A proton and an α – particle are accelerated with

Question

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

Deepak Patra · Accepted Answer

Hello Student,

Please find the answer to your question

. KEY CONCEPT :

eV = ½ mv ²_p and eV = ½ mv²_∝

V is is the potential difference

V_p = velocity of proton

v_∝ = velocity of ∝ - particle

m = mass of proton, mass of ∝ - particle = 4 m

⇒ v_p = √ 2eV / m, v_∝ = √2eV / 4m

Now when the particles enter in magnetic field, the force on proton is

ev _p B = mv²p/ r_p or r_p = mv_p / eB ⇒ r_∝= m / eB

√2eV / m = 1/B √ 2mV / e and r_∝ = 1/B √4mV /e

∴ r_p / r_∝ = 1/ √2

Thanks
Deepak patra
askIITians Faculty

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