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 an electron of kinetic energy $4.50 \times 10^{-29} \ J$ is $... \times 10^{-5} \ m$. (Nearest integer) Given: mass of electron is $9 \times 10^{-31} \ kg$,$h = 6.6 \times 10^{-34} \ J \ s$.

The de Broglie wavelength of an electron travelling with $20 \%$ of velocity of light is
$(h = 6.626 \times 10^{-34} \ J \ s; m_{e} = 9.1 \times 10^{-31} \ kg)$

The mass of an electron is $9.1 \times 10^{-31} \ kg$. If its $K.E.$ is $3.0 \times 10^{-25} \ J$,calculate its wavelength.

If $\lambda _L, \lambda _M$ and $\lambda _N$ are the wavelengths of an electron in $L, M, N$ energy levels of $H^{-}$ atom respectively,what is their decreasing order?

The de-Broglie wavelength of a particle with mass $1 \, g$ and velocity $100 \, m/sec$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module