Question

A narrow beam of singly charged potassium ions of kinetic energy 32 keV is injected into a region of width 1.00 cm with a magnetic field of strength 0.500 T, as shown in the figure. The ions are collected at a screen 95.5 cm 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 × 10−27) kg, where A is the mass number.
Figure

Answer 1

The separation between the points where these isotopes strike the screen is 67mm.

Explanation:

Single charged potassium ions’ kinetic energy = 32 keV

Magnetic region’s width = 1 cm

Strength of the magnetic field, B = 0.500 T

Distance between the region and the screen = 95.5 cm

Two isotopes atomic weights are 39 and 41.

Potassium ion mass = A (1.6 × $10^{-27}$) kg

Potassium ion that is singly-charged K-39 :

K-39 mass= 39 × 1.6 × $10^{-27}$ kg,

Charge, q = 1.6 × $10^{-19}$ C

Step 1:

According to the information provided, the singly-charged potassium ions has the narrow beam injected into a magnetic field region as

K.E = 32 keV

v = 4.05 × $10^{5}$

Throughout the motion, we observe that the horizontal velocity is persistent.

So, cross the magnetic field, the time taken, t = dv = 24.7 × $10^{-9}$ s

Step 2:

Now, in the magnetic field region, the acceleration:

F = qvB = ma

a = 5192 × $10^{8}$ $\frac{m}{s^2}$

The vertical direction velocity, $V_y$  = at

= 12824.24 $\frac{m}{s}$

To reach the screen, the time taken is

= $\frac{d}{v}=\frac{0.955}{4.05 \times 10^{5}}$

In this time, the vertically moved distance

= $V_y$ × t

= 3023.95×$10^{-5}$ m

Step 3:

Particle travels a vertical distance inside the magnetic field. This can be seen by using equation of motion

$v^{2}$ = 2aS

⇒ 15.83×$10^{-5}$ = S

Net display from line

= 3039.787 × $10^{-5}$ m.

For the potassium ion K-41 :

⇒ v = 3.94 × $10^{5}$ $\frac{m}{s}$

Step 4:

Likewise the  acceleration, is given as

a = 4805 × $10^{8}$ $\frac{m}{s^2}$

t = for the existing magnetic field, the time taken is = 25.4 × $10^{-9}$ sec.

$V y_{1}$= 12204.7× $10^{-9}$ $\frac{m}{s}$

To reach the screen, the time taken is 2423× $10^{-9}$ s.

Vertically moved distance = 2957.1× $10^{-5}$

Step 5:

The vertical distance moved inside magnetic field by the particle can be observed by using motion equation.

$V^{2}$ = 2aS

⇒ S = 15.49 × $10^{-5}$ m

Travelled net distance = 2972.68× $10^{-5}$ m.

Net gap between K-41 and K-39

=  67 mm.