Question

(a) An oxygen-16 ion with a mass of 2.66×10−26 kg travels at 5.00×106 m/s perpendicular to a 1.20-T magnetic field, which makes it move in a circular arc with a 0.231-m radius. What positive charge is on the ion? (b) What is the ratio of this charge to the charge of an electron? (c) Discuss why the ratio found in (b) should be an integer.

Answer 1

Answer:

(a) q = 4.8×10-¹⁹C.

(b) ratio 3:1

Explanation:

Given the radius R = 0.231m, B = 1.20T, M= 2.26×10-²⁶kg, v = 5.00×10⁶m/s

The ion experiences a magnetic force which by newton's second law is equal to the mass of the ion multiplied by its acceleration.

Fm = Ma

qvB = Ma = M×v²/R

Rearranging the equation

q = Mv²/ RvB = Mv/RB

q = 2.66×10-²⁶× 5×10⁶/(0.231×1.20)

q = 4.8×10-¹⁹ C

(b) qi/qe = (4.8×10‐¹⁹)/(1.6×10-¹⁹) = 3:1

(c) This ratio is an integer because the ion is simply an atom that is short of electrons or has less electrons than it should in its molecule/molecular structure.

Answer 2

Explanation:

This problem can be solved by using the formula

$r = \frac{mv}{qB}$

where m is the mass of the ion, v is the velocity, q is the charge, B is the strength of the magnetic field and r is the radius of the circular motion of the ion.

(a) Taking q from the equation and by replacing we have

$q = \frac{mv}{rB} = \frac{(2.66*10^{-26})(5*10^{6})}{(0.231)(1.20)} = 4.79*10^{-9} C$.

(b) $\frac{4.79*10^{-9}}{q_{e}} = \frac{4.79*10^{-9}}{1.6*10^{-19}} } = 2.99*10^{10}$  ≈ 3*10^(10) C

the ion has 2.99*10^(10) more charge than an electron .

(c) The ratio must be an integer because the electrons are indivisible.

I hope this is usefull for you,

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