First slide

Bohr Model of the Hydrogen atom.

Question

An electron in hydrogen atom makes a transition $n_{1} \to n_{2}$ where $n_{1} a n d n_{2}$ are principal quantum numbers of the two states. Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of $n_{1} a n d n_{2}$ are

Easy

A

$n_{1} = 6 and n_{2} = 2$
B

$n_{1} = 8 and n_{2} = 1$
C

$n_{1} = 8 and n_{2} = 2$
D

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

Solution

The time period T of an electron in a Bohr orbit of principal quantum number n is
$T = \frac{4 ε_{0}^{2} h^{3} n^{3}}{m e^{4}} T \propto n^{3} ∴ \frac{T_{1}}{T_{2}} = \frac{n_{1}^{3}}{n_{2}^{3}} As T_{1} = 8 T_{2}, the above relation gives {(\frac{n_{1}}{n_{2}})}^{3} = 8 or n_{1} = 2 n_{2} Thus the possible values of n_{1} and n_{2} are n_{1} = 2, n_{2} = 1; n_{1} = 4, n_{2} = 2; n_{1} = 6, n_{2} = 3, and so on.$

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