The kinetic energies of the photoelectrons ejected from a metal surface by light of wavelength `2000 Å` range from zero to `3.2 xx 10^(-19)` Joule. Then find the stopping potential (in V).

Answer

To find the stopping potential for the photoelectrons ejected from a metal surface by light of wavelength 2000 Å, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and stopping potential
The stopping potential (V) is related to the maximum kinetic energy (KE) of the photoelectrons. The relationship is given by:
\[ KE = q \cdot V \]
where \( q \) is the charge of the electron.

### Step 2: Identify the maximum kinetic energy
From the problem, we know that the maximum kinetic energy of the photoelectrons is given as:
\[ KE_{max} = 3.2 \times 10^{-19} \text{ Joules} \]

### Step 3: Determine the charge of an electron
The charge of an electron is:
\[ q = 1.6 \times 10^{-19} \text{ Coulombs} \]

### Step 4: Rearrange the equation to find the stopping potential
We can rearrange the equation to solve for the stopping potential (V):
\[ V = \frac{KE_{max}}{q} \]

### Step 5: Substitute the values into the equation
Substituting the values we have:
\[ V = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} \]

### Step 6: Calculate the stopping potential
Now, performing the division:
\[ V = \frac{3.2}{1.6} = 2 \text{ Volts} \]

### Final Answer
The stopping potential is:
\[ V = 2 \text{ Volts} \]

---