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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Atomic Physics

NEET 2015 · Physics · Atomic Physics

Consider third orbit of He ^ + (helium), using non-relativistic...

Consider third orbit of He+{He}^{+} (helium), using non-relativistic approach, the speed of electron in this orbit will be given constant K=9 \times109K{\text{=9 \times 10}}^{\text{9}}, Z=2Z =2 and hh (Planck's Constant)= 6.6×10−34J s\text{= 6}\text{.6} \times {\text{10}}^{- 34}\text{J s}

Options

  1. A2.92×106ms−12.92 \times {10}^{6}m{s}^{- 1}
  2. B1.46×106ms−11.46 \times {10}^{6}m{s}^{- 1}
  3. C0.73×106ms−10.73 \times {10}^{6}m{s}^{- 1}
  4. D3.0×108ms−13.0 \times {10}^{8}m{s}^{- 1}

Correct answer

B. 1.46×106ms−11.46 \times {10}^{6}m{s}^{- 1}

Step-by-step solution

Energy of electron He+H{e}^{+}
EnE_n=−13.6Z2n2eV;Z=2,n=3= - 13.6\dfrac{{Z}^{2}}{{n}^{2}}eV; Z = 2, n = 3
E3=−13.6(23)2×1.6×10−19J{E}_{3} = - 13.6{\left(\dfrac{2}{3}\right)}^{2} \times 1.6 \times {10}^{- 19}J
In Bohr's model energy, E=−KEE = - KE
E3=−KE3{E}_{3} = - K{E}_{3}
∴9.7×10−19=12mv2∴ 9.7 \times {10}^{- 19} = \dfrac{1}{2}m{v}^{2}
v=2×9.7×10−199.1×10−31v = \sqrt{\dfrac{2 \times 9.7 \times {10}^{- 19}}{9.1 \times {10}^{- 31}}}
Velocity of electron in third orbit of He+H{e}^{+} is 1.46×106ms−11.46 \times {10}^{6}m{s}^{- 1}.
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