The energy of an electron in the first Bohr o | vedclass

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(B) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For the $H$ atom,the first

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$54$

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(B) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For the $H$ atom,the first Bohr orbit $(n=1, Z=1)$ has energy $E_1 = -13.6 \ eV$.For the $Be^{3+}$ ion $(Z=4)$,the first excited state corresponds to $n=2$.Substituting these values into the formula: $E = -13.6 	imes \frac{4^2}{2^2} \ eV$.$E = -13.6 	imes \frac{16}{4} \ eV = -13.6 	imes 4 \ eV = -54.4 \ eV$.The magnitude of the energy is $|E| = 54.4 \ eV$.The nearest integer value is $54$.

The energy of an electron in the first Bohr orbit of the $H$ atom is $-13.6 \ eV$. The magnitude of the energy value of an electron in the first excited state of $Be^{3+}$ is $.......... \ eV$ $(nearest \ integer \ value)$.

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