Question

A singly charged 7Li ion has a mass of 1.16 10-26 kg. It is accelerated through a potential difference of 523 V and subsequently enters a uniform magnetic field of magnitude 0.370 T perpendicular to the ion's velocity. Find the radius of its path.

Answer 1

Answer:

$R=0.023m$

Explanation:

From the question we are told that:

Mass $m=1.16*10^{-26}$

Potential difference $V=523V$

Magnitude $m=0.370 T$

Generally the equation for Velocity is mathematically given by

$\frac{1}{2}mv^2=ev$

$v=\frac{2ev}{m}$

$v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}$

$v=12.22*10^4m/s$

Generally the equation for Force is mathematically given by

$F=qvBsin \theta$

Where

$qVB=m\frac{v^2}{R}$

$F=m\frac{v^2}{R}sin\theta$

Therefore

$R=\frac{mv}{qB sin \theta}$

$R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}$

$R=0.023m$