In hydrogen atom, if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E$, the ionization energy of hydrogen atom is

Question

Dr. Abubacker · Accepted Answer

(b) Energy $E = K\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ightbrack$ ($K$ = constant)
$n_{1} = 2$ and $n_{2} = 3$, so $E = \left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} ightbrack = K\left\lbrack \frac{5}{36} ightbrack$
For removing an electron, $n_{1} = 1$ to $n_{2} = \infty$
Energy $E_{1} = K\lbrack 1brack = \frac{36}{5}E = 7.2E$
$	herefore$ Ionization energy $= 7.2\ E$

Q. In hydrogen atom, if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E$, the ionization energy of hydrogen atom is

Explanation

Reviewed by Dr. Abubacker

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