An n-type semiconductor has impurity level `20 meV` below the conduction band. In a thermal collision, transferble enegry is `KT`. The value of `T` for which electrons start to jump in conduction bond is :

Options

A. 232 K
B. 348 K
C. 400 K
D. 600 K

Explanation

To find the temperature \( T \) at which electrons in an n-type semiconductor start to jump into the conduction band, we can follow these steps:

### Step 1: Understand the relationship between thermal energy and temperature
The thermal energy available to the electrons is given by the equation:
\[ E = kT \]
where:
- \( E \) is the energy (in electron volts) needed for the electrons to jump into the conduction band,
- \( k \) is the Boltzmann constant, approximately \( 8.62 \times 10^{-5} \, \text{eV/K} \),
- \( T \) is the temperature in Kelvin.

### Step 2: Identify the energy level
In this problem, the impurity level is \( 20 \, \text{meV} \) (milli-electron volts) below the conduction band. We convert this to electron volts:
\[ E = 20 \, \text{meV} = 20 \times 10^{-3} \, \text{eV} = 0.020 \, \text{eV} \]

### Step 3: Set up the equation
We want to find the temperature \( T \) at which the thermal energy \( kT \) equals the energy needed for the electrons to jump into the conduction band:
\[ kT = 0.020 \, \text{eV} \]

### Step 4: Solve for \( T \)
Rearranging the equation gives:
\[ T = \frac{E}{k} = \frac{0.020 \, \text{eV}}{8.62 \times 10^{-5} \, \text{eV/K}} \]

### Step 5: Calculate the temperature
Now, we can perform the calculation:
\[ T = \frac{0.020}{8.62 \times 10^{-5}} \approx 232.0 \, \text{K} \]

### Final Answer
Thus, the temperature \( T \) at which electrons start to jump into the conduction band is approximately:
\[ \boxed{232 \, \text{K}} \]