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A proton and an α – particle are accelerated with same potential difference and they enter in the region of constant magnetic field B perpendicular to the velocity of particles. Find the ratio of radius of curvature. Of proton to the radius of curvature of α – particle

A proton and an α – particle are accelerated with same potential difference and they enter in the region of constant magnetic field B perpendicular to the velocity of particles. Find the ratio of radius of curvature. Of proton to the radius of curvature of α – particle 

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
6 years ago
Hello Student,
Please find the answer to your question
. KEY CONCEPT :
eV = ½ mv 2p ­ and eV = ½ mv2
V is is the potential difference
Vp = velocity of proton
v = velocity of ∝ - particle
m = mass of proton, mass of ∝ - particle = 4 m
⇒ vp = √ 2eV / m, v = √2eV / 4m
Now when the particles enter in magnetic field, the force on proton is
ev p B = mv2p/ rp or rp = mvp / eB ⇒ r= m / eB
√2eV / m = 1/B √ 2mV / e and r = 1/B √4mV /e
∴ rp / r = 1/ √2
Thanks
Deepak patra
askIITians Faculty

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