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 ${n^{th}}$ orbit is proportional to ($n = $principal quantum number)
$\left( a \right)$ ${n^{ - 3}}$
$\left( b \right)$ ${n^{ - 4}}$
$\left( c \right)$ ${n^{ - 5}}$
$\left( d \right)$ ${n^{ - 2}}$
Answer
298.5k+ views
Hint By using the formula of the magnetic field that is $B = \dfrac{{{\mu _0}I}}{{2r}}$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,
$B = \dfrac{{{\mu _0}I}}{{2r}}$
And also it can be written as
$B = \dfrac{{{\mu _0}{I_n}}}{{2{r_n}}}$
Here,
$B$, is the magnetic field
${I_n}$, current in the ${n^{th}}$orbit
${r_n}$, the radius of the ${n^{th}}$ orbit
Complete Step By Step Solution
So first of all we will know the magnetic field is given by
$B = \dfrac{{{\mu _0}I}}{{2r}}$
And also it can be written as
$B = \dfrac{{{\mu _0}{I_n}}}{{2{r_n}}}$
And according to Bohr’s theory
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mk{e^2}}}$
Since here, all the terms are constant except the ${n^2}$
So we can say
$r{}_n \propto {n^2}$
Now let’s see the formula of frequency,
\[f = \dfrac{v}{{2\pi r}}\]
And further expanding the equation we get
\[ \Rightarrow \dfrac{{2\pi k{e^2}}}{{nh \times 2\pi r}}\]
So from here, we can see that all the other terms are constant except the two terms, that is
$f \propto \dfrac{1}{{nr}}$
Now on putting the value of $r$in the above equation we get
$ \Rightarrow f \propto \dfrac{1}{{n \times {n^2}}} \propto \dfrac{1}{{{n^3}}}$
Now as we know,
$I = \dfrac{q}{t}$
And also it can be written as,
$ \Rightarrow I = qf$
Therefore, ${I_n} \propto f$
And also it will be proportional to $\dfrac{1}{{{n^3}}}$
Now by using the magnetic field equation, we can write
$B \propto \dfrac{{{n^{ - 3}}}}{{{n^2}}}$
And it can be written as
$B \propto {n^{ - 5}}$
Therefore, the option $C$ 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.
Formula used:
Magnetic field,
$B = \dfrac{{{\mu _0}I}}{{2r}}$
And also it can be written as
$B = \dfrac{{{\mu _0}{I_n}}}{{2{r_n}}}$
Here,
$B$, is the magnetic field
${I_n}$, current in the ${n^{th}}$orbit
${r_n}$, the radius of the ${n^{th}}$ orbit
Complete Step By Step Solution
So first of all we will know the magnetic field is given by
$B = \dfrac{{{\mu _0}I}}{{2r}}$
And also it can be written as
$B = \dfrac{{{\mu _0}{I_n}}}{{2{r_n}}}$
And according to Bohr’s theory
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mk{e^2}}}$
Since here, all the terms are constant except the ${n^2}$
So we can say
$r{}_n \propto {n^2}$
Now let’s see the formula of frequency,
\[f = \dfrac{v}{{2\pi r}}\]
And further expanding the equation we get
\[ \Rightarrow \dfrac{{2\pi k{e^2}}}{{nh \times 2\pi r}}\]
So from here, we can see that all the other terms are constant except the two terms, that is
$f \propto \dfrac{1}{{nr}}$
Now on putting the value of $r$in the above equation we get
$ \Rightarrow f \propto \dfrac{1}{{n \times {n^2}}} \propto \dfrac{1}{{{n^3}}}$
Now as we know,
$I = \dfrac{q}{t}$
And also it can be written as,
$ \Rightarrow I = qf$
Therefore, ${I_n} \propto f$
And also it will be proportional to $\dfrac{1}{{{n^3}}}$
Now by using the magnetic field equation, we can write
$B \propto \dfrac{{{n^{ - 3}}}}{{{n^2}}}$
And it can be written as
$B \propto {n^{ - 5}}$
Therefore, the option $C$ 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.
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