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Standard XII

Physics

Moseley's Law

The electron ...

Question

The electron in hydrogen atom makes a transition from $n = n_{1} to n = n_{2}$ state. The time period of the electron in the initial state is eight times that in the final state. The possible values are

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

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

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

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

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Solution

The correct option is A
$n_{1} = 4, n_{2} = 2$

$T = \frac{2 π r}{v} and r \propto n^{2}, v \propto 1 / n \Rightarrow T \propto n^{3} \frac{T_{1}}{T_{2}} = {(\frac{n_{1}}{n_{2}})}^{3} 8 = {(\frac{n_{1}}{n_{2}})}^{3} o r \frac{n_{1}}{n_{2}} = 2$

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The electron in hydrogen atom makes a transition from n=n1 to n=n2 state. The time period of the electron in the initial state is eight times that in the final state. The possible values are

The correct option is A n1=4,n2=2 T=2πrv and r∝n2,v∝1/n⇒T∝n3T1T2=(n1n2)38=(n1n2)3 or n1n2=2

The electron in hydrogen atom makes a transition from $n = n_{1} to n = n_{2}$ state. The time period of the electron in the initial state is eight times that in the final state. The possible values are

The correct option is A
$n_{1} = 4, n_{2} = 2$

$T = \frac{2 π r}{v} and r \propto n^{2}, v \propto 1 / n \Rightarrow T \propto n^{3} \frac{T_{1}}{T_{2}} = {(\frac{n_{1}}{n_{2}})}^{3} 8 = {(\frac{n_{1}}{n_{2}})}^{3} o r \frac{n_{1}}{n_{2}} = 2$