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Standard X

Chemistry

Ionization Enthalpy

The electron ...

Question

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

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

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

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

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

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Solution

The correct option is B $n_{1} = 4, n_{2} = 2$
We have the equation $T α$ $n^{3}$ where $T$ is the time period.
Given $T_{n 1} = 8 T_{n 2}$
Hence $n_{1} = 2 n_{2}$
Therefore the possible value of $n_{1}$ and $n_{2}$ are $4$ and $2$ respectively.

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

The correct option is B n1=4,n2=2We have the equation Tαn3 where T is the time period.Given Tn1=8Tn2Hence n1=2n2Therefore the possible value of n1 and n2 are 4 and 2 respectively.

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

The correct option is B $n_{1} = 4, n_{2} = 2$
We have the equation $T α$ $n^{3}$ where $T$ is the time period.
Given $T_{n 1} = 8 T_{n 2}$
Hence $n_{1} = 2 n_{2}$
Therefore the possible value of $n_{1}$ and $n_{2}$ are $4$ and $2$ respectively.