The ratio of energy of the electron in ground state of hydrogen to the electron in first excited state of Be⁺³ is :

4 : 1

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1 : 4

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1 : 8

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8 : 1

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Solution

The correct option is B
1 : 4

$E_{n}$ = $- 13.6$ $\times$ $\frac{Z^{2}}{n^{2}}$

$\frac{E_{n (H)}}{E_{n (B e^{3 +}}}$ = $\frac{1}{\frac{4^{2}}{4}}$ = $\frac{1}{4}$ = $1 : 4$

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The ratio of energy of the electron in ground state of hydrogen to the electron in first excited state of Be+3 is :

The correct option is B 1 : 4 En= −13.6 × Z2n2 En(H)En(Be3+ = 1424 = 14 = 1:4

The ratio of energy of the electron in ground state of hydrogen to the electron in first excited state of Be⁺³ is :

The correct option is B
1 : 4

$E_{n}$ = $- 13.6$ $\times$ $\frac{Z^{2}}{n^{2}}$

$\frac{E_{n (H)}}{E_{n (B e^{3 +}}}$ = $\frac{1}{\frac{4^{2}}{4}}$ = $\frac{1}{4}$ = $1 : 4$