In a semiconductor crystal, the total number of electrons in the outer shell is 4N. At absolute zero, the number of energy states of valence and conduction band are respectively
4N and 4N
For N atoms of Si or Ge, the outer s and p levels give 2N + 6N = 8N available energy states in all.
These split into a valence band of 4N states and a conduction band of 4N states.
At absolute zero the 4N valence electrons completely fill the 4N valence states and the 4N conduction states stay empty, but the number of states in each band is 4N.
Hence, 4N and 4N.
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Important Questions from Modern Physics
- A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ $\Omega$.
- A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
- The average energy released per fission for the nucleus of ${}^{235}_{92}\text{U}$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure ${}^{235}_{92}\text{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is ________.
(Avogadro Number = $6 \times 10^{23} \text{ per mole}$) - The energy released when $\frac{7}{17.13} \text{ kg}$ of ${}^7_3\text{Li}$ is converted into ${}^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32} \text{ eV}$. The value of $\alpha$ is _________. (Nearest integer)
(Mass of ${}^7_3\text{Li} = 7.0183 \text{ u}$, mass of ${}^4_2\text{He} = 4.004 \text{ u}$, mass of proton = $1.008 \text{ u}$ and $1 \text{ u} = 931 \text{ MeV/}c^2$ and Avogadro number $= 6.0 \times 10^{23}$) - A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy $3.2$ Mev. The estimation of the radius of the nucleus of gold element $(Z = 79)$ by him is $\times 10^{-16}$ m
(Take $\frac{1}{4\pi \epsilon_0} = 9\times10^9 \frac{Nm^2}{C^2}$)
