Question

39. In a semiconductor, it is found that 4/5th of the total current is carried by electrons and the remaining 1/5th by the holes, and the electrons drift velocity is
2.5 times the drift velocity of holes. What is the ratio of the number densities of electrons and holes?
​

Answer 1

The ratio of the number densities of electrons and holes is 8:5

Explanation:

Let the current is i

Then the current due to electrons

$i_e=\frac{4}{5}i$

Current due to holes

$i_h=\frac{1}{5}i$

Given, drift velocity of electrons is 2.5 times the drift velocity of holes

Thus,

$v_e=2.5v_h$

We know that

For electrons

$i_e=n_ev_ee$

$\implies n_e=\frac{i_e}{v_ee}$

Where e is the charge on electrons and $n_e$ is number density

For electrons

$i_h=n_hv_he$

$\implies n_h=\frac{i_h}{v_he}$

Thus,

$\frac{n_e}{n_h}=\frac{i_e}{v_e}\times \frac{v_h}{i_h}$

$\implies \frac{n_e}{n_h}=\frac{4/5i}{2.5v_h}\times\frac{v_h}{1/5i}$

$\implies \frac{n_e}{n_h}=\frac{4}{2.5}=\frac{40}{25}=\frac{8}{5}$

$\implies n_e:n_h=8:5$

Hope this answer is helpful.

Know More:

Q: In a semiconductor, it is found that three quarter of current is being carried by electron and one quarter by holes. If at this temperature the drift speed of the electron is two and half times that of holes, determine the ratio of electrons to holes present .

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Q: If the ratio of the concentration of electron to that of holes in a semiconductor is 7/5 and the ratio of current is 7/4, then what is the ratio of their drift velocities?

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