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The electron in a hydrogen atom makes a transition n1n2, where n1 and n2 are the principal quantum numbers of the two energy states. Assume Bohr's model to be valid. The time period of the electron in the initial state is eight times that in the final state. What are the possible values of n1 and n2?

Answer
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Hint:Before calculating the n1 and n2 values, you should know the relation between time period and orbit of the electron. We all know the time period is equal to the revolution taking per frequency. So we can use this relation to obtain the values of n1 and n2.

Formula used:
T=1fn3
Where, T - is the time period,
f -frequency
n -number of orbitals


Complete step by step solution:
The period for the n1 th orbitals electron is written as,
T1=1f1n13
The period for the n2 th orbitals electron is written as,
T2=1f2n23
Hence, divide the two time period, we get
T1T2=n13n23
The time period for n1 orbital is 8 times greater than the n2 orbital. So we can write the relation as,
T1=8T2,
Substitute the relation into the equation means, we can get,
8=(n1n2)3
n1=2n2
The possible values of n1 and n2 are, n1=2;n2=1, n1=4;n2=2, n1=6;n2=3 and so on.

Note: From Bohr's model of the hydrogen atom, easily calculate the time period taken to complete the one rotation of the electron. The time period is calculated from the simple formula, the time period is equal to the total distance covered divided by the velocity of the electron. From this formula also we can easily calculate the time period of Bohr's orbit. The time it takes to complete one rotation to the electron is 1.6×1016 seconds.
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