Nuclear equations L3

^223₉₄Pu -> ^215₉₀Th + Z + Z

• Z = alpha

• Z = ^8₄Be

• Z = beta

• Z = positron

The binding energy per nucleon of Pm^147₆₁ is

• ((61(protons) + 147(neutrons)) - Pm mass)c²

• ((61(neutrons) + 86(protons)) - Pm mass)/147

• ((61(protons) + 86(neutrons)) - Pm mass)c²/147

• (61(protons) + 86(neutrons) - Pm mass)c²/147

The binding energy per nucleon of Pm^147₆₁ is

• ((61(1.672623 x 10^-27) + 147(1.674929 x 10^-27)) - 243.906111 x 10^-27)c²

• ((61(1.672623 x 10^-27) + 86(1.674929 x 10^-27)) - 243.906111 x 10^-27)c²/147

• ((61(243.906111 x 10^-27) + 86(1.674929 x 10^-27)) - 243.906111 x 10^-27)/147

• 61(1.672623 x 10^-27) + 86(1.674929 x 10^-27) - 243.906111 x 10^-27)c²/147

What is the mass defect for the K -> Ca + e + v decay

• Ca mass+ electron mass+ neutrino mass - K mass

• K mass - electron mass- neutrino mass - Ca mass

• Ca mass -e mass - neutrino mass - Ca mass

• K mass + electron mass+ neutrino mass - Ca mass

What is the mass defect for the K -> Ca + e + v decay

• 66.34121 x 10^-27+ 0.000911 x 10^-27+ 0 - 66.34446 x 10^-27

• (66.34446 x 10^-27+ 0.000911 x 10^-27+ 0) - 66.34121 x 10^-27

• ((66.34121 x 10^-27+ 0.000911 x 10^-27+ 0) - 66.34446 x 10^-27)c²

• (66.34121 x 10^-27+ 0.000911 x 10^-27+ 0) - 66.34446 x 10^-27

What is the mass defect for the K -> Ca + e + v decay

• 4.161 x 10 ^-13 kg

• -2.339 x 10 ^-30 kg

• -2.1051 x 10^-13 kg

Mass defect = final - initial

A negative mass defect means

• energy was absorbed

  endothermic

• energy was released

  exothermic

• you made a mistake

• you created antimatter

A nuclear accelerator can accelerate a proton to 700 000 V.

a) What speed is it travelling at, when it hits the target? charge of proton = 1.602 x 10^-19 C

b) What is the mass equivalent of this amount of kinetic energy?

mass of proton = 1.67262 × 10^-27 kg

c) Is the accelerated proton heavier, lighter or the same mass as a rest proton? Calculate the mass of the accelerated proton.

Answer explanation

a) kinetic Energy gained = 1/2mv² = the potential energy provided by the accelerator = qV

therefore v = (2qV/m)^0.5 =

b) change in mass of proton= Energy of proton/c² = qV/c²

c) This change in mass, makes the proton heavier the faster it goes, but it is a small amount of mass increase