A singly charged ion of unknow mass moves?

Question

A singly charged ion of unknown mass moves in a circle of radius 12.5 cm in a magnetic field of 1.2 T. The ion was accleerated through a potential difference of 7.0 kV before it entered the magnetic field. What is the mass of the ion?

Answer: 2.6x10^-35 kg

How do i solve this problem?

I have r = 12.5 cm = 0.125 m

B = 1.2 T

V = 7.0 kV = 70000 V

1/2mv^2 = qV

But I dont' know m and I don't if it's an electron or proton.

Electron Mass me = 9.1x10^-31 kg

Proton Mass mp = 1.67x10^-27 kg

q = 1.6x10^-19 C

v =sqrt[(2qV)/m] = sqrt[2(1.6x10^-19 C x 7000)/ (9.1x10^-31 kg)

= 4.96x10^7 m/s

m = rqB/v

= [0.125(1.6 x10^-19)(1.2)]/[4.96x10^7]

= 4.84 x10^-28 kg

What am i doing wrong?

Answer 1

First, the answer is 2.57E-25 kg (not E-35).

You need to solve for v first, using the equations for kinetic energy and centripetal force:

KE = VQ = mv^2/2

F(mag) = QvB = mv^2/r ==> QB = mv/r

VQ/(QB) = V/B= (mv^2/2)/(mv/r) = vr/2

v = 2V/(rB) = 93333 m/s

Now you can solve for m using either the KE or the F(mag)expression:

m = QrB/v = 2VQ/v^2 = 2.57464E-25 kg.

Answer 2

First, go to the following web site, read the explanation there under the heading "Circular Path from Magnetic Field," then read the explanation below:

http://hyperphysics.phy-astr.gsu.edu/Hba...

How do i solve this problem?

I have r = 12.5 cm = 0.125 m

B = 1.2 T

V = 7.0 kV = 70000 V

Woops! 7.0 kV is 7,000 V not 70000 V.

1/2mv^2 = qV

Yes, you need this to express velocity in terms of mass (unknown) and acceleration potential V.

But I dont' know m and I don't if it's an electron or proton.

Doesn’t matter what “it” is. You know "it" is a singly-charged ion, you just need to find its mass.

Electron Mass me = 9.1x10^-31 kg

Irrelevant.

Proton Mass mp = 1.67x10^-27 kg

Irrelevant.

q = 1.6x10^-19 C

Yes, you need to know this constant for unit charge. But it doesn’t matter whether ion is positive or negatively charged. The operative word in the problem is “singly” charged.

v =sqrt[(2qV)/m] = sqrt[2(1.6x10^-19 C x 7000)/ (9.1x10^-31 kg)

You were okay up to this point, but why are you substituting the mass of an electron for m, which is what you want to find? This is all about ANY moving charged object being deflected into a circle by a magnetic field applied at right angles to its direction of motion. The “ion” can be a massive as you please, it just makes the circle larger.

= 4.96x10^7 m/s NOT!

m = rqB/v

This is okay, but you need to express m in terms of the energy to which the ion was accelerated, 7.0 keV, instead of the velocity, v, that resulted from this acceleration.

What am i doing wrong?

You know that the force (at right angles to the ion velocity) applied by the magnetic field is F = qvB, and that this produces centripetal force mv²/r because the resulting (circular) motion is at right angles to BOTH the magnetic field and the ion velocity. From this relation you get m=rqB/v as shown above. Now all you need to do is express the velocity v in terms of the ion acceleration potential V.

Use the expression for kinetic energy, qV, acquired by accelerating the ion, as a function of the acquired velocity: qV=1/2 mv² and solve for velocity: v = √(2qV/m)

Substitute this expression for velocity into the expression for mass: m = rqB/√(2qV/m)

Square this equation to get m²=r²q²B²/2qV/m = r²B²qm/2V, which (after dividing both sides by m) yields: m=r²B²q/2V

Finally, substitute r = 12.5 x 10^(-2) m, B = 1.2 T, q = 1.6 x 10^(-19) C, V = 7 x 10^3 V and calculate m = 2.5714 x 10^(-25) kg.

The answer, correct to two significant figures, is m = 2.57 x 10^(-25) kg.

Answer 3

It's a singly charged particle.

It need not be a proton or electron.

(eg: it can be an Na+ ion or an Deuteron, etc)

qvB = mv^2/r                 --eqn1

(the force due to the magnetic field on a charged particle is q[v(bar)xB(bar))], which provides for the centripetal force for it to revolve in a circle.)

now, the initial speed(v) of the particle is attained by accelerating it in a  electric field (assume it being uniform), between points at a potential difference of V.

hence, qV = 1/2 mv^2              --eqn2

[Note that he magnetic force doesnot change the speed of the particle as  it is always perpendicular to it.]

manipulating the two eqns, we get,

given

{

V = 7000 volts,  [NOT 70000 v]

r = 12.5 x 10^-2 metres,

B = 1.2 T = 1.2 Weber*m^-2

}

v = 2V/rB = 93333.33 m/s

which gives,

m = (q r^2 * B^2)/(2V) = 2.571 * 10 ^ -25 kg....

check the calculations... (I'm not sure of the calcualtions either...)

Answer 4

this is all wrong, because you dont know what the charge is.

who said it was an electron or a proton for it to have a charge of 1.6x10^-19 C? an ion doesnt mean that ONLY ONE ELECTRON WAS GAINED OR LOST, IT COULD BE 2 OR 3.

infact, V=kq/r , so q=Vr/k = (7.0 kV *0.125 m)/ 9E9 = 9.72e -8 COLOUMBS.

ANOTHER THING IS THAT YOU ARE SUPPOSED TO USE THE MAGNETIC FEILD