In a Germanium crystal containing N atoms, the total number of outer electrons in the crystal is
4 N
Germanium (Z = 32) has configuration [Ar] \(3d^{10}4s^{2}4p^{2}\), so each atom has 4 outer (valence) electrons.
Total outer electrons = (number of atoms) \(\times\) (valence electrons per atom)
\(= N \times 4 = 4N\)
Hence, 4 N.
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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 $\frac{1}{4\pi \epsilon_0} = 9\times10^9 \frac{Nm^2}{C^2}$)
