Question

Rutherford studied alpha particles using the crossed-field technique Thomson had invented to study cathode rays. Assuming that $V_{alpha\approx }{V}_{cathoderay}$ (which turns out to be true), would the deflection of an alpha particle by a magnetic field be larger than, smaller than, or the same as the deflection of a cathode-ray particle by the same field? Explain.

Short answer

The deflection of the alpha particle is larger than cathode ray particle.

Step-by-step solution

Given information

We have give that B is same for both the particles and V is assumed to be same for both the particles.

Equation used for reason.

As. we know from the concept of electromagnetism.

R = $\frac{mv}{qB}$$--\left(1\right)$and F = $qvB$$--\left(2\right)$

Step 3:From  (1)  We can see that the radius depends on ratio of mass to charge (). The alpha particle has a charge of 2e and mass is 4u (where e is the charge of an electron and u is atomic mass unit) .For alpha particle  ()  =  = 2  --(3)For cathode ray particle ()  = = --(4)From (3) and (4) we can see that the ratio for alpha particle is greater as compared to cathode ray. So the alpha particle will follow a larger deflection --(5)As; F =qvB.So the charge for alpha particle is more as compared to the cathode ray . So alpha particle will experience a larger amount of force  --(6)

So from $\left(5\right)$and $\left(6\right)$;

We can conclude that alpha particle will undergo a larger force and will follow a large trajectory path with bigger radius And because of this it will deflect more