Ions having equal charges but masses of M and 2M are accelerated through the sam

Question

Ions having equal charges but masses of M and 2M are accelerated through the same potential difference and then enter a uniform magnetic field perpendicular to their path. If the heavier ions follow a circular arc of radius R, what is the radius of the arc followed by the lighter?

I know the answer is : R/2 (R/sqrt2)

Someone has already asked this same question, and it has been answered correctly. I do not understand the explanation that they gave though, which is what I need help with.

The above is the answer that was given (it was already cropped so part of the answer was cut off). Please explain the steps and formulas to me, and how r (is proportional to?) sqrt(m) means the answer is R/sqrt(2). I understand how they solved for v in the first equation, but not how substituting that for v into the r = mv/qB means r sqrt(m) => R/sqrt(2). I would have just asked as a comment on the same page, but I am unable to comment on other people's questions for some reason. Thanks for your help.

Explanation / Answer

r = mv/qB

r = (m/qB)*sqrt(2qV/m)

r = sqrt(2qV*m^2/(q^2*B^2*m))

r = sqrt(2Vm/(qB^2))

now V, B and q are constant so

r = k*sqrt(m)

r is proportional to sqrt m

which gives

r1/r2 = sqrt(m1/m2)

if r1 = R

then r2 = R*sqrt(m2/m1)

r2 = R*sqrt(M/2M)

r2 = R*sqrt(1/2)

r2 = R/sqrt(2)

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