Question

The value of Bohr radius for hydrogen atom is:
(A)-$$0.529\times {10}^{-8}\text{cm}$$
(B)-$$0.529\times {10}^{-10}\text{cm}$$
(C)-$$0.529\times {10}^{-6}\text{cm}$$
(D)-$$0.529\times {10}^{-12}\text{cm}$$

Answer 1

Hint: The radius of the orbit around the nucleus on which the electron moves as described in Bohr’s model is called Bohr’s radius. This model was devised for a single-electron atom such as hydrogen $$(\text{H}),\text{H}{\text{e}}^{+},\text{L}{\text{i}}^{2+}$$.

Complete step by step answer:
Bohr model was devised from the conclusion drawn from the gold foil experiment by Rutherford, wherein he concluded that the negative electrons are far away from the positive charge in the nucleus.
From the classical mechanical approach, he assumed the electrons to be orbiting around the nucleus with radius,$$r$$.
These electrons face centrifugal forces during their circular motion around the nucleus which forces them away from it. It is given by:
$$F_{centrifugal}=-m{v}^2/r$$
where m is mass of the electron and v is its velocity.
For a stable atom in which the electron moves in the orbit (having stable energy state) without emitting radiation. Thus, this centrifugal force is opposed by the coulombic attraction which draws the electron inward, experienced between the electron and nucleus called the centripetal force. It is given by:
$$F_{centripetal}=-Z{e}^2/r^2$$
Equating the two forces, $$F_{centrifugal}=F_{centripetal}$$
Rearranging the equation and solving for$$r$$, we get
$${\displaystyle \frac{-m{v}^2}{r}}={\displaystyle \frac{-Z{e}^2}{r^2}}$$

$$r={\displaystyle \frac{m{v}^2{r}^2}{Z{e}^2}}$$
Multiplying right hand side of the above equation by$$(m/m)$$,we get
$$r={\displaystyle \frac{m{v}^2{r}^2}{Z{e}^2}}\times {\displaystyle \frac{m}{m}}={\displaystyle \frac{{(mvr)}^2}{Z{e}^{2}m}}$$ --- (Equation 1)
Here $$(mvr)$$is the angular momentum of an electron.
For the quantization of the moving electron, Bohr’s second postulate was devised according to which the electron moves in orbit in which its angular momentum is equal to the integral multiple of$$h/2\pi$$.
Therefore, angular momentum of the$$n^{th}$$orbit is $$(mvr)=nh/2\pi$$
where $$h=Planc{k}^{\prime }sconstant=6.6\times {10}^{-34}Js$$
$$n=$$ permitted orbits on which electron revolve called principal quantum number
Now, substituting value of$$(mvr)=nh/2\pi$$,in Equation 1:
$$r={\displaystyle \frac{n^2{h}^2}{4{\pi }^{2}Z{e}^{2}m}}$$
For the hydrogen atom $$Z=1,n=1$$, we get
$$r={\displaystyle \frac{h^2}{4{\pi }^2{e}^{2}m}}={\displaystyle \frac{6.6\times {10}^{-34}\times 6.6\times {10}^{-34}}{4\times 3.14\times 3.14\times 1.6\times {10}^{-19}\times 1.6\times {10}^{-19}\times 9.1\times {10}^{-31}}}$$

$$r=0.59\times {10}^{-10}\text{m}\text{=}0.59\times {10}^{-8}\text{cm}$$

Therefore, the Bohr’s radius of hydrogen is option (A)- $$0.59\times {10}^{-8}\text{cm}$$.

Note: While calculating the units of the terms must be in the standard form and change of units like from metre to centimetre must be done with utmost care keeping the decimal places in mind.