Question

Kinetic energy of an electron in the second bohr orbit of a hydrogen atom is

Answer 1

The kinetic energy of an electron in the second Bohr orbit of s H atom is [a0 is Bohr radius],

$KE =  \frac{2\pi e^{4m}}{4h^{2} } \\= \frac{h^{2} e^{4m} }{2h^{2} } \\= \frac{\pi^{2}m }{2h^{2} } e^{4} \\= a_{0} = \frac{h^{2} }{4h^{2}e^{2}m\\}\\So, e_{4}=\frac{ h^{4}}{16\pi ^{4}m^{2} a^{2} } \\Hence, KE = \frac{h^{2} }{32\pi^{2} ma^{2} }$

Answer 2

Explanation:

According to Bohr, the angular momentum is given as follows.

                    mvr = $\frac{nh}{2 \pi}$

Squaring on both the sides, we get the equation as follows.

              $m^{2}v^{2}r^{2}$ = $\frac{n^{2}h^{2}}{(2)^{2} (\pi)^{2}}$

                            $v^{2}$ = $\frac{n^{2}h^{2}}{4(\pi)^{2}m^{2}r^{2}}$

Now, we multiply both sides by $\frac{1}{2}m$ as follows.

           $\frac{1}{2}mv^{2}$ = $\frac{1}{2}m \times \frac{n^{2}h^{2}}{4 (\pi)^{2} \times m^{2}r^{2}}$    

                        K.E = $\frac{n^{2}h^{2}}{8 (\pi)^{2} r^{2}m}$      ....... (1)

As r = radius of the second orbit of hydrogen atom. This means that, Z = 1 and n = 2.

Hence,     r = $a_{o} \times (\frac{n^{2}}{Z})$

                 $r^{2} = a^{2}_{o} \times (\frac{(2)^{2}}{1})$

                                 = $16 \times a^{2}_{o}$      .......... (2)

Therefore, putting equation (2) in equation (1) as follows.

              K.E = $\frac{n^{2}h^{2}}{8 (\pi)^{2} r^{2}m}$

                    = $\frac{(2)^{2}h^{2}}{8 (\pi)^{2} {16a^{2}_{o}}m}$

                    = $\frac{h^{2}}{32 {\pi^{2}} a^{2}_{o}m}$

 Thus, we can conclude that the kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\frac{h^{2}}{32 {\pi^{2}} a^{2}_{o}m}$.