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Q. Which is the correct relationship :

Solution:

$E _{1}( H )=-13.6 \times \frac{1^{2}}{1^{2}}=-13.6 \, eV ;$
$E _{2}\left( He ^{+}\right)=-13.6 \times \frac{2^{2}}{2^{2}}=-13.6\, eV$
$E _{3}\left( Li ^{2+}\right)=-13.6 \times \frac{3^{2}}{3^{2}}=-13.6 \, eV ;$
$E _{4}\left( Be ^{3+}\right)=-13.6 \times \frac{4^{2}}{4^{2}}=-13.6\, eV$
$\therefore E _{1}( H )= E _{2}\left( He ^{+}\right)$
$= E _{3}\left( Li ^{2+}\right)= E _{4}\left( Be ^{3+}\right)$