nformation about the current and the number densities of electrons and holes.

Let's denote the following:

According to the problem, the ratio of the number density of electrons to holes is:

$$\frac{n_e}{n_h} = 2:1$$

So we can write:

$$n_e = 2n_h$$

The total current density $$J$$ can be expressed as the sum of the current densities due to electrons and holes:

$$J = J_e + J_h$$

We also know the relationship for the current density for electrons and holes:

$$J_e = n_e e v_e$$

$$J_h = n_h e v_h$$

Given that $$\frac{1}{7}$$th of the total current is due to holes, we write:

$$J_h = \frac{1}{7} J$$

Therefore, the remaining current due to electrons is:

$$J_e = J - J_h = J - \frac{1}{7} J = \frac{6}{7} J$$

Now substituting the expressions for $$J_e$$ and $$J_h$$, we get:

$$n_e e v_e = \frac{6}{7} J$$

$$n_h e v_h = \frac{1}{7} J$$

Dividing these two equations, we have:

$$\frac{n_h e v_h}{n_e e v_e} = \frac{1/7 J}{6/7 J}$$

$$\frac{n_h v_h}{n_e v_e} = \frac{1}{6}$$

Substitute the ratio of the number densities $$\frac{n_h}{n_e} = \frac{1}{2}$$:

$$\frac{1}{2} \cdot \frac{v_h}{v_e} = \frac{1}{6}$$

$$\frac{v_h}{v_e} = \frac{1}{6} \cdot 2 = \frac{1}{3}$$

Therefore, the ratio of the drift velocity of holes to the drift velocity of electrons is:

$$\frac{v_h}{v_e} = \frac{1}{3}$$

So, the correct answer is Option D: $$\frac{1}{3}$$.

Semiconductor Devices And Logic Gates

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