Fe^{+} ions are accelerated 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.

#### Solution

Given:

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

Strength of the homogeneous magnetic field, *B* = 20.0 mT = 20 × 10^{−3} T

Mass numbers of the two isotopes are 57 and 58.

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

We know that the radius of the circular path described by a particle in a magnetic field,

`r = (mv)/(qB)`

For isotope 1,

`r_1= (m_1v_1)/(qB)`

For isotope 2,

`r_2 = (m_2v_2)/(qB)`

⇒`(r_1)/(r_2)=(m_2v_1)/(m_2v_2)`

As both the isotopes are accelerated via the same potential V, the K.E gained by the two particles will be same.

`qV = 1/2 m_1v_1^2 = 1/2m_2v_2^2`

`(m_1)/(m_2) = (v_1^2)/(v_2^2)`

⇒ `(r_1)/(r_2) = ((m_1)/(m_2))^{3/2}`

Also, r_{1} = `(mv_1)/(qB)`

= `(m_1 sqrt(2qv)/m_1)/(qB)`

= `1/B sqrt ((2m_1V)/q)`

= `sqrt(1000xx57xx1.6xx10^-27)/sqrt (1.6 xx 10^-19 xx 20xx10^-3)`

= `1.19 xx 10^-2m = 119 cm`

For the second isotope:

`As (r_1)/r_2 = ((m_1)/(m_2))^{3/2}`

`r_2 = ((m_2)/(m_1))^{3/2} r_1`

= `(58/57)3/2 xx 119 cm`

= 120 cm