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Standard XII

Physics

The De Broglie Explanation

The electron ...

Question

The electron in a hydrogen atom makes a transition $n_{1} \to n_{2}$ , where $n_{1}$ and $n_{2}$ are the principal quantum numbers of the two states. Assume the Bohr model as valid in this case. The frequency of the orbital motion of the electron in the initial state is $1 / 27$ of that in the final state. The possible values of $n_{1}$ and $n_{2}$ are

n_{1} = 6, n_{2} = 3

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

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

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n_{1} = 3, n_{2} = 1

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Solution

The correct option is D $n_{1} = 3, n_{2} = 1$
Frequency, $v \propto \frac{1}{n^{3}}$
Hence,
$\frac{v_{1}}{v_{2}} = \frac{n_{2}^{3}}{n_{1}^{3}}$
$\frac{1}{27} = \frac{n_{2}^{3}}{n_{1}^{3}}$
$n_{1} : n_{2} = 3 : 1$
Out of the possible values: $n_{1} = 3, n_{2} = 1$

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The correct option is D n1=3,n2=1Frequency, v∝1n3Hence,v1v2=n32n31127=n32n31n1:n2=3:1Out of the possible values: n1=3,n2=1

The correct option is D $n_{1} = 3, n_{2} = 1$
Frequency, $v \propto \frac{1}{n^{3}}$
Hence,
$\frac{v_{1}}{v_{2}} = \frac{n_{2}^{3}}{n_{1}^{3}}$
$\frac{1}{27} = \frac{n_{2}^{3}}{n_{1}^{3}}$
$n_{1} : n_{2} = 3 : 1$
Out of the possible values: $n_{1} = 3, n_{2} = 1$