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Question

The angular momentum of the electron revolving in 2nd orbit is

The correct answer is

\(\dfrac{h}{\pi}\)

Bohr's quantisation condition: \(L = \dfrac{nh}{2\pi}\).

For the second orbit, \(n = 2\).

\(L = \dfrac{2h}{2\pi} = \dfrac{h}{\pi}\)

Hence, \(\dfrac{h}{\pi}\).

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Important Questions from Modern Physics

  1. 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 ______ $\Omega$.
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    (Avogadro Number = $6 \times 10^{23} \text{ per mole}$)
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    (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 $= 6.0 \times 10^{23}$)
  5. 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
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