The KE and PE (in eV) of electron present in third Bohr’s orbit of hydrogen atom respectively are

–1.51, –3.02

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1.51, –3.02

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–3.02, 1.51

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1.51, –1.51

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Solution

The correct option is B 1.51, –3.02
Total energy of $3^{r d}$ shell $= \frac{- 13.6}{3^{2}} = - 1.51 e V$
$K E = - T E = 1.51 e V P E = 2 \times T E = - 3.02 e V$

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The KE and PE (in eV) of electron present in third Bohr’s orbit of hydrogen atom respectively are

The correct option is B 1.51, –3.02Total energy of 3rd shell =−13.632=−1.51 eV KE=−TE=1.51eVPE=2×TE=−3.02eV

The correct option is B 1.51, –3.02
Total energy of $3^{r d}$ shell $= \frac{- 13.6}{3^{2}} = - 1.51 e V$
$K E = - T E = 1.51 e V P E = 2 \times T E = - 3.02 e V$