JEE Main 2015 · 10 Apr · Shift 1 · Physics Q28

← JEE Main 2015 · 10 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 28 of 90

If one were to apply Bohr model to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a magnetic field 'B', the energy of the charged particle in the n $^{\text{th}}$ level will be :

n\left(\dfrac{hqB}{2\pi m}\right)

n\left(\dfrac{hqB}{4\pi m}\right)

n\left(\dfrac{hqB}{8\pi m}\right)

n\left(\dfrac{hqB}{\pi m}\right)

Answer and worked solution

Official answer: B

$mvr=\frac{nh}{2\pi}$ and $r=\frac{mv}{qB}$ , so $\frac{m^2v^2}{qB}=\frac{nh}{2\pi}$ . The energy is
$E=\frac12mv^2=\frac{nhqB}{4\pi m}.$

Topics:Bohr model and hydrogen spectrum Motion in magnetic and electric fields

← Previous question Next question →

JEE Main 2015 · 10 Apr · Shift 1, Physics question 28