For the quantization of the Bohr orbit,its ci | vedclass

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

Question

(A) According to the de Broglie hypothesis,a particle of mass $m$ moving with velocity $v$ has an associated wavelength $\lambda = \frac{h}{mv}$.For a

Vedclass · Accepted Answer

$= n\lambda$

Vedclass · Answer

(A) According to the de Broglie hypothesis,a particle of mass $m$ moving with velocity $v$ has an associated wavelength $\lambda = \frac{h}{mv}$.For an electron moving in a circular orbit of radius $r$,the circumference is $2\pi r$.Bohr's quantization condition states that the angular momentum is quantized as $mvr = \frac{nh}{2\pi}$,which rearranges to $2\pi r = \frac{nh}{mv}$.Substituting $\lambda = \frac{h}{mv}$ into the equation,we get $2\pi r = n\lambda$.Thus,the circumference of the orbit must be an integral multiple of the de Broglie wavelength for a stable orbit.

For the quantization of the Bohr orbit,its circumference will be equal to .......

Similar Questions

The wavelength of a moving electron-

The wavelength associated with an electron moving with a velocity of $10^8 \, cm/s$ is ........ $\mathop {\rm{A}}\limits^{\rm{o}} $.

The frequency of the de-Broglie wave of an electron in Bohr's first orbit of a hydrogen atom is . . . . . . . . . . $\times 10^{13} \ Hz$ (nearest integer).
[Given: $R_H$ (Rydberg constant) $= 2.18 \times 10^{-18} \ J$,$h$ (Planck's constant) $= 6.6 \times 10^{-34} \ J \cdot s$]

The kinetic energy of a proton wave is $500 \ eV$. Calculate its de Broglie wavelength. (Mass of proton $= 1.67 \times 10^{-27} \ kg$)

The wavelengths of electron waves in two orbits are in the ratio $3: 5$. The ratio of the kinetic energy of the electrons will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module