
The ratio of the magnetic dipole moment to the angular momentum of the electron in the first orbit of Hydrogen atom is:
$\begin{align}
& (A)\dfrac{e}{2m} \\
& (B)\dfrac{e}{m} \\
& (C)\dfrac{2m}{e} \\
& (D)\dfrac{m}{e} \\
\end{align}$
Answer
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Hint: We will first calculate the magnetic dipole moment of the electron in the first orbit of the Hydrogen atom using the formula that it is a product of current and area of orbit. Then, we shall calculate the angular momentum of the electron which is given as a product of: mass of electron, its velocity in the orbit and radius of orbit.
Complete answer:
Let us first define some terms which are to be used later.
Let the mass of the electron be denoted by $m$ .
Let the charge on the electron be denoted by $e$ .
Let the speed by which the electron is moving in the orbit be given by $v$ and the radius of orbit of the electron be denoted by $r$ .
Now, we will proceed ahead by calculating the magnetic dipole moment of the electron.
The magnetic dipole can be found using the following formula:
$\Rightarrow \mu =IA$
Here, we can calculate $I$ and $A$ as follows:
$\Rightarrow I=\dfrac{ev}{2\pi r}$
$\Rightarrow A=\pi {{r}^{2}}$
Putting these values in the above expression, the magnetic dipole comes out to be:
$\Rightarrow I=\left( \dfrac{ev}{2\pi r} \right)\left( \pi {{r}^{2}} \right)$
$\Rightarrow I=\dfrac{evr}{2}$ [Let this expression be equation number (1)]
Now, the orbital angular momentum can simply be given by the formula:
$\Rightarrow L=mvr$ [Let this expression be equation number (2)]
Now, on dividing equation number (1) by equation number (2), we get the ratio of magnetic dipole moment to the angular momentum of the electron as:
$\begin{align}
& \Rightarrow \dfrac{\mu }{L}=\dfrac{\dfrac{evr}{2}}{mvr} \\
& \Rightarrow \dfrac{\mu }{L}=\dfrac{e}{2m} \\
\end{align}$
Hence, the ratio of magnetic dipole moment to the angular momentum of the electron in the first orbit of the Hydrogen atom comes out to be $\dfrac{e}{2m}$ .
Hence, option (A) is the correct option.
Note:
We assumed the unknown terms in the beginning of our solution and after solving the problem, all the unnecessary terms phased out. So we shouldn’t be worried if the terms or the values of different terms haven’t been provided in the question. Rather, we should proceed step by step towards our solution, keeping in check we don’t concur any errors.
Complete answer:
Let us first define some terms which are to be used later.
Let the mass of the electron be denoted by $m$ .
Let the charge on the electron be denoted by $e$ .
Let the speed by which the electron is moving in the orbit be given by $v$ and the radius of orbit of the electron be denoted by $r$ .
Now, we will proceed ahead by calculating the magnetic dipole moment of the electron.
The magnetic dipole can be found using the following formula:
$\Rightarrow \mu =IA$
Here, we can calculate $I$ and $A$ as follows:
$\Rightarrow I=\dfrac{ev}{2\pi r}$
$\Rightarrow A=\pi {{r}^{2}}$
Putting these values in the above expression, the magnetic dipole comes out to be:
$\Rightarrow I=\left( \dfrac{ev}{2\pi r} \right)\left( \pi {{r}^{2}} \right)$
$\Rightarrow I=\dfrac{evr}{2}$ [Let this expression be equation number (1)]
Now, the orbital angular momentum can simply be given by the formula:
$\Rightarrow L=mvr$ [Let this expression be equation number (2)]
Now, on dividing equation number (1) by equation number (2), we get the ratio of magnetic dipole moment to the angular momentum of the electron as:
$\begin{align}
& \Rightarrow \dfrac{\mu }{L}=\dfrac{\dfrac{evr}{2}}{mvr} \\
& \Rightarrow \dfrac{\mu }{L}=\dfrac{e}{2m} \\
\end{align}$
Hence, the ratio of magnetic dipole moment to the angular momentum of the electron in the first orbit of the Hydrogen atom comes out to be $\dfrac{e}{2m}$ .
Hence, option (A) is the correct option.
Note:
We assumed the unknown terms in the beginning of our solution and after solving the problem, all the unnecessary terms phased out. So we shouldn’t be worried if the terms or the values of different terms haven’t been provided in the question. Rather, we should proceed step by step towards our solution, keeping in check we don’t concur any errors.
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