Question

Question 5 3 pts What is the final nuclide that results from a succession of two alpha emissions, followed by three beta emissions, beginning with 238 U ? 246 230 Am 95 230 Pa 91 91 249 ES 99

Answer 1

In nuclear decay, an alpha-particle is equivalent to a Helium atom. That is has a charge of +2 and mass of +4

So when an atom loses an alpha-particle, its mass decreases by 4 units and the atomic number decreases by 2 units.

So on loss of an alpha-particle, we get the atom with 2 unit less atomic number.

Similarly a beta-particle is equivalent to an electron with charge of -1 and mass 0

So on losing a beta-particle, the atomic number of the newly formed element increases by 1 unit while the mass remains the same.

What ever be the decay, thg e final element formed in the product will be as per the atomic number formed.

So for the given decay of 92 U 238-

238 + 3 216 + 1) (1) 90 atomer no. the total decrease is moon is = 42 - (2x) -(ex-) 392- 4+3 E 238 - 92 lée new element och atomerno-91) And total change is mass is (zry)-(3x0) 2.238-8 7230 (eeme element sit many 200) so the element with atomer no 91 and mana 230 es Postacle norim 220 pa - Then - Do 298 220 P + 2 (22") + 3 4 920