Question

The number of waves made by a bohr electron in an orbit of maximum magnetic quantum number $+ 2$ is $?$

Answer 1

First we have to know that the number of waves formed is equal to the principal quantum number or the number of orbits. The principal quantum number ( $n$) is one of four quantum numbers assigned to each electron in an atom to describe that electron's state. It is represented by $n$, where $n = 1,2,3,...$

Complete answer:
Principal quantum number describes the size of the orbital and the energy level.
Azimuthal quantum number describes the shape of the orbital. It is represented as ' $l$'. The value of $l$ ranges from $0$ to $\left( {n - 1} \right)$. For $l = 0,1,2,3,...$the orbitals are $s,p,d,f,...$
Magnetic quantum number describes the orientation of the orbitals. It is represented as ${m_1}$. The value of this quantum number ranges from $\left( { - l\;to\; + l} \right)$ . When l = 2, the value of ${m_1}$ will be $- 2, - 1,0,1,2$.
Given the maximum magnetic quantum number ${m_1} = + 2$. It means ${m_1}$ranges from $- 2$ to $2$..
So, $l = 2$.
For a value of $n$, $l$ ranges from $0$ to $\left( {n - 1} \right)$
Since for the given maximum magnetic quantum number ${m_1} = + 2$ we have
$l = n - 1$
So, when l=2, we get
$n = 3$.
Thus, the number of waves made by a Bohr electron in an orbit of the maximum magnetic quantum number $+ 2$ is $3$.

Note:
The set of numbers used to describe the position and energy of the electron in an atom are called Quantum numbers. There are four quantum numbers, namely, principal, azimuthal, magnetic and spin quantum numbers. Also, principal quantum numbers were proposed by Bohr to explain the hydrogen atomic spectrum.