1.

An electron in a H_2 atom makes a transition from n_(1) to n_(2). The time period of electron in the initial state is eight times that in the final state. Then ratio of n_(1)" to "n_(2):

Answer»

`1:2`
`2:1`
`4:1`
`8:1`

Solution :`T^(2) ALPHA R^(3), (T_(1)^(2))/(T_(2)^(2))=(r_(1)^(3))/(r_(2)^(3)), r_(1)/r_(2)=8^(2/3)=4`
Now `r alpha N^(2)`
`r_(1)/r_(2)=(n_(1))/(n_(2))^(2)`
`n_(1)/n_(2)=SQRT((r_(1))/r_(2))=sqrt(4/1)=2/1`


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