SATHEE: Atoms - Result Question 13

Atoms - Result Question 13

14. Consider $3^{rd}$ orbit of $H e^{+}$ (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) $0.73 \times 10^{6} m / s$

(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

$v_{n} = \frac{2 π K Z e^{2}}{n h}$

$v = (2.19 \times 10^{6} m / s) \frac{Z}{n}$

$v = (2.19 \times 10^{6}) \frac{2}{3} (Z = 2$ and $n = 3)$

$v = 1.46 \times 10^{6} m / s$

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Atoms - Result Question 13

14. Consider 3rd orbit of He+(Helium), using nonrelativistic approach, the speed of electron in this orbit will be [given K=9×109 constant, Z=2 and h (Plank’s Constant) .=6.6×10−34Js]

Answer:

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14. Consider $3^{rd}$ orbit of $H e^{+}$ (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) $. = 6.6 \times 10^{- 34} J s]$