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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. - Physics

Sum

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.

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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, r1 = `(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

Concept: Motion in Combined Electric and Magnetic Fields - Cyclotron
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 12 Magnetic Field
Q 41 | Page 233
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