According to Bohr's Model of hydrogen atom

Options

A. The de Broglie wavelength of electron in `n^(th)` energy state is proportional to n
B. The radius of `n^(th)` orbit of the atom is proportional to `n^2`
C. The magnitude of magnetic moment of electron in `n^(th)` orbit is proportional to n
D. The magnetic field at the nucleus of the atom due to motion of electron in `n^(th)` orbit is directly proportional to n.

Answer

A, B

Explanation

de Broglie wavelength of electron in `n^(th)` state is given by
`lambda=h/p=h/(mv)` and `v=e^2/(2v_0hn)`
`rArr lambda=((2h^2epsilon_0)/(me^2))n`
`rArr lambda prop n`
Thus statement a is correct.
The radius of `n^(th)` orbit of atom can be calculated as : Centripetal acceleration of electron = Coulomb.s force
`rArr (mv^2)/r=(Ze^2)/(4piepsilon_0r^2)`
Using `mvr=(nh)/(2pi)`, we get
`m/rxx(n^2h^2)/(4pi^2m^2r^2)=(Ze^2)/(4piepsilon_0r_0)`
`rArr r=(epsilon_0n^2h^2)/(pimZe^2)`
`therefore r prop n^2`
Thus statement (b) is also correct.
For magnitude of magnetic moment of electron in `n^(th)` orbit.
Consider electron revolving in `n^(th)` orbit of radius r and angular frequency `omega`. Thus ,
Magnetic moment, `M=(exxpir^2)/"time"`
`rArr M=(e pir^2)/((2pi)/omega)=(eomegar^2)/2`
Also angular momentum , `L=mr^2omega`
`rArr M/L =e/(2m)`
`rArr M=(eL)/(2m)`
Using `L=(nh)/(2pi)`, we get
`M=(nhe)/(4pim)`
`therefore M prop n`
Thus option (c) is incorrect
For magnetic field at nucleus of atom (calculated previously )
`B prop 1/n^5`
`therefore` option (d) is also incorrect
Hence , the correct options are (a) and (b)