The ground state energy of the hydrogen atom | vedclass

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Question

(A) For the hydrogen atom,the energy of the $n^{th}$ orbit is given by $E_n = \frac{-13.6}{n^2} \ eV$.For the second excited state,$n = 3$.Therefore,t

Vedclass · Accepted Answer

$+1.51 \ eV, -3.02 \ eV$

Vedclass · Answer

(A) For the hydrogen atom,the energy of the $n^{th}$ orbit is given by $E_n = \frac{-13.6}{n^2} \ eV$.For the second excited state,$n = 3$.Therefore,the total energy $E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.51 \ eV$.In any Bohr orbit,the kinetic energy $(K.E.)$ is equal to the negative of the total energy $(E)$: $K.E. = -E = -(-1.51 \ eV) = +1.51 \ eV$.The potential energy $(P.E.)$ is equal to twice the total energy $(E)$: $P.E. = 2E = 2 	imes (-1.51 \ eV) = -3.02 \ eV$.Thus,the kinetic energy is $+1.51 \ eV$ and the potential energy is $-3.02 \ eV$.

The ground state energy of the hydrogen atom is $-13.6 \ eV$. The kinetic and potential energy of the electron in the second excited state are,respectively:

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