Question

Draw the decay scheme of 42 K (Z=19), labeling the energies and probabilities (percentages) for each...

Draw the decay scheme of 42 K (Z=19), labeling the energies and probabilities (percentages) for each transition, using the information o f the Table below. The mass difference Δ of 42 K is - 35.02 MeV; the mass difference Δ of 42 C a (Z=20) is - 38.54 MeV.
Type of decay Half-life Major radiations, Energies (MeV), and frequency per disintegration (%)
β- 12.4 hr
β-: 3.52 max (82%), 2.00 max (18%)

: 1.52(18%)

(b) [4 marks] Calculate the activity of a 30 mCi source of 42K after 2.5 days.

0 0
Add a comment Improve this question Transcribed image text
Answer #1

a) K 42 is characterised by the emission of several beta particles with different end point energies.

The decay scheme of a radio active subtance is a graphical represention of all the transitions occuring in a decay,and their relationships.where the co-ordinate axis is energy increasing from bottom to top,and the abscissa is the proton number increasing from left to right,the arrows indicates the emitted particles.

194 40 (12.4 hr) +35.02 3.52 (824) Mev - 38.54 Meu 2.00 (18 20 Ca 42 (stable) De cay scheme f De Cay scheme a 19k 42 and 422=

b) Equation for calculating the activity of a radio active substance is

r= activity rate × 3.700×10^10 10atoms/s/ci

r=30mci × 3.700×10^10 10

=1.11×10^12 10

Therefore, r= 1.11×10^12 10 atoms/s/ci .

Add a comment
Know the answer?
Add Answer to:
Draw the decay scheme of 42 K (Z=19), labeling the energies and probabilities (percentages) for each...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT