Given, mass of 1 st ion, m1=4amu Mass of 2 nd ion, m2=16amu Charge on 1 st ion, q1=+2e Charge on 2 nd ion, q2=+3e Kinetic energy of both ion is same, KE1=KE2 The radius of path traced by a charged particle in magnetic field is given as r=
√2mKE
qB
⇒r∝
√m
q
⇒
r1
r2
=√
m1
m2
×(
q2
q1
) ⇒
r1
r2
=√
4
16
×(
+3e
2e
) =
1
2
×
3
2
=
3
4
⇒r1=
3
4
r2 ⇒r1<r2
Now, the deflection of ion in region with constant magnetic field as shown in figure is denoted by θ. From diagram, sinθ=
d
r
θ∝
1
r
[∵ d = constant] As, r1<r2 from above expression θ′1>θ2. Thus, deflection of lighter ion will be more than the deflection of heavier ion.