The kinetic energy of an electron in the seco | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Bohr's postulate,the angular momentum of an electron in the $n^{th}$ orbit is given by $m v r_n = \frac{n h}{2 \pi}$.For the second o

Vedclass · Accepted Answer

$\frac{h^2}{32 \pi^2 m a_0^2}$

Vedclass · Answer

(C) According to Bohr's postulate,the angular momentum of an electron in the $n^{th}$ orbit is given by $m v r_n = \frac{n h}{2 \pi}$.For the second orbit $(n = 2)$,the radius is $r_2 = n^2 a_0 = 2^2 a_0 = 4 a_0$.Substituting these values: $m v (4 a_0) = \frac{2 h}{2 \pi} = \frac{h}{\pi}$.Solving for velocity $v$: $v = \frac{h}{4 m \pi a_0}$.The kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v^2$.Substituting $v$: $KE = \frac{1}{2} m \left( \frac{h}{4 m \pi a_0} \right)^2 = \frac{1}{2} m \cdot \frac{h^2}{16 m^2 \pi^2 a_0^2} = \frac{h^2}{32 m \pi^2 a_0^2}$.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [$a_0$ is Bohr radius]:

Similar Questions

The work functions of two metals ($M_A$ and $M_B$) are in the $1:2$ ratio. When these metals are exposed to photons of energy $6 \ eV$,the kinetic energy of liberated electrons of $M_A:M_B$ is in the ratio of $2.642:1$. The work functions (in $eV$) of $M_A$ and $M_B$ are respectively.

Which observations are not explained by the electromagnetic theory of physics?

If the ratio of the area of two orbits of an $H$ atom is $4 : 1$,then the ratio of the frequency of the $e^-$ in these two orbits is:

Write down the postulates of Bohr's model for the hydrogen atom.

The expression for Bohr's radius of an atom is

Vedclass Test Series

Exam Paper Generator

Online Exam Module