The quantity we are looking for is simply the density of free electrons and we can write it as the product of the number of cadmium atoms in the same volume and the factor two. The former is expressed via Avogadro’s number, molar mass, and density so we have that

$$
rho_{free}=frac{2times N_A rho}{M}
$$

Finally, we have that the number of the free electrons is obtained after plugging the values as

$$
rho_{free}=frac{2times 6.023times 10^{23}times 8.65}{112}=boxed{9.26times 10^{22}frac{textrm{e}^-}{textrm{cm}^{3}}}
$$

So the answer is B.

$$
textrm{ B) }rho_{free}=9.26times 10^{22}frac{textrm{e}^-}{textrm{cm}^{3}}
$$

In this problem, we are given a transistor with known base and collector currents. We want to know the current gain which is defined as

$$
G=frac{I_C}{I_B}
$$

In this problem, we are given a transistor with known base and collector currents. We want to know the current gain which is defined as

$$
G=frac{I_C}{I_B}
$$

and see how much the collector current changes when the base current increases by 5 $mu$A.

The collector current increases then as

$$
I_C=Gtimes I_B=190times 5times 10^{6}=950times 10^{-6}textrm{ A}boxed{approx1times 10^3}
$$

so the correct answer is B.

In this problem, we are given a transistor with known gain and collector current. We want to know the base current which is defined as

$$
I_B=frac{I_C}{G}
$$

B) $19times 10^{-6}$ A

The correct statement would be that added electrons make $n$-type semiconductor and $p$-type semiconductor is made from silicon when it is doped with Arsenic. So the correct answer is B.

The conductivity of silicon doubles every 8 degrees so the initial and the final currents are related as

$$
I_f=2^{n}I_i
$$

where $n=frac{Delta T}{8}$.

$$
textrm{B) }I_f=33times 10^{-6}textrm{ A}
$$

The equation that describes this process is Ohm’s law applied on the diode circuit

$$
V_b=IR+V_d
$$

$$
V_b=7.5 textrm{ V}
$$