The ratio of areas between the electron orbits for the first excited state to the ground state for the hydrogen atom is

2 : 1

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4 : 1

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8 : 1

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16 : 1

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Solution

The correct option is D $16 : 1$
$r \propto n^{2} \Rightarrow π r^{2} \propto n^{4}$
So area $\propto n^{4} \Rightarrow$ Ratio is $2^{4} : 1 \Rightarrow 16 : 1$ .

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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 :

1.6 \times 10^{- 16} s .

s^{- 1}

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