Question

In the Bohr's model for the hydrogen-like atoms, how does the orbital radius depend upon the principal quantum numbern ? Z Is the atomic number.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

The electron of charge e− orbiting a nucleus carrying a positive charge Ze+ will undergo centripetal acceleration due to the electrostatic force of attraction. If the electron's mass is taken as m, and its speed as v, we can write -
$\frac{m{v}^2}{r}=\frac{Z{e}^2}{4\pi {ϵ}_0{r}^2}$
$\Rightarrow$ $r=\frac{Z{e}^2}{4\pi {ϵ}_{0}m{v}^2}$ ....(1)

Here, for different values of v $ϵ$ $(0,\infty$), we can have different values for r $ϵ$ $(0,\infty$) - but we know from Bohr's postulates that not all values of r are allowed. So there must be some dependence on a discrete variable (a variable that can take only certain values). This discreteness can be obtained by applying Bohr's third postulate - angular momentum can only be integer multiples of $\frac{h}{2\pi }$. From mechanics, we know that the angular momentum of a point mass m moving in a circle of radius r with a speed v is equal to mvr. We can write -
$mvr=n\frac{h}{2\pi }$
$\Rightarrow v=\frac{nh}{2\pi mr}$ ....(2)
Substituting the value of v from equation (2) in equation (1), we obtain the discreteness in the orbital radius that we were looking for –
$r=\frac{Z{e}^2}{4\pi {ϵ}_{0}m{v}^2}=\frac{Z{e}^2}{4\pi {ϵ}_{0}m{\left(\frac{nh}{2\pi mr}\right)}^2}=\frac{Z{e}^2}{4\pi {ϵ}_{0}m}\times \frac{4{\pi }^2{m}^2{r}^2}{h^2{n}^2}.$
Cancelling r from both numerators and rearranging,
$r=\frac{h^2{ϵ}_0}{\pi m{e}^2}\frac{n^2}{Z}.$
The term in the brackets is just a bunch of constants. For Z=1, putting all S.I values of these constants gives us the orbital radius r for n=1 in the hydrogen atom, also called the Bohr radius (denoted by $a_0$), numerically equal to 52.9 pm or 0.529 A ̇.