Physics MCQs for NEET — Practice Questions with Answers

The NCERT states, 'In extrinsic semiconductors, because of the abundance of majority current carriers, the minority carriers produced thermally have more chance of meeting majority carriers and thus getting destroyed. Hence, the dopant, by adding a large number of current carriers of one type, which become the majority carriers, indirectly helps to reduce the intrinsic concentration of minority carriers.'

The passage confirms, 'the donor energy level $E_D$ is slightly below the bottom $E_C$ of the conduction band and electrons from this level move into the conduction band with very small supply of energy.' This small energy difference makes it easy for electrons to transition.

The NCERT explicitly states for p-type semiconductors, and generally applies, 'Note that the crystal maintains an overall charge neutrality as the charge of additional charge carriers is just equal and opposite to that of the ionised cores in the lattice.'

The text states, 'At room temperature, most of the donor atoms get ionised but very few (~10^12) atoms of Si get ionised.' This implies a significantly lower number of intrinsic Si atoms ionising compared to donor atoms contributing to free charge carriers.

The NCERT states, 'With very small supply of energy an electron from the valence band can jump to the level $E_A$ and ionise the acceptor negatively. (Alternately, we can also say that with very small supply of energy the hole from level $E_A$ sinks down into the valence band. Electrons rise up and holes fall down when they gain external energy).'

The NCERT explicitly provides the relationship: 'The electron and hole concentration in a semiconductor in thermal equilibrium is given by $n_e n_h = n_i^2$ (14.5).'

The NCERT states, 'The semiconductor’s energy band structure is affected by doping. In the case of extrinsic semiconductors, additional energy states due to donor impurities ($E_D$) and acceptor impurities ($E_A$) also exist.'

The donor impurity atoms, being pentavalent, have an extra electron that is weakly bound and can easily be donated to the conduction band, making them the source of majority carriers (electrons) in n-type semiconductors, as implied by $E_D$ being slightly below $E_C$.

The text states, 'With very small supply of energy an electron from the valence band can jump to the level $E_A$ and ionise the acceptor negatively.' When an electron leaves the valence band to fill an acceptor level, it leaves behind a hole in the valence band. So, the holes in the valence band are due to acceptance of electrons by acceptor impurities.

The NCERT text states that stopping distance depends on initial velocity ($v_0$) and braking capacity (deceleration, $-a$). While reaction time contributes to the total stopping distance (reaction distance + braking distance), the stopping distance specifically refers to the distance traveled after brakes are applied, which is primarily influenced by initial velocity and deceleration. The formula provided, $d_s = -v_0^2 / (2a)$, does not include mass. Although mass affects the deceleration for a given braking force, the question asks about factors directly influencing the stopping distance given a deceleration.