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Supose the potential energy between elec...

Supose the potential energy between electron and proton at a distance r is given by `-(Ke^(2))/(3r^(3))` . Applicatiojn of Bohr's theroy of hydrogen atom in this case shows that

A

energy in the `n^(th)` ornits is proportional to `n^(6)`

B

energy is proportional to `m^(-3)` (m = mass of electron)

C

energy in the nth orbit is proportional to `n^(-2)`

D

energy is proportional to `m^(3)` (mass of electron)

Text Solution

Verified by Experts

The correct Answer is:
A, B
`U = - (ke^(2))/(3r^(3)), F = (-delU)/(delr) = (ke^(2))/(r^(4)) = (mv^(2))/(r)`
`mv^(2)r^(3) = ke^(2), m^(2)v^(2)r^(2) = (n^(2)h^(2))/(4pi^(2))`
`r = (m)/(n^(2))` (constant), Also `V^(2) prop (n^(6))/(m^(4))`
`U = (ke^(2))/(3((m^(3))/(n^(6)))), K.E. = 1/2 mv^(2) prop (n^(6))/(m^(3))`
`U prop (n^(6))/(m^(3)), T.E. = U + K.E. prop (n^(6))/(m^(3))`
then total enegy `prop (n^(6))/(m^(3))`
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