For an electron in the nth Bohr orbit of hydrogen atom, what will be the ratio of radius of orbit to its de-Broglie wavelength

\frac{n}{2 π}

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\frac{n^{2}}{2 π}

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\frac{1}{2 π n}

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\frac{1}{2 π n^{2}}

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Solution

The correct option is A $\frac{n}{2 π}$
We have $m v r = \frac{n h}{2 π},$
$\Rightarrow \frac{m}{\frac{h}{m v}} = \frac{n}{2 π}$

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For an electron in the nth Bohr orbit of hydrogen atom, what will be the ratio of radius of orbit to its de-Broglie wavelength

The correct option is A n2πWe have mvr=nh2π, ⇒mhmv=n2π

The correct option is A $\frac{n}{2 π}$
We have $m v r = \frac{n h}{2 π},$
$\Rightarrow \frac{m}{\frac{h}{m v}} = \frac{n}{2 π}$