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  • Consider orbit of <img alt="\mathrm{He}^{+}" src="https://entrancecorner.oncodeco

Consider 3^{\text {rd }} orbit of \mathrm{He}^{+}(Helium), using non-relativistic approach, the speed of electron in this orbit will be \mathrm{[given, \mathrm{K}=9 \times 10^{\circ} constant, \mathrm{Z}=2 and \mathrm{h} (Planck \: constant) =6.6 \times 10^{-34} \mathrm{Js} ]}
 

Option: 1

2.92 \times 10^6 \mathrm{~m} / \mathrm{s}
 


Option: 2

1.46 \times 10^6 \mathrm{~m} / \mathrm{s}
 


Option: 3

0.73 \times 10^8 \mathrm{~m} / \mathrm{s}


Option: 4

3.0 \times 10^8 \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

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

\mathrm{=-13.6 \times \frac{Z^2}{n^2} \mathrm{eV} }

\mathrm{E_3 =-13.6 \times \frac{4}{9} \mathrm{eV} }

\mathrm{=-13.6 \times \frac{4}{9} \times 1.6 \times 10^{-19} \mathrm{~J}}

In Bohr's model, \mathrm{E_3=-\mathrm{KE}_3}

\mathrm{\therefore \quad 9.7 \times 10^{-19} \mathrm{~J} =\frac{1}{2} m_{\mathrm{e}} v^2 }

\mathrm{v =\sqrt{\frac{2 \times 9.7 \times 10^{-19}}{9.1 \times 10^{-31}}}=1.46 \times 10^6 \mathrm{~m} / \mathrm{s}}

Hence option 2 is correct.
 

Posted by

Gaurav

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