The ratio of potential energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is

2:-1

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2:1

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1:2

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1:-2

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Solution

The correct option is B 2:1
$P . E = \frac{- K e^{2}}{r_{n}}$

$T . E = \frac{- K e^{2}}{2 r_{n}}$
where $r_{n}$ is the radius of $n^{t h}$ orbit

$\frac{P . E}{T . E} = 2$

P.E:T.E = 2:1

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The ratio of potential energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is

The correct option is B 2:1P.E=−Ke2rn T.E=−Ke22rn where rn is the radius of nth orbit P.ET.E=2 P.E:T.E = 2:1

The correct option is B 2:1
$P . E = \frac{- K e^{2}}{r_{n}}$

$T . E = \frac{- K e^{2}}{2 r_{n}}$
where $r_{n}$ is the radius of $n^{t h}$ orbit

$\frac{P . E}{T . E} = 2$

P.E:T.E = 2:1