Question

In a hydrogen atom, the electron makes a transition from \[n = 2\] to \[n = 1\]. The magnetic field produced by the circulating electron at the nucleus
A. Decreases 16 times
B. Increases 4 times
C. Decreases 4 times
D. Increases 32 times

Answer 1

Hint:We use the Bohr’s postulate to find the speed of electrons and the radius of the nth orbit. Then we calculate the equivalent magnetic field at the center of the orbit.
Formula used:
\[B = \dfrac{{{\mu _0}e{v_n}}}{{4\pi {r^2}}}\]
where B is the magnitude of the magnetic field at the center of the circle of radius r due to circulating electrons with speed \[{v_n}\].
\[{v_n} = \sqrt {\dfrac{{{K_e}{e^2}}}{{{m_e}r}}} \]
where \[{v_n}\] is the speed of electrons in nth orbit.

Complete step by step solution:
The magnitude of the magnetic field produced at the center of the circular orbit due to revolution of the electron is given as,
\[B = \dfrac{{{\mu _0}e{v_n}}}{{4\pi {r^2}}}\]
Here, \[{v_n}\] is the speed of the electron in the nth orbit of radius r.

According to Bohr’s postulate for hydrogen atom, the speed of the electron in nth orbit is given as,
\[{v_n} = \sqrt {\dfrac{{{K_e}{e^2}}}{{{m_e}r}}} \\ \]
And the radius of the nth orbit is given as \[{r_n} = {a_0}{n^2}\]
Here, \[{a_0}\] is the radius of the ground state orbit.
Putting the expression for radius in speed expression, we get
\[{v_n} = \dfrac{1}{n}\sqrt {\dfrac{{{K_e}{e^2}}}{{{m_e}{a_0}}}} \]

Putting back in the expression for the magnetic field, we get
\[B = \dfrac{{{\mu _0}e\dfrac{1}{n}\sqrt {\dfrac{{{K_e}{e^2}}}{{{m_e}{a_0}}}} }}{{4\pi {{\left( {{a_0}{n^2}} \right)}^2}}} \\ \]
On simplifying, we get
\[B = \dfrac{{{\mu _0}e\dfrac{1}{n}\sqrt {\dfrac{{{K_e}{e^2}}}{{{m_e}{a_0}}}} }}{{4\pi a_0^2{n^4}}} \\ \]
\[\Rightarrow B = \left( {\dfrac{{{\mu _0}{e^4}}}{{4\pi }}\sqrt {\dfrac{{{K_e}}}{{{m_e}a_0^5}}} } \right)\dfrac{1}{{{n^5}}} \\ \]
\[\Rightarrow B = C\left( {\dfrac{1}{{{n^5}}}} \right)\]
Here, C is a constant equal to
\[\left( {\dfrac{{{\mu _0}{e^4}}}{{4\pi }}\sqrt {\dfrac{{{K_e}}}{{{m_e}a_0^5}}} } \right) \\ \]
It is given that an electron makes a transition from the first excited state to the ground state.
\[{n_1} = 2\]
\[\Rightarrow {n_2} = 1\]
So, the magnetic field at the center of the orbit for respective value of n is,
\[{B_1} = \dfrac{C}{{{2^5}}}\]
\[\Rightarrow {B_1} = \dfrac{C}{{32}}\]

And,
\[{B_2} = \dfrac{C}{{{1^5}}}\]
\[\Rightarrow {B_1} = C\]
Hence, the ratio of the magnetic field is,
\[\dfrac{{{B_1}}}{{{B_2}}} = \dfrac{{\left( {\dfrac{C}{{32}}} \right)}}{C} \\ \]
\[\Rightarrow \dfrac{{{B_1}}}{{{B_2}}} = \dfrac{1}{{32}} \\ \]
\[\therefore {B_2} = 32{B_1}\]
So, the final magnitude of the magnetic field is 32 times the initial magnetic field at the center of the orbit. Hence the magnitude of the magnetic field increases by 32 times.

Therefore, the correct option is D.

Note: We assume the magnetic field is uniform inside the circular orbit due to circulating electrons. But, the magnitude of the magnetic field is inversely proportional to the distance from the point of motion of the electron. So, as we move to the center the magnitude of the magnetic field decreases.