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Hydrogen atom is in its n^(th) energy st...

Hydrogen atom is in its `n^(th)` energy state. If de-Broglie wavelength of the electrons is `lamda`, then

A

`lamda prop (1)/(n^(2))`

B

`lamda prop (1)/(n)`

C

`lamda prop n^(2)`

D

`lamda = n`

Text Solution

Verified by Experts

The correct Answer is:
D
Angular momentum of electron in `n^(th)` orbit of hydrogen is
`L= mvr =(nh)/(2pi)`
de-Brogile wavelength, `lamda =(h)/(mv) rArr (mv)/(h) =(1)/(lamda)`
`rArr" "(r )/(lamda) = (n)/(2pi) rArr lamda =(2pi r)/(n)`
Now, `r prop n^(2)` (Bohr radius, `r=0.529 xx n^(2) Å)`
`rArr" "lamda prop n`
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