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 (V) 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 maximum kinetic energy (KE_max) of the photoelectrons can be expressed in terms of the stopping potential (V) using the equation:
\[ KE_{max} = qV \]
where:
- \( KE_{max} \) is the maximum kinetic energy of the photoelectrons,
- \( q \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) Coulombs),
- \( V \) is the stopping potential in volts.

### Step 2: Substitute the known values
From the problem, we know:
\[ KE_{max} = 3.2 \times 10^{-19} \text{ Joules} \]
Now, substituting the known values into the equation:
\[ 3.2 \times 10^{-19} = (1.6 \times 10^{-19}) \cdot V \]

### Step 3: Solve for the stopping potential (V)
To find V, we can rearrange the equation:
\[ V = \frac{KE_{max}}{q} \]
Substituting the values:
\[ V = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} \]

### Step 4: Calculate the value of V
Now, perform the division:
\[ V = \frac{3.2}{1.6} = 2 \text{ Volts} \]

### Final Answer
The stopping potential is \( V = 2 \text{ Volts} \).

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