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Bohr's Model of a Hydrogen Atom

The time peri...

Question

The time period for revolution of electron in a Hydrogen atom state $n_{1}$ is $T_{1}$ and time period for revolution of electron in higher orbit $n_{2}$ is $T_{2}$ . If $\frac{T_{2}}{T_{1}} = 27$ . Which values of $n_{1}$ and $n_{2}$ are not possible?

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

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

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n_{1} = 3, n_{2} = 9

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n_{1} = 4, n_{2} = 12

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Solution

The correct option is B $n_{1} = 3, n_{2} = 6$
We know, period of revolution $T \propto \frac{n^{3}}{Z^{2}}$
where n is the orbit number and Z is the atomic number.
Since, Z = 1 for Hydrogen
So, $\frac{T_{2}}{T_{1}} = {(\frac{n_{2}}{n_{1}})}^{3} = 27$ where $n_{1}$ and $n_{2}$ are the orbit numbers.

$\Rightarrow \frac{n_{2}}{n_{1}} = 3$

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Bohr's Model of a Hydrogen Atom

Standard XII Chemistry

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The time period for revolution of electron in a Hydrogen atom state n1 is T1 and time period for revolution of electron in higher orbit n2 is T2. If T2T1=27. Which values of n1 and n2 are not possible?

The correct option is B n1=3,n2=6We know, period of revolution T∝n3Z2 where n is the orbit number and Z is the atomic number. Since, Z = 1 for Hydrogen So, T2T1=(n2n1)3=27 where n1 and n2 are the orbit numbers. ⇒n2n1=3

The time period for revolution of electron in a Hydrogen atom state $n_{1}$ is $T_{1}$ and time period for revolution of electron in higher orbit $n_{2}$ is $T_{2}$ . If $\frac{T_{2}}{T_{1}} = 27$ . Which values of $n_{1}$ and $n_{2}$ are not possible?