will follow the principles of photoelectric effect and electrostatics.

### Step 1: Calculate the energy of the incident photons
The energy of the photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m} \)

Calculating:
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{200 \times 10^{-9}} = 9.939 \times 10^{-19} \, \text{J} \]

### Step 2: Convert the energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{9.939 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 6.21 \, \text{eV} \]

### Step 3: Calculate the maximum kinetic energy of the emitted electrons
The maximum kinetic energy \( K.E. \) of the emitted electrons can be calculated using the equation:
\[ K.E. = E - \phi \]
where \( \phi \) is the work function. Given \( \phi = 4.7 \, \text{eV} \):
\[ K.E. = 6.21 \, \text{eV} - 4.7 \, \text{eV} = 1.51 \, \text{eV} \]

### Step 4: Calculate the potential \( V \) acquired by the sphere
The potential \( V \) of the sphere can be expressed as:
\[ V = \frac{kQ}{R} \]
where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant)
- \( Q = n \cdot e \) (where \( n \) is the number of emitted electrons and \( e = 1.6 \times 10^{-19} \, \text{C} \))
- \( R = 1 \, \text{cm} = 0.01 \, \text{m} \)

Substituting \( Q \):
\[ V = \frac{(9 \times 10^9)(n \cdot 1.6 \times 10^{-19})}{0.01} \]
\[ V = 1.44 \times 10^{12} n \]

### Step 5: Calculate the potential energy \( U \) of the emitted electrons
The potential energy \( U \) is given by:
\[ U = n \cdot e \cdot V = n \cdot (1.6 \times 10^{-19}) \cdot (1.44 \times 10^{12} n) \]
\[ U = 2.304 \times 10^{-7} n^2 \]

### Step 6: Set the potential energy equal to the maximum kinetic energy
From conservation of energy:
\[ U = K.E. \]
\[ 2.304 \times 10^{-7} n^2 = 1.51 \times 1.6 \times 10^{-19} \]
Calculating the right side:
\[ 1.51 \times 1.6 \times 10^{-19} = 2.416 \times 10^{-19} \]
Setting the equations equal:
\[ 2.304 \times 10^{-7} n^2 = 2.416 \times 10^{-19} \]
Solving for \( n^2 \):
\[ n^2 = \frac{2.416 \times 10^{-19}}{2.304 \times 10^{-7}} \approx 1.048 \times 10^{-12} \]
\[ n \approx 1.024 \times 10^6 \]

### Step 7: Express \( n \) in the required form
The maximum number of photoelectrons emitted is \( n \approx 1.024 \times 10^6 \), which can be expressed as \( A \times 10^e \) where \( A \approx 1.024 \) and \( e = 6 \).

### Conclusion
Thus, the value of \( z \) is \( 6 \).