The angular momentum of an electron in the hy | vedclass

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

Question

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}

Vedclass · Accepted Answer

$1.51$

Vedclass · Answer

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}{2\pi}$,comparing the two expressions,we get $n = 3$.The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For $n = 3$,$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 	ext{ eV}$.The kinetic energy $K$ of an electron in an orbit is equal to the magnitude of its total energy,so $K = |E_n| = 1.51 	ext{ eV}$.

The angular momentum of an electron in the hydrogen atom is $\frac{3h}{2\pi}$. Here $h$ is Planck's constant. The kinetic energy of this electron is.....$eV$

Similar Questions

Which state of triply ionised Beryllium $(Be^{+++})$ has the same orbital radius as that of the ground state of hydrogen?

The ratio of the velocity of the electron in the first Bohr orbit to that in the second Bohr orbit of a hydrogen atom is

The potential energy of an electron in an orbit of a hydrogen atom is $-6.8 \text{ eV}$. The de Broglie wavelength of the electron in this orbit is (where $r_0$ is the Bohr radius).

The magnetic moment $(m_{orb})$ of a revolving electron around the nucleus varies with the principal quantum number $(n)$ as

When a hydrogen atom absorbs a photon of wavelength $60 \ nm$,the atom undergoes photo-ionization. What will be the maximum kinetic energy of the emitted electron in $eV$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module