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A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses. In one design for such an instrument, ions with mass m and charge q are accelerated through a potential difference V. They then enter a uniform magnetic field that is perpendicular to their velocity and are deflected in a semicircular path of radius R. A detector measures where the ions complete the semicircle and from this it is easy to calculate R.

Derive the equation for calculating the mass of the ion from measurements of B, V, R, and q.

What potential difference V is needed so that singly ionized 12C atoms will have R=50.0 cm in a0.150-T magnetic field?

Do you think that this beam separation is sufficient for the two ions to be distinguished?

Respuesta :

There's a part of the question missing and it says ;

"Suppose the beam consists of a mixture of 12C and 14C ions. If V and B have the same values as in

part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam

separation is sufficient for the two ions to be distinguished?"

Answer:

A) m = qB²R²/2V

B) V = 22.5904 KV

C) Separation = 4cm

This separation is sufficient for easy separation of the two ions.

Explanation:

A) The speed of the ions when they enter the region of a uniform magnetic field is found by energy

conservation;

qV = (1/2)mv²

v =√(2qV/m)

When the ion enters this region, it will feel a magnetic force

Fb = qvB

Now, this force is always perpendicular to v and therefore, the acceleration is always perpendicular to v. This

is a clear characteristic of circular motion, and thus, the ion will start going in a circle once it enters the region of uniform B. Furthermore, B is always perpendicular to v so the magnitude is given as;

Fb = qvB = ma = m v ²/ R

Hence, qvB = m v ²/ R

qBR/m = v

From earlier, we saw that;

v =√(2qV/m)

Thus, qBR/m = √(2qV/m)

Now, let's make m the subject of the formula;

q²B²R²/m² = 2qV/m

qB²R² = 2Vm

m = qB²R²/2V

B) If we rearrange the equation of m we just got to make V the subject, we get;

V = qB²R²/2m

q is charge on proton =

1.6 x 10^(-19)C

B= 0.15 T

R= 50cm = 0.5m

Now, from the question, the mass of Carbon-12 is 12u and if it’s singly ionized it obtains a charge e and so, m = 12 x 1.66 x 10^(-27) kg.

Thus, the necessary potential is :

V = (1.6 x 10^(-19) x 0.15² x 0.5²)/(2 x 12 x 1.66 x 10^(-27)) = 22,590.4 V or 22.5904 KV

C) Carbon-14 has a mass of 14 u. Again rearranging the formula V = qB²R²/2m to make R the subject;

R = √(2mV)/qB²

Thus,the differences in radii of the

two semicircles will be ;

Radius of C-14 - Radius of C-12

Thus;

ΔR = √(2(m-C14)V)/qB²) - √(2(m-C12)V)/qB²)

ΔR = [√(2 x 14 x 1.66 x 10^(-27) x 22,590.4)/1.6 x 10^(-19) x 0.15²)] - [√(2 x 12 x 1.66 x 10^(-27) x 22,590.4)/1.6 x 10^(-19) x 0.15²)]

= 0.54 - 0.5 = 0.04m or 4cm

This beam separation is sufficient for easy separation of the two ions.

Whenever the magnetic field is perpendicular to the velocity, magnetic pressure is produced.

[tex]\to F =qvB\ \sin 90^{\circ} =qvB[/tex]

The path is then circular in this case [tex]F=\frac{mv^2}{R}\\\\[/tex]  

Therefore,

 [tex]\to \frac{mv^2}{R} =qvB\\\\ \to v=\frac{qBR}{m}................ (1) \\\\[/tex]

The charged particle is now propelled by electric potential, resulting in kinetic.

 [tex]\to \frac{1}{2} mv^2 = qvB \\\\\to v=\frac{qBR}{m}...............(2)\\\\[/tex]

using equations (1) and (2)  

[tex]\to \frac{qBR}{m} =\sqrt{\frac{2qV}{m}}[/tex]

squaring both sides  

[tex]\to \frac{q^2B^2R^2}{m^2} ={\frac{2qV}{m}}\\\\ \tom= \frac{qB^2R^2}{2V}\\\\[/tex]

Calculating the potential difference:  

[tex]\to V= \frac{qB^2R^2}{2m}\\\\[/tex]

        [tex]=\frac{1.6\times 10^{-19} \ C\times (0.150\ T)^2 \times (50.0\times 10^{-2} \ m)^2}{2 \times (12\times 1.67 \times 10^{-27} \ kg)} \\\\= 2.246 \times 10^4\ \ volts[/tex]

As just a result, this beam divergence should enable the ions to be separated.

Learn more:

brainly.com/question/15290407

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