Question

The kinetic energy of an electron in the second bohr

Answer 1

Explanation:

The kinetic energy of an electron in the second Bohr orbit of s H atom is [a0 is Bohr radius],

KE = \frac{2\pi e^{4m}}{4h^{2} } \\= \frac{h^{2} e^{4m} }{2h^{2} } \\= \frac{\pi^{2}m }{2h^{2} } e^{4} \\= a_{0} = \frac{h^{2} }{4h^{2}e^{2}m\\}\\So, e_{4}=\frac{ h^{4}}{16\pi ^{4}m^{2} a^{2} } \\Hence, KE = \frac{h^{2} }{32\pi^{2} ma^{2} }

Answer 2

Answer

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