Question

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

Answer 1

Hint: The Bohr radius is based on the Bohr model of the atom.

Complete step by step answer:
Bohr’s radius is named after the Danish scientist Niels Bohr. Bohr found that atoms consist of small, dense nuclei with positive electric charge, around which negatively charged electrons orbit in circular paths.
Now, we will find the value of Bohr’s radius for a hydrogen atom.
According to Bohr’s theory we can write that,
     \[mvr = \dfrac{{nh}}{{2\pi }}\]
Now, we can write the above given equation as
     \[v = \dfrac{{nh}}{{2\pi mr}}\]
So, we can write that
${v^2} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}{r^2}}}$ ……..(1)

We can say that in the case of atoms, centripetal force is equal to the electrostatic force. So, we can write that in the form of equation as
\[\dfrac{{m{v^2}}}{r} = \dfrac{{KZ{e^2}}}{{{r^2}}}\]
From this, we can say that
${v^2} = \dfrac{{KZ{e^2}}}{{rm}}$ ………….(2)

Now, we will compare equation (1) and (2). So, we will get
\[\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}{r^2}}} = \dfrac{{KZ{e^2}}}{{mr}}\]
We can write that
\[r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}KZ{e^2}m}}\]

Now, we know that r in the above equation is the Bohr radius which is to be found.
K and Z are equal to 1 in case of hydrogen atom.
e is the charge on an electron and it is $4.8 \times {10^{10}}esu$.
m is the mass of the electron and it is equal to $9.1 \times {10^{ - 28}}g$.
n is equal to 1 as hydrogen’s valence orbital has a primary quantum number of 1.
h is the Planck’s constant and its value is $6.62 \times {10^{ - 27}}erg \cdot s$
Putting all these available values into the equation of Bohr’s radius, we get
\[r = \dfrac{{{{(1)}^2}(6.62 \times {{10}^{ - 27}})}}{{(4){{(3.14)}^2}(1)(1){{(4.8 \times {{10}^{10}})}^2}(9.1 \times {{10}^{ - 28}})}}\]
\[r = 0.529 \times {10^{ - 8}}cm = 0.529\mathop A\limits^ \circ \]
So, the correct answer is “Option A”.

Note: The Bohr model of the atom was replaced by the Quantum Mechanics model based upon the Schrodinger equation in the 1920’s. However, the Bohr radius is still a useful constant because it represents the smallest mean radius normally attainable by a neutral atom.