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(C) According to Bohr's theory,the time period $T$ of an electron in an orbit with principal quantum number $n$ is given by $T \propto n^3$.Let $T_1$

Vedclass · Accepted Answer

$n_1 = 2n_2$

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(C) According to Bohr's theory,the time period $T$ of an electron in an orbit with principal quantum number $n$ is given by $T \propto n^3$.Let $T_1$ be the time period in the initial state $n_1$ and $T_2$ be the time period in the final state $n_2$.Thus,$T_1 = k n_1^3$ and $T_2 = k n_2^3$,where $k$ is a constant.The ratio of the time periods is given by $\frac{T_1}{T_2} = \frac{n_1^3}{n_2^3}$.Given that the initial time period is eight times the final time period,we have $T_1 = 8T_2$,which implies $\frac{T_1}{T_2} = 8$.Substituting this into the ratio equation:$8 = \frac{n_1^3}{n_2^3}$Taking the cube root on both sides:$2 = \frac{n_1}{n_2}$Therefore,$n_1 = 2n_2$.

An electron in a hydrogen atom makes a transition from $n_1$ to $n_2$. If the time period of the electron in the initial state is eight times that in the final state,then:

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