An α-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field B→, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them. ??
Answers
Answer:
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Explanation:
Mass of alpha particle = mα
Charge on alpha particle = qα
Velocity of alpha particle = vα
Mass of a proton = mp
Charge on a proton = qp
Velocity of proton = vp
Since the KE of alpha particle and proton are equal.
∴12mαvα2=12mp·vp2
⇒mαvα2mα=mp·vp2mp⇒mαvαmp·vp=mαmp12
∴12mαvα2=12mp·vp2
⇒mαvα2mα=mp·vp2mp⇒mαvαmp·vp=mαmp12
Radius of circular = R = mvqB
∴ Ratio of radii of alpha particle to proton
=RαRp=mα·vαqαB·qpBmp·vp=mαmp12·qpqαUsing mαmp=4 and qpqα=12∴RαRp=41212=1
follow me !
Mass of alpha particle = mα
Charge on alpha particle = qα
Velocity of alpha particle = vα
Mass of a proton = mp
Charge on a proton = qp
Velocity of proton = vp
Since the KE of alpha particle and proton are equal.
∴12mαvα2=12mp·vp2
⇒mαvα2mα=mp·vp2mp⇒mαvαmp·vp=mαmp12
∴12mαvα2=12mp·vp2
⇒mαvα2mα=mp·vp2mp⇒mαvαmp·vp=mαmp12
Radius of circular = R = mvqB
∴ Ratio of radii of alpha particle to proton
=RαRp=mα·vαqαB·qpBmp·vp=mαmp12·qpqαUsing mαmp=4 and qpqα=12∴RαRp=41212=1