The statement "At absolute zero temperature, it behaves like a conductor" is wrong for a semiconductor.

Explanation:

• At absolute zero (0 K), intrinsic semiconductors (like silicon and germanium) do not have free charge carriers (electrons and holes) because all the electrons occupy the valence band, and there is no thermal energy available to promote them to the conduction band.
• Therefore, intrinsic semiconductors behave like insulators at absolute zero, not conductors.

When a conducting wire of Copper and Germanium is cooled from room temperature to 80K:

• Copper (metal): Its resistance will decrease because metals have lower resistance at lower temperatures due to reduced electron scattering.
• Germanium (semiconductor): Its resistance will increase because the number of charge carriers (electrons and holes) decreases at lower temperatures, leading to higher resistance.

Thus, Copper's resistance decreases, while Germanium's resistance increases when cooled to 80K

Both N-type and P-type germanium are electrically neutral because, while they have an imbalance of charge carriers (electrons or holes), the overall number of positive and negative charges remains equal.

- **N-type**: Donor atoms add extra electrons (negatively charged carriers), but the atoms themselves become positively charged ions. This ensures charge neutrality.

- **P-type**: Acceptor atoms create holes (positively charged carriers) by accepting electrons, but the atoms become negatively charged ions. This balances the charge.

In both cases, the total positive and negative charges cancel each other out, keeping the material electrically neutral.

Resistance of conductor increases with increasing temperature and resistance of semiconductor decreases with increasing temperature.

The resistivity (ρ) of a pure (intrinsic) semiconductor is given by the formula:

$$\rho = \frac{1}{q (n_i) (\mu_n + \mu_p)}$$

ρ=q(ni​)(μn​+μp​)1​

Where:

• 
  $\rho = 0.5 \, \Omega\text{m}$ 

  $\rho =0.5\Omega \text{m}$ (resistivity),

• 
  $q = 1.6 \times 10^{-19} \, \text{C}$ 

  q=1.6×10−19C (charge of an electron),

• 
  $n_i$ 

  ni​ = intrinsic carrier concentration (to be calculated),

• 
  $\mu_n = 0.39 \, \text{m}^2/\text{V-s}$ 

  μn​=0.39m2/V-s (electron mobility),

• 
  $\mu_p = 0.19 \, \text{m}^2/\text{V-s}$ 

  μp​=0.19m2/V-s (hole mobility).

Rearranging the formula to solve for

$n_i$

ni​:

$$n_i = \frac{1}{q \rho (\mu_n + \mu_p)}$$

​

Substitute the values:

$$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.39 + 0.19)}$$

​

$$n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.58)}$$

$$n_i=\frac{1}{4.64\times 10^{-20}}=2.16\times 10^{19}{\text{m}}^{-3}$$

Thus, the intrinsic carrier concentration

$n_i$

ni​ is

$2.16 \times 10^{19} \, \text{m}^{-3}$

Majority charge carrier in N-type semiconductor is electrons and in P-type semiconductor is holes

In N type semiconductor number of electrons is more than number of holes

An extrinsic semiconductor with a large number of current carriers is still electrically neutral overall because the total number of positive and negative charges remains balanced.

A p-type semiconductor is created by doping a tetravalent semiconductor (like Ge) with a trivalent impurity such as Boron.

The mobility of electrons (\(mu_e\)) is greater than the mobility of holes (\(mu_h\)). Since current is proportional to mobility for the same carrier concentration and electric field, the current in the \(n\)-type semiconductor is greater than that in the \(p\)-type.