Maximum number of electrons possible with spin quantum number $+\frac{1}{2}$ with principal quantum number $n=4$ in an atom is

  • A
    $16$
  • B
    $9$
  • C
    $4$
  • D
    $25$

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$A$ hydrogen atom has only one electron,so mutual repulsion between electrons is absent. However,in multielectron atoms,mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

For both $2p$ and $3s$ orbitals,the value of $(n + l)$ is equal to $3$. Which of these has lower energy and why?

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