This discussion is now closed.

Check out other Related discussions

Radioactivity???

Ugh. I hate these questions...really hoping someone can help me out here!

The activity of a piece of radioactive material is 4.3 x 10^5 Bq at time t=0. The number of undecayed atoms in the material at time t=0 is 7.9 x 10^15. Calculate:
a) i) the activity after 4.0 half lives have elapsed
ii) the number of undecayed atoms after 4.0 half lives (4 marks)

b) the decay constant (3 marks)

c) the half life (2 marks)

And all I can say is UGH!!!

Reply 1

habosh

13

*starbuck*

Ugh. I hate these questions...really hoping someone can help me out here!

The activity of a piece of radioactive material is 4.3 x 10^5 Bq at time t=0. The number of undecayed atoms in the material at time t=0 is 7.9 x 10^15. Calculate:
a) i) the activity after 4.0 half lives have elapsed
ii) the number of undecayed atoms after 4.0 half lives (4 marks)

b) the decay constant (3 marks)

c) the half life (2 marks)

And all I can say is UGH!!!

a)i)2.3x 10^5/16
7.9x10^15-(7.9 x10^15/16)
b)lambda =A/N

c)t=ln2/lambda

Reply 2

Golden Maverick

4

The activity of a piece of radioactive material is 4.3 x 10^5 Bq at time t=0. The number of undecayed atoms in the material at time t=0 is 7.9 x 10^15. Calculate:
a) i) the activity after 4.0 half lives have elapsed
(4.3 x 10^5) x 0.5^4.0 = 1.687.. x 10^4

ii) the number of undecayed atoms after 4.0 half lives
(4 marks)
(7.9 x 10^15) x 0.5^4.0 = 4.937.. x 10^14

b) the decay constant (3 marks)
lambda (decay constant) = A/N
= (4.3 x 10^5) / (7.9 x 10^15)
= 5.443.. x 10^-11

c) the half life (2 marks)
half life = ln(2) / lambda
= 0.6931.. / (5.443.. x 10^-11)
= 1.273.. x 10^10 seconds

:smile:

Reply 3

habosh

13

Golden Maverick

The activity of a piece of radioactive material is 4.3 x 10^5 Bq at time t=0. The number of undecayed atoms in the material at time t=0 is 7.9 x 10^15. Calculate:
a) i) the activity after 4.0 half lives have elapsed
(4.3 x 10^5) x 0.5^4.0 = 1.687.. x 10^4

ii) the number of undecayed atoms after 4.0 half lives
(4 marks)
(7.9 x 10^15) x 0.5^4.0 = 4.937.. x 10^14b) the decay constant (3 marks)
lambda (decay constant) = A/N
= (4.3 x 10^5) / (7.9 x 10^15)
= 5.443.. x 10^-11

c) the half life (2 marks)
half life = ln(2) / lambda
= 0.6931.. / (5.443.. x 10^-11)
= 1.273.. x 10^10 seconds

:smile:

does that represent the number left after 4 halflives..oh dumb me :redface:

Reply 4

Golden Maverick

4

habosh

does that represent the number left after 4 halflives..oh dumb me :redface:

I think so, I don't quite see what you've done there :confused:

I might be wrong, been a while since I did this.

Reply 5

*starbuck*

OP

1

Aaah yep I follow b and c thank you! :smile:

For a though, I did divided by 2^4 rather than multiply by 0.5^4 but it should give the same answer? It does for part ii) but I get 2.6875 x 10^4 for part i). :confused:

Reply 6

habosh

13

*starbuck*

Aaah yep I follow b and c thank you! :smile:

For a though, I did divided by 2^4 rather than multiply by 0.5^4 but it should give the same answer? It does for part ii) but I get 2.6875 x 10^4 for part i). :confused:

yeah cause he raised the whole 1/2 to the power 4 I guess you are right,it's been long time since I'vedone this too,I'm just trying to have a little practice for synoptics :frown:

Reply 7

Golden Maverick

4

*starbuck*

Aaah yep I follow b and c thank you! :smile:

For a though, I did divided by 2^4 rather than multiply by 0.5^4 but it should give the same answer? It does for part ii) but I get 2.6875 x 10^4 for part i). :confused:

Sorry, typo. I did mean 2.6875 x 10^4 :redface:

Reply 8

*starbuck*

OP

1

Golden Maverick

Sorry, typo. I did mean 2.6875 x 10^4 :redface:

Hehe no worries...thank you!!! Now I have no excuse to avoid doing my Biology report. :rolleyes:

Reply 9

Jason_Hyde

This is a homework question and i can not get my head round it!!! :confused:

If anyone can explain it to me i would be very greatful.
Here it is...

Patassium-42 decays with a half-life of 12 hours. when potassium-42 decays it emits beta- particles and gamma rays. one freshly prepared source has an activity of 3.0x10^7 Bq.

(a) to determine the dose recieved by a scientist working with the source the number of gamma ray photons incident on each cm^2 of the body has to be known.
one in every five decaying nuclei produces a gamma ray photon. a scientist is intially working 1.50m form the fresh source with no shielding. show that at this time approximately 21 gamma ray photons per second are incident on each cm^2 of the scientist's body. (2 marks)

(b) the scientist returns 6 hours later and works at the same distance from the source.
(i) calculate the new number of gamma ray photons incident per secound on each cm^2 of the sceintist's body. (3 marks)

(ii) at what distance from the source could the scientist now work and recieve the original dose of 21 photons per cm^2. (2 marks)

(c) explain why it is not necessary to consider the beta particle emission when determining the dose of radiation the scientist recieves. (2 marks)

Many thanks to anyone who can help me!!!

Radioactivity???

Related discussions

Latest

Trending

Trending

Articles for you