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Standard X

Chemistry

Bohr's Model of Atom

An electron i...

Question

An electron in hydrogen atom jumps from $n_{1}$ state to $n_{2}$ state, where $n_{1}$ and $n_{2}$ represent the quantum number of two states. The time period of revolution of electron in initial state is $8$ times that in final state. Then the ratio of $n_{1}$ and $n_{2}$ is.

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Solution

Time period $\propto n^{3}$
$\frac{T_{1}}{T_{2}} = \frac{n_{1}^{3}}{n_{2}^{3}}$
${\frac{n_{1}}{n_{2}}}^{3} = 8$
$\frac{n_{1}}{n_{2}} = 2$

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An electron in hydrogen atom jumps from n1 state to n2 state, where n1 and n2 represent the quantum number of two states. The time period of revolution of electron in initial state is 8 times that in final state. Then the ratio of n1 and n2 is.

Time period ∝n3T1T2=n31n32n1n23=8n1n2=2

An electron in hydrogen atom jumps from $n_{1}$ state to $n_{2}$ state, where $n_{1}$ and $n_{2}$ represent the quantum number of two states. The time period of revolution of electron in initial state is $8$ times that in final state. Then the ratio of $n_{1}$ and $n_{2}$ is.

Time period $\propto n^{3}$
$\frac{T_{1}}{T_{2}} = \frac{n_{1}^{3}}{n_{2}^{3}}$
${\frac{n_{1}}{n_{2}}}^{3} = 8$
$\frac{n_{1}}{n_{2}} = 2$