Chapter 3, Problem 3.19P

To determine

The velocity of the ground state electrons in hydrogen and in helium.

Answer to Problem 3.19P

Hydrogen: $v=2.2\times {10}^{6}m/\sec$

Helium: $v=4.49\times {10}^{6}m/\sec$

Explanation of Solution

Formula used:

The equation which is used to find the radii of the electronic orbits is:

  $r=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}Z{k}_0}$

Calculation:

To find the velocity of the hydrogen electron:

  $\begin{array}{l}n=1\\ h=6.625\times {10}^{-34}J\sec \\ m_{electron}=9.11\times {10}^{-31}kg\\ Z=1\\ k_0=9\times {10}^{9}N{m}^2/C^2\\ e=4.8\times {10}^{-10}sC=4.8\times {10}^{-10}sC\times ((\frac{1C}{3\times {10}^{9}Sc}))=1.6\times {10}^{-19}C\end{array}$

  $r=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}Z{k}_0}$

  $r=\frac{1^2\times {(6.625\times {10}^{-34}J\sec )}^2}{4{\pi }^2(9.11\times {10}^{-31}kg)\times {(1.62\times {10}^{-19}C)}^2\times 1\times 9\times {10}^{9}N{m}^2/C^2}$

  $r=5.17\times {10}^{-11}m$

The velocity is determined by:

  $v=\frac{nh}{2\pi rm}=\frac{1\times 6.625\times {10}^{-34}J\sec }{2\pi \times 5.17\times {10}^{-11}m\times 9.11\times {10}^{-31}kg}$

  $=2.2\times {10}^{6}m/\sec$

For helium, the same method is used.

  $\begin{array}{l}n=1\\ h=6.625\times {10}^{-34}J\sec \\ m_{electron}=207\times 9.11\times {10}^{-31}kg=1.89\times {10}^{-28}kg\\ Z=1\\ k_0=9\times {10}^{9}N{m}^2/C^2\\ e=4.8\times {10}^{-10}sC=4.8\times {10}^{-10}sC\times ((\frac{1C}{3\times {10}^{9}Sc}))=1.6\times {10}^{-19}C\end{array}$

  $r=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}Z{k}_0}$

  $r=\frac{1^2\times {(6.625\times {10}^{-34}J\sec )}^2}{4{\pi }^2(9.11\times {10}^{-31}kg)\times {(1.62\times {10}^{-19}C)}^2\times 2\times 9\times {10}^{9}N{m}^2/C^2}$

  $=2.58\times {10}^{-11}m$

The velocity is determined by:

  $v=\frac{nh}{2\pi rm}=\frac{1\times 6.625\times {10}^{-34}J\sec }{2\pi \times 2.58\times {10}^{-11}m\times 9.11\times {10}^{-31}kg}$

  $=4.49\times {10}^{6}m/\sec$

Conclusion:

The velocity of the hydrogen electron is v $=2.2\times {10}^{6}m/\sec$ and the velocity of the helium electron is v $=4.49\times {10}^{6}m/\sec$