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lorasvet [3.4K]
3 years ago
7

A) A mass spectrometer has a velocity selector that allows ions traveling at only one speed to pass with no deflection through s

lits 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
Physics
1 answer:
KengaRu [80]3 years ago
6 0

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

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