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Question: According to Bohr's theory, the time-averaged magnetic field at the center (i.e. nucleus) of a hydro...

According to Bohr's theory, the time-averaged magnetic field at the center (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth{n^{th}} orbit is proportional to (n=n = principal quantum number)
(a)\left( a \right) n3{n^{ - 3}}
(b)\left( b \right) n4{n^{ - 4}}
(c)\left( c \right) n5{n^{ - 5}}
(d)\left( d \right) n2{n^{ - 2}}

Explanation

Solution

Hint By using the formula of the magnetic field that is B=μ0I2rB = \dfrac{{{\mu _0}I}}{{2r}}and also by using the formula of frequency. And the making relation between both the formula and we will get the result in which the magnetic field is being proportional to the orbit.
Formula used:
Magnetic field,
B=μ0I2rB = \dfrac{{{\mu _0}I}}{{2r}}
And also it can be written as
B=μ0In2rnB = \dfrac{{{\mu _0}{I_n}}}{{2{r_n}}}
Here,
BB, is the magnetic field
In{I_n}, current in the nth{n^{th}}orbit
rn{r_n}, the radius of the nth{n^{th}} orbit

Complete Step By Step Solution
So first of all we will know the magnetic field is given by
B=μ0I2rB = \dfrac{{{\mu _0}I}}{{2r}}
And also it can be written as
B=μ0In2rnB = \dfrac{{{\mu _0}{I_n}}}{{2{r_n}}}
And according to Bohr’s theory
rn=n2h24π2mke2{r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mk{e^2}}}
Since here, all the terms are constant except the n2{n^2}
So we can say
rnn2r{}_n \propto {n^2}
Now let’s see the formula of frequency,
f=v2πrf = \dfrac{v}{{2\pi r}}
And further expanding the equation we get
2πke2nh×2πr\Rightarrow \dfrac{{2\pi k{e^2}}}{{nh \times 2\pi r}}
So from here, we can see that all the other terms are constant except the two terms, that is
f1nrf \propto \dfrac{1}{{nr}}
Now on putting the value of rrin the above equation we get
f1n×n21n3\Rightarrow f \propto \dfrac{1}{{n \times {n^2}}} \propto \dfrac{1}{{{n^3}}}
Now as we know,
I=qtI = \dfrac{q}{t}
And also it can be written as,
I=qf\Rightarrow I = qf
Therefore, Inf{I_n} \propto f
And also it will be proportional to 1n3\dfrac{1}{{{n^3}}}
Now by using the magnetic field equation, we can write
Bn3n2B \propto \dfrac{{{n^{ - 3}}}}{{{n^2}}}
And it can be written as
Bn5B \propto {n^{ - 5}}

Therefore, the option CC will be the correct option.

Note In the atoms of materials, electrons are around the nuclei creating tiny currents that create tiny magnetic fields. In most materials the tiny magnetic field vectors all point in different directions and cancel out but there are a few metals where all of these little magnetic field vectors line up to create a magnetic field that is noticeable on the scale of us humans.