Question

7.
If the orbital radius of the electron in a hydrogen atom is 4.7x10-1'm. Compute
the kinetic energy of the electron in hydrogen atom.​

Answer 1

The kinetic energy of the electron in hydrogen atom is 21.40×$10^{-42}$.

Given,

radius = 4.7×$10^{-11}$

m = 9.11 x $10^{-31}$ kg

Find,

Kinetic Energy = ?

Formula used,

Kinetic energy is tendency of any object or particles in motion which depends on its mass and not only on the motion. This kind of motion can be rotational or translational.

K.E = $\frac{1}{2}mv^{2}$

Since,

$\frac{mv^{2} }{r} = mg\\\frac{v^{2} }{r} = g\\\\{v^{2} = rg$

As,

$\frac{1}{2}mv^{2} = \frac{1}{2}mrg$

Substituting values in formula,

$= \frac{1}{2} (9.11[10^{-31}][4.7[10^{-11}]] \\\\= 21.40[10^{-42}]$

Therefore, the kinetic energy of the electron in hydrogen atom is 21.40×$10^{-42}$.