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3 years ago

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A mass spectrometer accelerates doubly ionized atoms of charge 2e over a potential difference V before they enter a uniform magn

etic field B which is perpendicular to the direction of motion of the ions. If d is the radius of the ions' path in the magnetic field, what is the mass M of one ion? Express your answer ONLY in terms of V, B, e and d. The potential V is low enough that no relativistic corrections are needed

1 answer:

saul85 [17]3 years ago

3 0

Answer:

(B² x e x d²) / V = M

Explanation:

Under electric field by potential gradient if K is the kinetic energy imparted to

the charged particle

K = 1/2 M u² = V x 2e

When charged particle enters the magnetic field with velocity u , it travels on circular path with radius d . So

centripetal force = magnetic force

M u² / d = B x 2e x u

u = (B x 2e x d) / M

Putting this value of u in the relation above

m (B² x 4e² x d²) / 2M² = V x 2e

(B² x 4e² x d²) / V x 2e = 2 M

(B² x e x d²) / V = M

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<u>For process 1:</u>

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

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We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

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We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

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Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

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3 years ago

Read 2 more answers