MCQ
The ratio of the energy of the electrons in ground state of hydrogen to the electrons in first excited state of $Be^{3+}$ is :
- ✓$1:4$
- B$1:8$
- C$1:16$
- D$16:1$
$E=E_{0} \frac{z^{2}}{n^{2}}$
hydrogen,
$E=E_{0} \frac{z^{2}}{n^{2}}=E_{0} \frac{1^{2}}{1^{2}}$
$E=E_{0}$
$B e^{3+}$
$E=E_{0} \frac{z^{2}}{n^{2}}=E_{0} \frac{4^{2}}{2^{2}}$
$E=4 E_{0}$
Ratio $=\frac{E_{0}}{4 E_{0}}$
1: 4
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