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.

Answer 1

Kinetic energy of singly-charged potassium ions = 32 keV
Width of the magnetic region = 1.00 cm
Magnetic field's strength, B = 0.500 T
Distance between the screen and the region = 95.5 cm
Atomic weights of the two isotopes are 39 and 41.
Mass of a potassium ion = A (1.6 × 10⁻²⁷) kg
For a singly-charged potassium ion K-39 :
Mass of K-39 = 39 × 1.6 × 10⁻²⁷ kg,
Charge, q = 1.6 × 10⁻¹⁹ C
As per the question, the narrow beam of singly-charged potassium ions is injected into a region of magnetic field.
As
K.E = 32 keV
$\frac{1}{2}m{v}^2=32\times {10}^3\times 1.6\times {10}^{-19}$
$\frac{1}{2}$ × 39 × (1.6 × 10⁻²⁷) × v² = 32 × 10³ × 1.6 × 10⁻¹⁹
v = 4.05 × 10⁵
We know that throughout the motion, the horizontal velocity remains constant.
So, the time taken to cross the magnetic field,
t = $\frac{d}{v}=\frac{0.01}{4.05\times {10}^5}$
= 24.7 × 10⁻⁹ s
Now, the acceleration in the magnetic field region:
F = qvB = ma
$$\begin{gathered} a=\frac{qvB}{m} \\ =\frac{1.6\times {10}^{-19}\times 4.05\times {10}^5}{39\times 1.6\times {10}^{-27}} \end{gathered}$$
= 5192 × 10⁸ m/s²
Velocity in the vertical direction, v_y = at
= 5193.53 × 10⁸ × 24.7 × 10⁻⁹
= 12824.24 m/s
Time taken to reach the screen
$=\frac{d}{v}=\frac{0.955}{4.05\times {10}^5}$
= 0.000002358 s.

Distance moved vertically in this time
= v_y × t
= 12824.24 × 2358 × 10⁻⁹
= 3023.95×10^-5 m
Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion
v² = 2aS
⇒ (12824.24)² = 2 × 5192 × 10⁸ × S
⇒ 15.83×10⁻⁵ = S
Net display from line
= 15.83×10⁻⁵+3023.95×10^-5
= 3039.787×10⁻⁵ m.
For the potassium ion K-41l :
$\frac{1}{2}$ × 41 × 1.6 × 10⁻²⁷ v²= 32 × 10³ × 1.6 × 10⁻⁹
⇒ v = 3.94×10⁵ m/s
Similarly, acceleration,
a = 4805 × 10⁸ m/s²
t = time taken for exiting the magnetic field
= 25.4 × 10⁻⁹ sec.
v_y1= at (vertical velocity)
= 4805 × 10⁸ × 25.4 × 10⁻⁹
= 12204.7× 10⁻⁹ m/s
Time to reach the screen
= 2423× 10⁻⁹ s.

Distance moved vertically
= 12204.7 × 2423 × 10⁻⁹
= 2957.1× 10⁻⁵
Now,
Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion
v² = 2aS
(12204.7)² = 2 × 4805 × 10⁸ S
⇒ S = 15.49× 10^-5 m
Net distance travelled
= 15.49× 10^-5 +2957.1× 10⁻⁵
= 2972.68× 10^-5 m.
Net gap between K-39 and K-41
= 3039.787×10⁻⁵− 2972.68× 10^-5
= 67 mm.