ATAR Notes: Forum

HSC Stuff => HSC Science Stuff => HSC Subjects + Help => HSC Physics => Topic started by: louisaaa01 on November 09, 2019, 08:45:23 pm

Title: Energy Released when Forming Helium
Post by: louisaaa01 on November 09, 2019, 08:45:23 pm: Hi,

A few of my friends and I were discussing the attached question and we're a bit confused as to how to find the mass defect for part b.

The worked solution (attached) says that four protons combine to form a helium nucleus, and calculate the mass defect and thus energy released accordingly. However, four protons don't combine to form just a helium nucleus - to balance the equation it appears that two positrons should also be released, so we don't know whether this should also be included in the mass defect equation.

Alternatively, we've been taught that mass defect is the mass of the constituents - the mass of the nucleus, which seems to suggest that we do the mass of 2 protons and 2 neutrons - the mass of the nucleus, then calculate energy using E=mc². However, this renders a different answer.

Which approach is correct?

Thank you.
Title: Re: Energy Released when forming Helium
Post by: not a mystery mark on November 09, 2019, 09:29:48 pm: Quote from: louisaaa01 on November 09, 2019, 08:45:23 pm
Hi,

A few of my friends and I were discussing the attached question and we're a bit confused as to how to find the mass defect for part b.

The worked solution (attached) says that four protons combine to form a helium nucleus, and calculate the mass defect and thus energy released accordingly. However, four protons don't combine to form just a helium nucleus - to balance the equation it appears that two positrons should also be released, so we don't know whether this should also be included in the mass defect equation.

Alternatively, we've been taught that mass defect is the mass of the constituents - the mass of the nucleus, which seems to suggest that we do the mass of 2 protons and 2 neutrons - the mass of the nucleus, then calculate energy using E=mc². However, this renders a different answer.

Which approach is correct?

Thank you.

I feel the question is a bit vague when compared to their sample response. Logically the two positrons are from the initial hydrogen nuclei so I'd naturally think it's safer to assume that you include the two positrons in the calculation - but the question even acknowledges this by saying no marks penalised for whatever option. I think the HSC will make sure they're more specific (hopefully lol).

This was my process:
$4_{1}^{1}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+2_{+1}^{0}\textrm{e}$
$m_{defect} = 4\cdot m_{hydrogen}\ - m_{helium}\ -2\cdot m_{positron}$
$m_{defect} = (4\cdot 1.673\times 10^{-27}) - (6.648\times 10^{-27}) -2(\cdot 9.109\times 10^{-31})$
$m_{defect} = 4.22\times 10^{-29}\ kg$

Idk if this helps or not, but that's my thinking when doing this question.
Title: Re: Energy Released when forming Helium
Post by: louisaaa01 on November 09, 2019, 09:46:27 pm: Quote from: not a mystery mark on November 09, 2019, 09:29:48 pm
I feel the question is a bit vague when compared to their sample response. Logically the two positrons are from the initial hydrogen nuclei so I'd naturally think it's safer to assume that you include the two positrons in the calculation - but the question even acknowledges this by saying no marks penalised for whatever option. I think the HSC will make sure they're more specific (hopefully lol).

This was my process:
$4_{1}^{1}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+2_{+1}^{0}\textrm{e}$
$m_{defect} = 4\cdot m_{hydrogen}\ - m_{helium}\ -2\cdot m_{positron}$
$m_{defect} = (4\cdot 1.673\times 10^{-27}) - (6.648\times 10^{-27}) -2(\cdot 9.109\times 10^{-31})$
$m_{defect} = 4.22\times 10^{-29}\ kg$

Idk if this helps or not, but that's my thinking when doing this question.

Thanks for your response! It's good being able to clarify this method.

Do you have any idea why we don't use the traditional mass of constituent nuclides - mass of nucleus for this question?