Question

A narrow beam of singly-charged potassium ions of kinetic energy 32keV is injected into a region of width 1.00cm having a magnetic field of strength 0.500T as shown in figure (34-E16). The ions are collected at a screen 95.5cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion = A (1.6 x 10^-27kg where A is the mass number.

Answer 1

For K – 39 : m = 39 × 1.6 × 10^–27kg, B = 5 × 10^–1T, q = 1.6 × 10^–19C, K.E = 32KeV.

Velocity of projection : = (1/2) × 39 × (1.6 × 10^–27) v² = 32 × 10³ × 1.6 × 10^–27 => v = 4.050957468 × 10⁵

Through out ht emotion the horizontal velocity remains constant.

[Time taken to cross the magnetic field]

Accln. In the region having magnetic field = qvB/m

5193.535216 × 10⁸m/s²

V(in vertical direction) = at = 5193.535216 × 10⁸ × 24 × 10^–9 = 12464.48452m/s.

Total time taken to reach the screen = 

Time gap = 2383 × 10^–9 – 24 × 10^–9 = 2358 × 10^–9 sec.

Distance moved vertically (in the time) = 12464.48452 × 2358× 10^–9 = 0.0293912545 m

V² = 2as => (12464.48452)² = 2 × 5193.535216 × 10⁸ × S => S = 0.1495738143 × 10^–3m.

Net displacement from line = 0.0001495738143 + 0.0293912545 = 0.0295408283143m

For K – 41 : (1/2) × 41 × 1.6 × 10^–27v = 32 × 10³ 1.6 × 10^–19 => v = 39.50918387 m/s.

= 4818.193154 × 10⁸m/s²

t = (time taken for coming outside from magnetic field)

V = at (Vertical velocity) = 4818.193154 × 10⁸ × 10⁸ 25 × 10^–9 = 12045.48289m/s.

(Time total to reach the screen)

Time gap = 2442 × 10^–9 – 25 × 10^–9 = 2417 × 10^–9

Distance moved vertically = 12045.48289 × 2417 × 10^–9 = 0.02911393215

Now, V² = 2as => (12045.48289)² = 2 × 4818.193151 × S => S = 0.0001505685363 m

Net distance travelled = 0.0001505685363 + 0.02911393215 = 0.0292645006862

Net gap between K– 39 and K– 41 = 0.0295408283143 – 0.0292645006862

= 0.0001763276281 m ≈ 0.176mm