Q. 42

# A narrow beam of

Given -

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
m1 =39 and m2 =41

.Mass of a potassium ion = A (1.6 × 10−27) kg

For a singly-charged potassium ion K-39, separation between the points can be calculated as follows –

Mass of K-39 = 39 × 1.6 × 10−27 kg

Charge, q = 1.6 × 10−19 C

Given in the question that the narrow beam of singly-charged potassium ions is injected into a region of magnetic field.

As, given that kinetic energy K.E is

K.E = 32 keV   We know that throughout the motion, the horizontal velocity remains constant.

So, the time taken to cross the magnetic field,   Now, the acceleration in the magnetic field region-    Now, velocity in the vertical direction can be found from Newton’s 2nd law as Since, initial velocity is 0, Substituting the values-  Time taken by the ion to reach the screen-   Now, distance moved by the ion vertically in this given time -   Vertical distance travelled by the particle inside magnetic field

can be calculated by using 3rd equations of motion – where,

v= final velocity

u=initial velocity

a= acceleration acting on the ion

S=distance travelled

Since initial velocity is zero,   Now, time taken t,  Now,for the potassium ion K-41   And, acceleration a is given by- t = time taken by the ion to exit the magnetic field is given by - From newton’s law velocity in vertical direction-   Time to reach the screen, t - .

Now distance moved in vertical direction is-  Now,
Vertical distance travelled by the particle inside magnetic field can be found out by using 3rd equation of motion given by where,

v= final velocity

u=initial velocity

a= acceleration acting on the ion

S=distance travelled

Since initial velocity is zero,   Net distance travelled by K-41 potassium ion  Net gap between K-39 and K-41 potassium ions is given by-

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