Question

The ratio of the energy of the electron in the ground state of hydrogen to the electron in first excited state of $B{e}^{3+}$ is

• A. $1:4$
• B. $1:8$
• C. $1:16$
• D. $16:1$

Answer 1

The correct option is A $1:4$

Formula,

$E=E_0\frac{z^2}{n^2}$

hydrogen,

$E=E_0\frac{z^2}{n^2}=E_0\frac{1^2}{1^2}$

$E=E_0$

$B{e}^{3+},$

$E=E_0\frac{z^2}{n^2}=E_0\frac{4^2}{2^2}$

$E=4E_0$

$Ratio=\frac{E_0}{4E_0}$

$1:4$

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