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The ground state energy in J, of hydrogen atom is – x. The minimum energy in J, required to promote on electron from n = 1 to n = 2 in ${He}^{+}$ is

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$\frac{4 x}{3}$

$3 x$

$\frac{3 x}{4}$

$\frac{x}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} E_{1} = \frac{- {Xz}^{2}}{n^{2}} = \frac{- X 1^{2}}{1} \\ E_{2} = \frac{- X 4}{4} = - X \\ ΔE = E_{2} - E_{1} = - X + 4 X = 3 X \end{array}$

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