Consider orbit of (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given constant, and (Planck's constant) = ] [2015]

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Q.
Consider $3^{rd}$ orbit of ${He}^{+}$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $K = 9 \times 10^{9}$ constant, $Z = 2$ and $h$ (Planck's constant) = $6.6 \times 10^{- 34}$ $J$ $s$ ] [2015]

1 $0.73 \times 10^{6} m/s$

2 $3.0 \times 10^{8} m/s$

3 $2.92 \times 10^{6} m/s$

4 $1.46 \times 10^{6} m/s$

Ans.
(4)

Energy of electron in ${He}^{+}$ $3^{rd}$ orbit

$E_{3} = - 13.6 \times \frac{Z^{2}}{n^{2}} eV = - 13.6 \times \frac{4}{9} eV$

$= - 13.6 \times \frac{4}{9} \times 1.6 \times 10^{- 19} J ≃ - 9.7 \times 10^{- 19} J$

As per Bohr’s model,

Kinetic energy of electron in the $3^{rd}$ orbit $= - E_{3}$

$∴$ $9.7 \times 10^{- 19} = \frac{1}{2} m_{e} v^{2} \Rightarrow v = \sqrt{\frac{2 \times 9.7 \times 10^{- 19}}{9.1 \times 10^{- 31}}} = 1.46 \times 10^{6}$ ${ms}^{- 1}$

Solution

Q.
Consider $3^{rd}$ orbit of ${He}^{+}$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $K = 9 \times 10^{9}$ constant, $Z = 2$ and $h$ (Planck's constant) = $6.6 \times 10^{- 34}$ $J$ $s$ ] [2015]

1 $0.73 \times 10^{6} m/s$

2 $3.0 \times 10^{8} m/s$

3 $2.92 \times 10^{6} m/s$

4 $1.46 \times 10^{6} m/s$

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Solution

Q. Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K=9×109 constant, Z=2 and h (Planck's constant) = 6.6×10-34 J s] [2015]

1 0.73×106m/s

2 3.0×108m/s

3 2.92×106m/s

4 1.46×106m/s

Ans. (4) Energy of electron in He+ 3rd orbit E3=-13.6×Z2n2 eV=-13.6×49 eV =-13.6×49×1.6×10-19 J≃-9.7×10-19 J As per Bohr’s model, Kinetic energy of electron in the 3rd orbit =-E3 ∴ 9.7×10-19=12mev2⇒v=2×9.7×10-199.1×10-31=1.46×106 ms-1

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Q.
Consider $3^{rd}$ orbit of ${He}^{+}$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $K = 9 \times 10^{9}$ constant, $Z = 2$ and $h$ (Planck's constant) = $6.6 \times 10^{- 34}$ $J$ $s$ ] [2015]

1 $0.73 \times 10^{6} m/s$

2 $3.0 \times 10^{8} m/s$

3 $2.92 \times 10^{6} m/s$

4 $1.46 \times 10^{6} m/s$