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Atoms and Nuclei question

2015 · Q131
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Atoms and Nuclei question

2015 · Q131

NEETPhysicsAtoms and NucleiMCQ+4 / −1
Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 ×\times× 109 constant, Z = 2 and h (Planck's Constant) = 6.6 ×\times× 10−-−34 J s]
  1. A
    0.73 ×\times× 106 m/s
  2. B
    3.0 ×\times× 108 m/s
  3. C
    2.92 ×\times× 106 m/s
  4. D
    1.46 ×\times× 106 m/s
View written solutionFree

Correct answer: D

Energy of electron in He+ 3rd orbit

E3 = -13.6 ×\times× 49eV{4 \over 9}eV94​eV

= -13.6 ×\times× 49{4 \over 9}94​ ×\times× 1.6 ×\times× 10-19 J

= - 9.7 ×\times× 10-19 J

According to Bohr’s model,

Kinetic energy of electron in the 3rd orbit = – E3

∴\therefore∴ 9.7 ×\times× 10-19 = 12mev2{1 \over 2}{m_e}{v^2}21​me​v2

v = 2×9.7×10199.1×10−31\sqrt {{{2 \times 9.7 \times {{10}^{19}}} \over {9.1 \times {{10}^{ - 31}}}}} 9.1×10−312×9.7×1019​​ = 1.46 ×\times× 106 m/s

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