Question

Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.

Answer 1

Maximum angular Speed of electron $\omega = {4.142\times 10^{17}\frac{rad}{sec}$

Explanation:

Hydrogen atom exists in three dimensions, we realize that the electron is a sort of spherical standing wave surrounding the nucleus. Because the wave should be continuous as we travel around the sphere, we should think of it is having the same limits  at both "ends". This shows that the circumference of the sphere should be equal to an integer times of the wavelength:-

$2 \pi r = \frac{nh}{mv}$

where, $r = \textrm{radius of the orbit} = 0.53\times 10^{-10} m$

$\textrm{m = mass of the electron} = 9.1\times10^{-31} Kg$

$\textrm{h = planck's constant} = 6.67\times10^{-34}m^{2}\frac{Kg}{s}$

$\textrm{v = speed of the electron = r }\omega$

$\textrm{n = angular momentum quantum number = 1}$ for hydrogen

On arranging the equation, we get

$\omega = \frac{nh}{2\pi \times m\times r^{2}}$

$\omega = \frac{1\times 6.67\times10^{-34}}{2\pi \times9.1\times10^{-31}\times0.53\times 10^{-10}}$

Angular speed $\omega = {4.142\times 10^{17}}$

Answer 2

Angular velocity is $4.08 \times 10^1^6 \frac {rad}{s}$

Explanation:

• Angular Velocity = Relative to one another point how fast a body revolves

Angular Velocity is $\omega = \frac {v}{r}$

$\omega = \frac {2.16 \times 10^6 }{0.529 \times 10_{-10}} \frac {Z^2}{n^3}$

which is maximum for n = 1 and given Z = 1

So, $\omega = \frac {2.16 \times 10^6 }{0.529 \times 10_{-10}}$

= $4.08 \times 10^1^6 \frac {rad}{s}$