Question

calculate angular momentum and kinetic energy of electron in first orbit of hydrogen atom​

Answer 1

Hope this answer helps you

Answer 2

Answer:

The angular momentum and kinetic energy of an electron in the first orbit of the hydrogen atom​ are $h/2\pi ,0.02*10^{-16}J$ respectively

Explanation:

• The angular momentum $(L)$  gives the quantity of rotation in a rotating body

• From Bohr's postulate for allowed orbits, angular momentum is given by the formula

         $L=n\frac{h}{2\pi }$

         where,

         $n$- the principal quantum number

         $h=6.62*10^{-34}$(Planck's constant)

• The kinetic energy of electrons in an orbit is

         $KE=\frac{Ze^{2} }{8\pi\varepsilon _0 r}$

        where

        $Z$-atomic number

        $e=1.6*10^{-19} C$(charge of an electron)

        $r$-radius of the orbit

        $\varepsilon _0 =8.8*10^{-12}$

from the question,

for the first orbit of a hydrogen atom

⇒

the principal quantum number $n=1$

the atomic number of H-atom $Z=1$

the radius of the first orbit $r=0.5*10^{-10}m$

⇒

the angular momentum $L=h/2\pi$

the kinetic energy of the electron $KE=\frac{(1.6*10^{-19} )^{2} }{8*3.14*8.8*10^{-12}*0.5*10^{-10} } =0.02*10^{-16}J$