Question

A) A mass spectrometer has a velocity selector that allows ions traveling at only one speed to pass with no deflection through slits at the ends. While moving through the velocity selector, the ions pass through a 60,000-N/C E? field and a 0.0500-T B? field. The quantities v?, E?, and B? are mutually perpendicular. Determine the speed of the ions that are not deflected.B) After leaving the velocity selector, the ions continue to move in the 0.0500-T magnetic field. Determine the radius of curvature of a singly charged lithium ion, whose mass is 1.16

Answer 1

Answer:

(A). The speed of the ions is $1.2\times10^{6}\ m/s$

(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$

Explanation:

Given that,

Electric field = 60000 N/C

Magnetic field = 0.0500 T

(A). We need to calculate the velocity

For no deflection

$F_{E}=F_{B}$

$Eq=Bqv$

$v = \dfrac{E}{B}$

$v=\dfrac{60000}{0.0500}$

$v=1.2\times10^{6}\ m/s$

(B). We need to calculate the radius

Using magnetic force balance by centripetal force

$Bqv=\dfrac{mv^2}{r}$

$r=\dfrac{mv^2}{Bqv}$

Put the value into the formula

$r=\dfrac{1.16\times10^{-26}\times(1.2\times10^{6})^2}{0.0500\times1.6\times10^{-19}}$

$r=2.0\times10^{6}\ m$

Hence, (A). The speed of the ions is $1.2\times10^{6}\ m/s$

(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$