Structure of Atom - Result Question 16
####16. In hydrogen atom, energy of first excited state is $-3.4 eV$. Find out $K E$ of the same orbit of Hydrogen atom
[2002]
(a) $+3.4 eV$
(b) $+6.8 eV$
(c) $-13.6 eV$
(d) $+13.6 eV$
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Solution:
- (a) Suppose the nucleus of hydrogen atom have charge of one proton $+e$. The electron revolves in an orbit of radius $r$ around it. Therefore the centripetal force is supplied by electrostatic force of attraction between the electron and nucleus i.e.
$\frac{m v^{2}}{r}=\frac{Z e^{2}}{r^{2}}$
or $m v^{2}=\frac{Z e^{2}}{r}$
or $\frac{1}{2} m v^{2}=\frac{1}{2} \frac{Z e^{2}}{r}=K . E$
now total energy $(E_n)=K . E+$ P.E
in first excited state
$E=\frac{1}{2} m v^{2}+[-\frac{Z e^{2}}{r}]$
$=+\frac{1}{2} \frac{Z e^{2}}{r}-\frac{Z e^{2}}{r}$
$-3.4 eV=-\frac{1}{2} \frac{Z e^{2}}{r}$
$\therefore K . E=\frac{1}{2} \frac{Z e^{2}}{r}=+3.4 eV$
For Bohr orbit, kinetic energy $=-$ Total energy and Potential energy $=2 \times$ Total energy.
