Q. 415.0( 1 Vote )

# Fe^{+}ions are acceleration through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A(1.6 × 10^{–27}) kg where A is the mass number.

Answer :

Given-

Potential difference through which the Fe^{+}, *V =* 500 V

Strength of the homogeneous magnetic field

*B* = 20.0 mT= 20 × 10^{−3} T

Mass numbers of the two isotopes ,

m_{1} =57 and m_{2} =58.

Mass of an ion = *A* (1.6 × 10^{−27}) kg

The radius of the circular path described by a particle in a magnetic field,

where,

*m* is the mass of a proton

v= velocity of the particle

B = magnetic force

q= charge on the particle = C

For calculating the radius of isotope 1-

For isotope 2,

Since isotopes are accelerated from the same potential V, the Kinetic energy gained by the two particles will be same for both the particles.

We, know the force developed by potential difference –

where

v = applied potential

q=charge

Also,we have

From (1)

For calculating the radius of 2 isotope-

From (2)

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