Atoms Ques 24
24. The total energy of an electron in the first excited state of hydrogen atom is about -3.4 $eV$. Its kinetic energy in this state is
[2005]
(a) $3.4 eV$
(b) $6.8 eV$
(c) $-3.4 eV$
(d) $-6.8 eV$
Show Answer
Answer:
Correct Answer: 24.(a)
Solution:
(a) K.E. $=\frac{Z^{2}}{n^{2}}(13.6 eV)$
Mechanical energy $=\frac{-Z^{2}}{n^{2}}(13.6 eV)$
$\therefore$ K.E. in 2 nd orbital for hydrogen
$=-$ Mechanical energy
$=\frac{(1)^{2}}{(2)^{2}}(13.6)=+3.4 eV$
BETA
