(1)α -ray particles make uniform circular motion in a uniform magnetic field. Let its radius be r, and according to Newton's second law, there is qvB=mv2r.

Substituting the data, the result is: r = 0.89 m.

(2) (1) No.2 trace is facing the opening of the lead box with radioactive source, indicating that the ray forming No.2 trace moves in a straight line at a uniform speed, that is, it is not affected by electric field force and is not charged, so No.2 trace is formed by γ -ray irradiation.

(2) When α -ray particles come out of the radioactive source and pass through a horizontal uniform electric field to make a negative film, they are subjected to a constant electric field force, which is perpendicular to the vertical initial velocity direction, so they should move in a uniform variable speed curve.

Let the vertical distance between the lead box and the negative film be d, the electric field intensity be e, the initial velocity of charged rays when they are emitted from the radioactive source be v0, the mass be m, the amount of charge they carry be q, and the movement time in the electric field be t, then the movement of particles in the electric field is as follows:

Vertical direction d=v0t, horizontal lateral displacement x= 12qEmt2.

Solution: x=qEd22mv20

Therefore, the lateral displacement of alpha ray and beta ray is as follows.

xαxβ=qαqβ？ mβmα？ (vβvα)2 = 2 1× 14× 1840×( 10. 1)2 = 5 184

The results show that the deflection of alpha ray is very small, so the third trace should be formed by alpha ray.

Answer: (1) The radius of the arc trajectory formed by the alpha ray emitted by the radioactive source in this magnetic field is 0.89 m;

(2)( 1)γ -ray irradiation forms the No.2 trace;

(2) α particles do similar flat throwing motion after coming out of the lead box; The third exposure mark is formed by alpha-ray irradiation.