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Q.
The energy of electron in first Bohr’s orbit of $H$-atom is $-13.6\, eV$. What will be its potential energy in $n = 4^{th}$ orbit.
Structure of Atom
Solution:
Energy of nth orbit of H-atom
$= -\frac{2\pi^{2} me^{4} k^{2}}{h^{2}} \times\frac{1}{n^{2}}$
Energy of nth orbit of H-atom
$=-\frac{2\pi^{2} me^{4} k^{2}}{h^{2}}$
$=-13.6 ev \left(given\right) $
Energy of fourth Bohr’s orbit of H-atom $=-\frac{2\pi^{2} me^{4} k^{2}}{h^{2}} \times\frac{1}{4^{2}}$
$=13.6 \times\frac{1}{16}eV =-0.85 ev $
PE of electron in nth orbit $=2\times E_{n} $
So P.E. of electron in 4th orbit $= 2 \times \left(-0.85\right) = -1.70 eV$