Given that the mobility of electrons in Ge is 0.4 metre square/volt sec and electronic charge is 1.6 X10-19 C. How many donor atom (per m3) have in semiconductor of conductivity 500 mho/m?

To determine the number of donor atoms per cubic meter in a semiconductor with a given conductivity, we can use the relationship between conductivity, charge carrier concentration, and mobility. The formula we will use is:

Conductivity Formula

The conductivity (\( \sigma \)) of a semiconductor can be expressed as:

\( \sigma = n \cdot q \cdot \mu \)

Where:

• \( \sigma \) = conductivity (in mho/m)
• n = charge carrier concentration (in m^-3)
• q = charge of an electron (in coulombs)
• \( \mu \) = mobility of the charge carriers (in m²/V·s)

Given Values

From your question, we have:

• Conductivity (\( \sigma \)) = 500 mho/m
• Mobility (\( \mu \)) = 0.4 m²/V·s
• Charge of an electron (\( q \)) = \( 1.6 \times 10^{-19} \) C

Calculating Charge Carrier Concentration

We can rearrange the conductivity formula to solve for \( n \):

\( n = \frac{\sigma}{q \cdot \mu} \)

Now, substituting the known values into the equation:

\( n = \frac{500}{(1.6 \times 10^{-19}) \cdot (0.4)} \)

Performing the Calculation

First, calculate the denominator:

\( (1.6 \times 10^{-19}) \cdot (0.4) = 6.4 \times 10^{-20} \) C·m²/V·s

Now, substitute this back into the equation for \( n \):

\( n = \frac{500}{6.4 \times 10^{-20}} \)

Calculating this gives:

\( n \approx 7.8125 \times 10^{21} \) m^-3

Conclusion

Thus, the number of donor atoms per cubic meter in the semiconductor is approximately \( 7.81 \times 10^{21} \) m^-3. This high concentration of donor atoms indicates a heavily doped n-type semiconductor, which is typical for materials like germanium when used in electronic applications.