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Standard XII

Physics

Hydrogen Spectra

The electron ...

Question

The electron in a hydrogen atom makes a transition from $n_{1}$ to $n_{2}$ state. The time period of electron in $n_{1}$ is $8$ times in $n_{2}$ . The possible values of $n_{1}$ and $n_{2}$ are

n_{1} = 8, n_{2} = 1

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n_{1} = 4, n_{2} = 2

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n_{2} = 2, n_{2} = 4

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n_{1} = 1, n_{2} = 8

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Solution

The correct option is B $n_{1} = 4, n_{2} = 2$
Since $T \propto n^{3}$ therefore, $T_{1} / T_{2} = n_{1}^{3} / n_{2}^{3}$
Since $T_{1} / T_{2} = 8$
Therefore, $n_{1} / n_{2} = 2$ only set of values which satisfy this is
$n_{1} = 4$
$n_{2} = 2$

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The electron in a hydrogen atom makes a transition from n1 to n2 state. The time period of electron in n1 is 8 times in n2. The possible values of n1 and n2 are

The correct option is B n1=4,n2=2Since T∝n3 therefore, T1/T2=n31/n32Since T1/T2=8Therefore, n1/n2=2 only set of values which satisfy this isn1=4n2=2

The electron in a hydrogen atom makes a transition from $n_{1}$ to $n_{2}$ state. The time period of electron in $n_{1}$ is $8$ times in $n_{2}$ . The possible values of $n_{1}$ and $n_{2}$ are

The correct option is B $n_{1} = 4, n_{2} = 2$
Since $T \propto n^{3}$ therefore, $T_{1} / T_{2} = n_{1}^{3} / n_{2}^{3}$
Since $T_{1} / T_{2} = 8$
Therefore, $n_{1} / n_{2} = 2$ only set of values which satisfy this is
$n_{1} = 4$
$n_{2} = 2$