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For a doubly ionized Li-atom

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Angular momentum of an electron in 3rd orbit is $\frac{3 h}{2 π}$

Energy of electron in 2nd excited state is -13.6 eV

Speed of electron in 3rd orbit is $\frac{c}{137}$ , where c is speed of light

Kinetic energy of electron is 2nd excited state is half of the magnitude of the potential energy

answer is A, B, C, D.

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Detailed Solution

(1) $L = \frac{nh}{2 π}$ (2) $E = \frac{- 13 .6 Z^{2}}{n^{2}}$

(3) $v = \frac{Zc}{137 n}$ (4) $K = - \frac{U}{2}$

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