The energy of an electron in the first Bohr's | vedclass

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For a h

Vedclass · Accepted Answer

$-3.4$

Vedclass · Answer

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For a hydrogen atom,$Z = 1$,so $E_n = -13.6 / n^2 \ eV$.For the ground state $(n = 1)$,$E_1 = -13.6 \ eV$.For the first excited state $(n = 2)$,$E_2 = -13.6 / 2^2 = -13.6 / 4 = -3.4 \ eV$.For the second excited state $(n = 3)$,$E_3 = -13.6 / 3^2 = -13.6 / 9 \approx -1.51 \ eV$.Comparing these values with the given options,$-3.4 \ eV$ is a possible energy value for the excited state.Therefore,the correct option is $A$.

The energy of an electron in the first Bohr's orbit of $H$ atom is $-13.6 \ eV$. The possible energy value$(s)$ of the excited state$(s)$ for an electron in Bohr's orbits of hydrogen is (are): ............. $eV$

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