Orbits of a particle moving in a circle are s | vedclass

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

Question

(C) According to the condition given,the circumference of the orbit is an integer multiple of the de-Broglie wavelength:$2 \pi r = n \lambda$Since $\l

Vedclass · Accepted Answer

$n^{1/2}$

Vedclass · Answer

(C) According to the condition given,the circumference of the orbit is an integer multiple of the de-Broglie wavelength:$2 \pi r = n \lambda$Since $\lambda = \frac{h}{p} = \frac{h}{mv}$,we have:$2 \pi r = \frac{nh}{mv} \implies mvr = \frac{nh}{2 \pi}$For a charged particle moving in a magnetic field $B$,the magnetic Lorentz force provides the centripetal force:$qvB = \frac{mv^2}{r} \implies mv = qBr$Substituting $mv = qBr$ into the quantization condition:$(qBr)r = \frac{nh}{2 \pi} \implies qBr^2 = \frac{nh}{2 \pi}$Solving for $r^2$:$r^2 = \frac{nh}{2 \pi qB}$Thus,$r = \sqrt{\frac{nh}{2 \pi qB}}$,which implies $r \propto n^{1/2}$.

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field,the radius of the $n^{th}$ orbital will therefore be proportional to

Similar Questions

$A$ subatomic particle of mass $10^{-30} \ kg$ is moving with a velocity $2.21 \times 10^6 \ m/s$. Under the matter wave consideration,the particle will behave closely like . . . . . . . $(h = 6.63 \times 10^{-34} \ J \cdot s)$

Consider two particles of different masses. In which of the following situations will the heavier of the two particles have a smaller de-Broglie wavelength?

What does the de Broglie equation for an electron represent?

The de-Broglie wavelength of a body of mass $1 \ kg$ moving with a velocity of $2000 \ m/s$ is:

An electron,accelerated by a potential difference $V$,has de Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $4V$,its de Broglie wavelength will become $\frac{\lambda}{n}$ where $n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module