Kangaroo Posted May 13, 2013 Posted May 13, 2013 Hi there, I am hoping someone could help me through this: Q) During the radioactive decay of cobalt-60, a gamma ray with 1.33 MeV of energy is released. Calculate the decrease in mass (in kg) of the cobalt-60 nucleus as a result of this emission. A) 2.36*10^-30 I will ask the teacher tomorrow and reply back here with the strategy if no one knows this. Thank you.
swansont Posted May 13, 2013 Posted May 13, 2013 Hi there, I am hoping someone could help me through this: Q) During the radioactive decay of cobalt-60, a gamma ray with 1.33 MeV of energy is released. Calculate the decrease in mass (in kg) of the cobalt-60 nucleus as a result of this emission. A) 2.36*10^-30 I will ask the teacher tomorrow and reply back here with the strategy if no one knows this. Thank you. What have you done so far? We're not going to do you HW for you, but we'll help you figure it out.
Kangaroo Posted May 13, 2013 Author Posted May 13, 2013 What have you done so far? We're not going to do you HW for you, but we'll help you figure it out. Sure, I know E=MC^2 has something to do with it. As I have solved the joules for the mass of 3.48*10^-28kg by doing 3.1*10^-11/1.602*10^-13=193.5 MeV So I thought If I rearrange the equation to E*C^2=M however that seems not to work. Thank you.
pwagen Posted May 13, 2013 Posted May 13, 2013 (edited) When you use E=mc^2, do you convert the energy to joule? Edit: Sorry, I can't read. Edited May 13, 2013 by pwagen
swansont Posted May 13, 2013 Posted May 13, 2013 Sure, I know E=MC^2 has something to do with it. As I have solved the joules for the mass of 3.48*10^-28kg by doing 3.1*10^-11/1.602*10^-13=193.5 MeV So I thought If I rearrange the equation to E*C^2=M however that seems not to work. Thank you. I don't see where 193.5 MeV comes into it. You have a 1.33 MeV change in energy. You need to convert that to a mass difference.
pwagen Posted May 13, 2013 Posted May 13, 2013 So I thought If I rearrange the equation to E*C^2=M however that seems not to work.I think you'd need to rearrange it a bit differently. Something like [latex]m=E/c^2[/latex].
Kangaroo Posted May 14, 2013 Author Posted May 14, 2013 I think you'd need to rearrange it a bit differently. Something like [latex]m=E/c^2[/latex]. You are correct thank you.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now