2Is wcC Fa11 2037 6. What quantity of energy would be 234.0409 annu (3.886 x 10-

Question

2Is wcC Fa11 2037 6. What quantity of energy would be 234.0409 annu (3.886 x 10-22 g produced as one atom of plutonium-238 undergoes alpha decay? amu (3.953 x 10g) and the nuclide mass of uranium-234 is 8). Alpha particle mass is 6.64465 x 1024 g. The speed of light is 2.998 x1o particle ma b.5.0 x10-12 j c.7.0 x1010 d.2.6x 10*1.1 x10 probable mode of radioactive decay for the radioactive nuclide . Which of the following is the most 40 C1? (2) a. beta emission b. gamma emission c. neutron emission d alpha emission e.position emission 235 o2 Uabsorbs a neutron, fission occ n, fission occurs. One possible fission pathway is s When 0

Explanation / Answer

Ans. #6. Total loss in mass during decay = Mass of 238P – Mass of 234U

                                                = 3.953 x 10-22 g - 3.886 x 10-22 g

                                                = 6.700 x 10-24 g

The total loss in mass in in form in form of alpha particle (not converted into energy) plus the mass converted into energy.

Now,

Net loss in mass (that converted into energy) = Total loss in mass – Mass of a-particle

                                                = 6.700 x 10-24 g – 6.64465 x 10-24 g

                                                = 5.5350 x 10-26 g                                         ; [1 g = 10-3 kg]

                                                = 5.5350 x 10-29 kg

# The conversion of mass into energy is given by-

            E = mc2          - equation 1

                        Where, E = energy in joule ; m = mass loss in kg ; c = speed of light

Putting the values in equation 1-

Or, E = (5.5350 x 10-29 kg) x (2.998 x 108 m s-1)2

            Or, E = 4.975 x 10-12 J

Hence, correct option is – b. 5.0 x 10-12 J

#7. Correct option. a. beta emission.

40Cl17 ---- b emission ---------> 40Cl18

Atomic number = number of protons = 17

Atomic mass = Number of protons + Number of neutron

So, number of neutrons = Atomic mass – Atomic number = 40 -17 = 23

            Difference in number of p and n = 6

# Note that, number of neutrons (23) >> number of protons (17).

When number of neutrons is relatively much higher than that of proton, the nuclide is unstable. So, the nuclide generally undergo b-emission that converts a neutron into a proton plus electrons. The overall atomic number is increased by 1 but mass number remains constant.

Number of neutrons in 40Ar18 = 22

            Difference in number of p and n = 4

The resultant nuclide is stable because lesser n/p ratio than it was in in 40Cl

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