Atoms - Result Question 29
30. 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) $-6.8 eV$
[2004]
(a) predicts the same emission spectra for all types of atoms
(b) assumes that the angular momentum of electrons is quantised
(c) uses Einstein’s photoelectric equation
(d) predicts continuous emission spectra for atoms
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Answer:
Correct Answer: 30. (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
$=\frac{(1)^{2}}{(2)^{2}}(13.6)=+3.4 eV$
