The square of angular speed of the electron in the n th orbit of Bohr-s hydrogen atom is-A- Directly proportional to nB- Inversely proportional to -nC- Inversely proportional to n6D- Inversely proportional to n 2

The correct option is C Inversely proportional to n6 ω=v/r Further, v 1/nand r n^2, hence ω(1/n^3) ...

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Bohr's Postulates

The square of...

Question

The square of angular speed of the electron in the $n^{t h}$ orbit of Bohr’s hydrogen atom is:

Directly proportional to n

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Inversely proportional to

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Inversely proportional to n²

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Inversely proportional to n⁶

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The time taken by the electron in one complete revolution in the $n^{t h}$ Bohr's orbit of the hydrogen atom is:

R

η

v

A B (a q) ⇋ A^{+} (a q) + B^{-} (a q)

A B (a q) + B^{-} (a q) ⇋ A B_{2}^{-} (a q)

K_{1}

K_{2}

A^{+}

A B_{2}^{-}

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