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**Hint:**Study the Somerfield Model. The value of k represents the length of the minor axis of the ellipse which cannot be taken equal to zero. A single orbital can contain a maximum of two electrons of opposite spin according to the Pauli Exclusion Principle.

**Complete step by step answer:**

The Sommerfeld Atomic Model explains the fine spectrum of hydrogen atoms. According to the model, the orbits can be both circular and elliptical.

$\frac{n}{k}=$ Length of major axis/length of minor axis OR $\frac{k}{n}=\frac{b}{a}$ where b – minor axis and a – major axis

So, k cannot be zero because then the length of the minor axis ‘b’ will become zero. So, the ellipse will not be formed and it will represent only a straight line i.e. the electron passes through the nucleus. So, given l=k and k starts from $1$

So, for the ‘K’ shell, $l=1$because$n=1$. No. of orbitals are given as - $\left (2l+1 \right)$

$=2\times 1+1$

$=3$

Each orbital contains 2 electrons so the no. of maximum electrons for ‘k’ shell are$3\times 2=6$electrons

For ‘L’ shell, $l=2, 1$and $n=2$, No. of orbitals are $\left (2l+1 \right)$

$=2\times 2+1$ $=2\times 2+1$

$=5$(for $l=2$) and $3$for $l=1$

So, maximum no. of electron in L shell are\ [\left (5\times 2=10 \right)+\left (3\times 2=6 \right)\]

For ‘M’ shell, $l=3, 2, 1 $, no. of orbitals are $\left (2\times 3+1 \right) = 7$(for $l=3$)

For $l=2$, no. of orbitals are $5$

For $l=1$, no. orbitals are $3$

So, we get for M shell, maximum no. of electrons are $6+10+14=30$

**For ‘N’ shell, $l=4, 3, 2, 1$ so maximum no. of electrons are $48$.**

**Note:**

The postulates of Sommerfeld models are- orbits may be both circular and elliptical. When the path is elliptical, then there are two axes- major and minor axes. When the length of major and minor axes becomes equal then orbit is circular. One should study the different types of quantum number, their importance and the relation between them.

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