In class we learned that you can detect The Big C (molecular weight of 303 grams/mol) by mass spectrometry. The mass spec has a magnetic field of 0.3 T and the singly charged ions in the sample are moving at 50,000 m/s. How far from the entrance in m do the Florida Snow ions strike the spectrometer detector after travelling through a semicircle trajectory?

Answer :

mavila18

Answer:

1.04m

Explanation:

the distance is determined by the diameter of the trajectory.

you can find the radius of the trajectory by using the following formula:

[tex]r=\frac{m_cv}{qB}[/tex]

mc : mass of the Big C = 0.303kg/mol/(6.02*10^{23}/mol)=5.03*10^{-25}kg

v: velocity of the ion = 50000m/s

B: magnetic constant = 0.3T

q: 1.6*10^{-19}C

By replacing you obtain:

[tex]r=\frac{(5.03*10^{-25}kg)(50000m/s)}{(1.6*10^{-19}C)(0.3T)}=0.52m[/tex]

the diameter wiil be:

d=2r=1.04m

the ion strikes the detector from 1.04m to the entrance of the spectrometer

Cricetus

The Florida Snow ions strike the spectrometer detector after travelling through a semicircle trajectory will be "1.04 m" far.

Spectrometer

According to the question,

Big C's mass, [tex]m_c[/tex] = [tex]\frac{0.303}{6.02\times 10^{23}}[/tex]

                             = 5.03 × 10⁻²⁵ kg

Ion's velocity, v = 50000 ms

Magnetic constant, B = 0.3 T

Charge, q = 1.6 × 10⁻¹⁹ C

We know that,

→ r = [tex]\frac{m_c v}{qB}[/tex]

or,

The radius, r = [tex]\frac{5.03\times 10^{-25}\times 50000}{1.6\times 10^{-19}\times 0.3 }[/tex]

                     = 0.52 m or,

Diameter, d = 2 × r

                     = 2 × 0.52

                     = 1.04 m

Thus the above approach is correct.  

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