Potential energy is $- 27.2 e V$ in second orbit of $H e +$ then calculate, double of total energy in first excited state of hydrogen atom:-

- 13.6 e V

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- 54.4 e V

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- 6.8 e V

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- 27.2 e V

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Solution

The correct option is C $- 6.8 e V$
Energy(Helium) $= - 13.6 \times \frac{4}{n^{2}}$
$P E = - 2 K E .$
$T E = P E \times 2 = - \frac{27.22}{2 \times 2} = - 6.8 e V$
Hence,
option $C$ is correct answer.

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Potential energy is −27.2eV in second orbit of He+ then calculate, double of total energy in first excited state of hydrogen atom:-

The correct option is C −6.8eVEnergy(Helium) =−13.6×4n2PE=−2KE.TE=PE×2=−27.222×2=−6.8eVHence,option C is correct answer.

Potential energy is $- 27.2 e V$ in second orbit of $H e +$ then calculate, double of total energy in first excited state of hydrogen atom:-

The correct option is C $- 6.8 e V$
Energy(Helium) $= - 13.6 \times \frac{4}{n^{2}}$
$P E = - 2 K E .$
$T E = P E \times 2 = - \frac{27.22}{2 \times 2} = - 6.8 e V$
Hence,
option $C$ is correct answer.