Physics MCQs for NEET — Practice Questions with Answers

(a) ${\mathrm{\lambda }}_{\max }=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.14\times 1.6\times {10}^{-19}}=10888$Å

(d) Electrons require low energy for mobility as they suffer low resistance during their motion as compared to holes.

(a) Because electrons needed less energy to move.

(c) Number density of atoms in silicon specimen = $5\times {10}^{28} \mathrm{atom}/{\mathrm{m}}^3$ = $5\times {10}^{22} \mathrm{atom}/{\mathrm{cm}}^3$. Since one atom of indium is doped in $5\times {10}^7$ Si atom. So number of indium atoms doped per ${\mathrm{cm}}^3$ of silicon.

$\mathrm{n}=\frac{5\times {10}^{22}}{5\times {10}^7}=1\times {10}^{15} \mathrm{atom}/{\mathrm{cm}}^3$

(a) The probability of electrons to be found in the conduction band of an intrinsic semiconductor

$\mathrm{P}\left(\mathrm{E}\right)=\frac{1}{1+{\mathrm{e}}^{{\displaystyle \frac{\left(\mathrm{E}-{\mathrm{E}}_{\mathrm{F}}\right)}{\mathrm{kT}}}}}$   ; where k = Boltzmann's constant
Hence, at a finite temperature, the probability decreases exponentially with increasing band gap.

(b) In half wave rectifier ${\mathrm{V}}_{\mathrm{dc}}=\frac{{\mathrm{V}}_0}{\mathrm{\pi }}=\frac{10}{\mathrm{\pi }}$

(a) As we know current density J=nqv

$$\begin{gathered} \Rightarrow {\mathrm{J}}_{\mathrm{e}}={\mathrm{n}}_{\mathrm{e}}{\mathrm{qv}}_{\mathrm{e}} \mathrm{and} {\mathrm{J}}_{\mathrm{h}}={\mathrm{n}}_{\mathrm{h}}{\mathrm{qv}}_{\mathrm{h}} \\ \Rightarrow \frac{{\mathrm{J}}_{\mathrm{e}}}{{\mathrm{J}}_{\mathrm{h}}}=\frac{{\mathrm{n}}_{\mathrm{e}}}{{\mathrm{n}}_{\mathrm{h}}}\times \frac{{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{v}}_{\mathrm{h}}}\Rightarrow \frac{3/4}{1/4}=\frac{{\mathrm{n}}_{\mathrm{e}}}{{\mathrm{n}}_{\mathrm{h}}}\times \frac{5}{20}\Rightarrow \frac{{\mathrm{n}}_{\mathrm{e}}}{{\mathrm{n}}_{\mathrm{h}}}=\frac{6}{5} \end{gathered}$$

(b) As value of n increases, energy gap decreases due to increasing $Z_{eff.}$ on valence shell.

(b) A derivation for mean free path of gas.

• If v is the velocity of the moving molecule then in one second it will collide with all moleculeswith in a distance σ between the centres.
• In one second it sweeps a volume πσ²v where any other molecule will collide with it.
• If n is the total number of molecules per unit volume, then nπσ²v is number of collisions a molecule suffers in one second.
• If v is the distance traversed by molecule in one second then mean free path is given by
       λ = total distance traversed in one second /no. of collision suffered by the molecules
        =v/πσ²vn
        =1/πσ²n
• This expression was derived with the assumption that all the molecules are at rest except the one which is colliding with the others.

(a) Molar mass of moist air is lesser than dry air.