A uniform magnetic field B exists in a region | vedclass

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

Question

(A, B, C, D) For a charged particle in a magnetic field, the radius of the circular path is given by $r = \frac{mv}{qB}$.Applying Bohr's quantization

Vedclass · Accepted Answer

Transition frequency $\omega$ between two successive levels is independent of $n$

Vedclass · Answer

(A, B, C, D) For a charged particle in a magnetic field, the radius of the circular path is given by $r = \frac{mv}{qB}$.Applying Bohr's quantization condition for angular momentum: $mvr = \frac{nh}{2\pi} = n\hbar$.Substituting $v = \frac{qBr}{m}$ into the quantization condition: $m \left( \frac{qBr}{m} ight) r = n\hbar \implies qBr^2 = n\hbar \implies r_n = \sqrt{\frac{n\hbar}{qB}}$. Thus, $r_n \propto \sqrt{n}$. (Statement $A$ is true).Now, $v_n = \frac{qBr_n}{m} = \frac{qB}{m} \sqrt{\frac{n\hbar}{qB}} = \sqrt{\frac{n q B \hbar}{m^2}}$. (Statement $B$ is true).The energy of the $n^{	ext{th}}$ level is $E_n = \frac{1}{2}mv_n^2 = \frac{1}{2}m \left( \frac{n q B \hbar}{m^2} ight) = n \left( \frac{q B \hbar}{2m} ight)$. Thus, $E_n \propto n$. (Statement $C$ is true).The transition frequency $\omega$ is given by $\Delta E = \hbar \omega$. For successive levels, $\Delta E = E_{n+1} - E_n = \frac{q B \hbar}{2m}$. Therefore, $\omega = \frac{\Delta E}{\hbar} = \frac{qB}{2m}$, which is independent of $n$. (Statement $D$ is true).

$A$ uniform magnetic field $B$ exists in a region. An electron of charge $q$ and mass $m$ moving with velocity $v$ enters the region in a direction perpendicular to the magnetic field. Considering Bohr angular momentum quantization, which of the following statement$(s)$ is/are true?

Similar Questions

An ion beam of specific charge $5 \times 10^7 \ C/kg$ enters a uniform magnetic field of $4 \times 10^{-2} \ T$ with a velocity $2 \times 10^5 \ m/s$ perpendicularly. The radius of the circular path of the ions in meters will be:

$A$ block of mass $m$ and charge $q$ is released on a long smooth inclined plane. $A$ magnetic field $B$ is constant,uniform,horizontal,and parallel to the surface as shown. Find the time from the start when the block loses contact with the surface.

An electron moving towards the east enters a magnetic field directed towards the north. The force on the electron will be directed

In Thomson's experiment,if the value of $q/m$ is the same for all positive ions striking the photographic plate,then the trace would be

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module