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Characteristics of Bohr's Atomic Model

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 to be valid. The time period of the electron in the initial state is 8 times that in the final state. The possible values of $n_{1}$ and $n_{2}$ are

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

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

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

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

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Solution

The correct option is D $n_{1} = 6, n_{2} = 3$
We know that, time period, $T = \frac{2 π r}{v} = \frac{2 π r}{\frac{n h}{(2 π m r)}} = (\frac{4 π^{2} m}{n h}) r^{2}$
Now, $r = (\frac{h^{2} ϵ_{0}}{π m e^{2}}) n^{2}$
$∴ T = (\frac{4 h^{3} ϵ_{0}^{2}}{m e^{4}}) \times n^{3}$
For the two orbits,
$\frac{T_{1}}{T_{2}} = {(\frac{n_{1}}{n_{2}})}^{3} \Rightarrow \frac{8 T_{2}}{T_{2}} = {(\frac{n_{1}}{n_{2}})}^{3} \Rightarrow \frac{n_{1}}{n_{2}} = 2$
Hence, the correct choice is (d).

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The electron in a hydrogen atom makes a transition n1→n2 where n1 and n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is 8 times that in the final state. The possible values of n1 and n2 are

The correct option is D n1=6, n2=3We know that, time period, T=2πrv=2πrnh(2πmr)=(4π2mnh)r2 Now, r=(h2ϵ0πme2)n2 ∴T=(4h3ϵ20me4)×n3 For the two orbits, T1T2=(n1n2)3⇒8T2T2=(n1n2)3⇒n1n2=2 Hence, the correct choice is (d).