Atoms - Result Question 13
14. Consider $3^{\text{rd }}$ orbit of $He^{+}$(Helium), using nonrelativistic approach, the speed of electron in this orbit will be [given $K=9 \times 10^{9}$ constant, $Z=2$ and $h$ (Plank’s Constant) $.=6.6 \times 10^{-34} J s]$
[2015]
(a) $1.46 \times 10^{6} m / s$
(b) $0.73 \times 10^{6} m / s$
(c) $3.0 \times 10^{8} m / s$
(d) $2.92 \times 10^{6} m / s$
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Answer:
Correct Answer: 14. (a)
Solution:
- (a) Speed of electron in nth orbit
$v_n=\frac{2 \pi KZe^{2}}{nh}$
$v=(2.19 \times 10^{6} m / s) \frac{Z}{n}$
$v=(2.19 \times 10^{6}) \frac{2}{3}(Z=2$ and $n=3)$
$v=1.46 \times 10^{6} m / s$
