If, in a hydrogen atom, radius of nth Bohr orbit is `r_(n)` frequency of revolution of electron in nth orbit is `f_(n)` and area enclosed by the nth orbit is `A_(n)` , then which of the pollowing graphs are correct?

Options

A.
B.
C.
D.

Answer

A, B, C

Explanation

Since a hydrogen atom `r_(n) prop n^(2)`, therefore, graph between `r_(n)` and `n` will be a parabola through origin and having increasing slop. Therefore, option ( c) is correct . Snce `r_(n) prop n^(2)`, therefore `r_(n)// r_(1) = n^(2)`
Hence. Log `(r_(n)// r_(1)) = 2 loh n`. It means, graph between log `(r_(n)// r_(1))` and log `n` will be a straigth line passing through origin and having positive slop `(tan theta = 2)`. Therefore option (b)is also correct.If radius of an orbit is equal to `r` , then area enclosed by it will be equal to `A = pi r^(2)`.

Since `r_(n) prop n^(2)`, therefore `A_(n) prop n^(4)`
Hence. `(A_(n))/ A_(1) = n^(4)` or loh n.
It means, graph between log `(A_(n)// `A_(1)`)` and log `n` will be a straigth line passing through origin and having positive slop `(tan theta = 4)`. Therefore option ( c)is also correct.
If frequency of ravolution of electrons is `f` , then its angular velocity will be equal to`omega = 2 pi f`. Hence, its angular momentum will be equal to`omega = m r^(2) omega`. But according to Bohr's theory, it is equal to `nh//2 pi`, therefore .
`m r^(2) (2 pi f) = (nh)/(2 pi)` or `f = (nh)/(4 pi^(2) m r^(2))`
Since `r prop n^(2)`, therefore `f prop (1)/(n^(3))`
Hence, `(f_(n))/(f_(1)) = (1)/(n^(3))` or log `((f_(n))/(f_(1))) = 3 log n`
It means, graph between log `(f_(n)// f_(1))` and log `n` will be a straigth line passing through origin and having negative slop `(tan theta = - 3)`. Hence , it will be as shiown in figure. Hence the potion (d) is wrong.