1.

A sample of radioactive material has an alpha emitter (X) and a beta emitter (D). The energy of decay reactions are Q_(1) and Q_(2) respectively. ._(Z)^(A)Xrarr._(Z-2)^(A-4)Y+._(2)^(4)He+Q_(1) ._(Z')^(A')Drarr._(Z'+1)^(A')B+._(-1)^(0)e+vecv+Q_(2) The radiation coming out from the sample is passed through two parallel slits S_(1)and S_(2) to get a narrow parallel beam to alpha and beta particles. This beam is allowed to enter perpendicularly into a region of inform magnetic field. The particles after taking a semicircular path strike the screen at a point a above the line L and, below the line L particles are found to strike the screen everwhere from O to b. the distance Oa and Ob and r_(1) and r_(2) respectively. Find the ratio (r_(1))/(r_(2)). Take (Q_(1))/(Q_(2))=k and assume that mass of alpha particles is 4 time the mass of a proton which is eta times the mass of an electron.

Answer»


Answer :`(r_(1))/(r_(2))=sqrt(pik((A-4)/(A)))`


Discussion

No Comment Found

Related InterviewSolutions