## In a mass spectrometer used for commercial purposes, uranium ions of mass 3.76 X 10^(-25) kg and charge 3.5 X 10^(-19) C are separated from

Question

In a mass spectrometer used for commercial purposes, uranium ions of mass 3.76 X 10^(-25) kg and charge 3.5 X 10^(-19) C are separated from a larger sample. The atoms start from rest and are first accelerated through a potential difference of 110 kV and, moving to the right, they pass into a region with a magnetic field. Under the action of the magnetic field, the particle travels in a circle with a radius of 0.8 m. After completing half of the circle, the ions pass through a small 1.0 mm X 1.0 cm vertical slit and into a test tube.

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6 months 2021-08-05T01:10:32+00:00 1 Answers 7 views 0

## Answers ( )

a. 0.394 T b. 0.255 A c. 1.309 × 10⁸ J

Explanation:

Here is the complete question

A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are accelerated through a potential difference of 109 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.31 m. After traveling through 180° and passing through a slit of width 0.752 mm and height 0.991 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 1.12 mg of material per hour, calculate (b) the current (in A) of the desired ions in the machine and (c) the thermal energy (in J) produced in the cup in 1.31 h.

Solution

a. The magnitude of the (perpendicular) magnetic field in the separator

The kinetic energy of the uranium ions = electric potential energy

¹/₂mv² = qV

v = √(2qV/m) where v = speed of uranium ions, q = uranium ion charge = 3.2 × 10⁻¹⁹ C , m = mass of uranium ions = 3.92 × 10⁻²⁵ kg and V = 109 kV = 1.09 × 10⁵ V

v = √(2qV/m) = √(2 × 3.2 × 10⁻¹⁹ C × 1.09 × 10⁵ V/3.92 × 10⁻²⁵ kg)

v = 4.22 × 10⁵ m/s

Also, the magnetic force on the uranium ions equals the centripetal force when it passes through the magnetic field.

Bqv = mv²/r

B = mv/rq   where B = magnetic field strength and r = radius of circle = 1.31 m

B = m(√(2qV/m))/rq

B = √(2mV/q)/r

B = √(2 × 3.92 × 10⁻²⁵ kg × 1.09 × 10⁵ V/3.2 × 10⁻¹⁹ C)/1.31 m

B = √0.26705/1.31

B = 0.394 T

(b) the current (in A) of the desired ions in the machine

Since a mass m of 3.92 × 10⁻²⁵ kg of uranium ions carries a charge q of 3.2 × 10⁻¹⁹ C, then 1.12 mg per hour = 1.12 × 10⁻³ kg/h. In 1.31 h, our mass is M = 1.12 × 10⁻³ kg/h × 1.31 h = 1.47 × 10⁻³ kg carries a charge of Q of

m/q = M/Q

Q = Mq/m

Q = 1.47 × 10⁻³ kg × 3.2 × 10⁻¹⁹ C/3.92 × 10⁻²⁵ kg

Q = 1200 C

The current i = Q/t where t = time = 1.31 h = 1.31 × 60 × 60 s = 4716 s

i = 1200/4716

i = 0.2545 A

i ≅ 0.255 A

(c) the thermal energy (in J) produced in the cup in 1.31 h.

The thermal energy produced in the cup equals the kinetic energy lost by the uranium ions hitting the cup in 1.31 h.

E = ¹/₂Mv² = ¹/₂ × 1.47 × 10⁻³ kg × (4.22 × 10⁵ m/s)²

E = 1.309 × 10⁸ J