JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 2)
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m$$-$$3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :
$$2\,\Omega $$m
4$$\,\Omega $$m
0.4 $$\,\Omega $$m
0.2 $$\,\Omega $$m
Explanation
For semiconductor,
Conductivity, $$\sigma $$ = ne q $$\mu $$e + nh q $$\mu $$h
given that semiconductor is n-type. So contribution of holes is ignored.
$$ \therefore $$ nh q $$\mu $$h = 0
$$ \therefore $$ $$\sigma $$ = ne q $$\mu $$e
Resistivity, $$\rho $$ = $${1 \over \sigma }$$
= $${1 \over {{n_e}q{\mu _e}}}$$
= $${1 \over {{{10}^{19}} \times 1.6 \times {{10}^{ - 19}} \times 1.6}}$$
= $${1 \over {1.6 \times 1.6}}$$
= 0.4 $$\Omega $$ m
Conductivity, $$\sigma $$ = ne q $$\mu $$e + nh q $$\mu $$h
given that semiconductor is n-type. So contribution of holes is ignored.
$$ \therefore $$ nh q $$\mu $$h = 0
$$ \therefore $$ $$\sigma $$ = ne q $$\mu $$e
Resistivity, $$\rho $$ = $${1 \over \sigma }$$
= $${1 \over {{n_e}q{\mu _e}}}$$
= $${1 \over {{{10}^{19}} \times 1.6 \times {{10}^{ - 19}} \times 1.6}}$$
= $${1 \over {1.6 \times 1.6}}$$
= 0.4 $$\Omega $$ m
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