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) For a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is given by $r_n = n^2 a_0$,where $a_0$ is the Bohr radius.For the second orbit $(n = 2)

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is given by $r_n = n^2 a_0$,where $a_0$ is the Bohr radius.For the second orbit $(n = 2)$,$r_2 = 2^2 a_0 = 4 a_0$.The velocity of an electron in the $n^{th}$ orbit is $v_n = \frac{nh}{2\pi m r_n}$.Substituting $r_n = n^2 a_0$,we get $v_n = \frac{nh}{2\pi m (n^2 a_0)} = \frac{h}{2\pi m n a_0}$.For $n = 2$,$v_2 = \frac{h}{2\pi m (2) a_0} = \frac{h}{4\pi m a_0}$.The kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v_2^2$.$KE = \frac{1}{2} m \left( \frac{h}{4\pi m a_0} \right)^2 = \frac{1}{2} m \left( \frac{h^2}{16\pi^2 m^2 a_0^2} \right) = \frac{h^2}{32\pi^2 m a_0^2}$.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is = ....... (Bohr radius $a_0$)

Similar Questions

For a hydrogen $(H)$ atom,what will be the wavelength (in $nm$) associated with the transition of an electron from the infinite energy level to the first energy level? (Rydberg constant = $1.097 \times 10^{7} \ m^{-1}$)

The maximum energy is present in any electron at

The ionization energy of a hydrogen atom is $13.6 \, eV$. What is the energy required (in $eV$) to excite the electron from the ground state to the first excited state?

The total energy of the electron of $H^{-}$ atom in the second quantum state is $-E_2$. The total energy of the $He^{+}$ atom in the third quantum state is

Describe the hydrogen spectrum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module