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.

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.

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.

When you use E=mc^2, do you convert the energy to joule?

 

Edit: Sorry, I can't read.

Edited May 13, 201312 yr by pwagen

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.

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].

I think you'd need to rearrange it a bit differently. Something like [latex]m=E/c^2[/latex].

You are correct thank you.