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BITSAT 2019 Paper

Section: Physics

Question : 25 of 150

Marks: +1, -0

The angular speed of the electron in the nth orbit of Bohr hydrogen atom is

Solution:

Form the Bohr's theory,

Speed of an electron in nth orbit,

v \propto n \ldots (1)

Radius of the orbit,

r \propto n^{2} \ldots (2)

Time period,

T = \;\frac{2\pi r}{v}

2\pi

is a constant.

Hence

{}^T \propto \;\frac{r}{v} \ldots

(3)

From

(1), (2)

and

(3)

T \propto \;\frac{n^{2}}{\frac{1}{n}}

\Rightarrow T \propto n^{3}

Angular speed,

\omega = \;\frac{2\pi}{T}

\Rightarrow \omega \propto \;\frac{1}{n^{3}}

Hence, The angular speed of the electron in nth orbit of Bohr hydrogen is inversely proportional to

n^{3}

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