Question
In hydrogen atom, energy of first excited state is $$ - 3.4\,eV.$$ Then, $$KE$$ of same orbit of hydrogen atom is
A.
$$ + 3.4\,eV$$
B.
$$ + 6.8\,eV$$
C.
$$ - 13.6\,eV$$
D.
$$ + 13.6\,eV$$
Answer :
$$ + 3.4\,eV$$
Solution :
$$\eqalign{
& \because \,\,{\text{Total energy }}\left( {{E_n}} \right) = KE + PE \cr
& {\text{In first excited state}} \cr
& = \frac{1}{2}m{v^2} + \left[ { - \frac{{Z{e^2}}}{r}} \right] \cr
& = + \frac{1}{2}\frac{{Z{e^2}}}{r} - \frac{{Z{e^2}}}{r} \cr
& {\text{Energy of first excited state is}}\,\,3.4\,eV \cr
& - 3.4\,eV = - \frac{1}{2}\frac{{Z{e^2}}}{r} \cr
& \therefore \,\,\,\,\,KE = \frac{1}{2}\frac{{Z{e^2}}}{r} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = + 3.4\,eV \cr} $$