Question

Find the velocity (ms ⁻¹ ) of an electron in the first Bohr orbit of radius $A_0$. Also, Find the De Broglie wavelength (In 'm'). Find the orbital angular momentum of the 2p orbital of the hydrogen atom in units of $\frac{h}{2\pi }$.

Answer 1

Calculation of velocity, De Broglie wavelength, and angular momentum

Step 1: Calculation of velocity
For H and H-like particles, velocity in the nth orbit
${\mathrm{v}}_{\mathrm{n}}=2.188\times {10}^6\times \frac{\mathrm{Z}}{\mathrm{n}}{\mathrm{ms}}^{-1}$
For H-atom, Z = 1 and for 1st orbit n = 1
$\therefore \mathrm{v}=2.188\times {10}^6{\mathrm{ms}}^{-1}$

Step 2: Calculation of De Broglie wavelength
According to De Broglie's hypothesis
$\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$

here, $$\begin{gathered} \mathrm{\lambda }\mathrm{reperesents}\mathrm{wavelength} \\ \mathrm{h}\mathrm{represents}\mathrm{planck}\text{'}\mathrm{s}\mathrm{constant} \\ \mathrm{m}\mathrm{represents}\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{electron} \\ \mathrm{v}\mathrm{represents}\mathrm{the}\mathrm{velocity} \end{gathered}$$
$$\begin{gathered} \mathrm{\lambda }=\frac{6.626\times {10}^{-34}{\mathrm{kgm}}^2{\mathrm{s}}^{-1}}{9.1\times {10}^{-31}\mathrm{kg}\times 2.188\times {10}^6{\mathrm{ms}}^{-1}} \\ \mathrm{\lambda }=3.33\times {10}^{-10}\mathrm{m} \end{gathered}$$

Step 3: Calculation of Orbital angular momentum
Orbital angular momentum $=\sqrt{\mathrm{l}(\mathrm{l}+1)}\frac{\mathrm{h}}{2\mathrm{\pi }}$
For 2p orbital,${\displaystyle l=1}$
${\displaystyle \therefore }$ Orbital angular momentum =$\sqrt{1(\mathrm{l}+1)}\frac{\mathrm{h}}{2\mathrm{\pi }}$
Orbital angular momentum = $\sqrt{2}\frac{\mathrm{h}}{2\mathrm{\pi }}$

Hence, the velocity of an electron in the first Bohr orbit is $2.188\times {10}^{6}m{s}^{-1}$ the wavelength is $3.33\times {10}^{-10}m$ and the orbital angular momentum of the 2p orbital of the Hydrogen atom in units of $\frac{h}{2\pi }$ is $\sqrt{2}\frac{h}{2\mathrm{\pi }}$.