Based on the atomic masses given in the table below, take apart analpha particle

Question

Based on the atomic masses given in the table below, take apart analpha particle (4He nucleus), in sequence, by removing aproton, then a neutron, and finally separating the remaining protonand neutron. Calculate the energy (work) required for eachstep.

4He 4.002 60 u 2H 2.014 10 u 3H 3.016 05 u 1H 1.007 83 u n 1.008 67 u - -
(a) Energy required to remove a proton
1 MeV
(b) Energy required to remove a neutron
2 MeV
(c) Energy required to separate the remaining proton andneutron
3 MeV
(d) For an alpha particle, what is the total binding energy?
4 MeV
(e) For an alpha particle, what is the binding energy pernucleon?
5 MeV
(f) Do your answers in parts (d) and (e) match any of the answersin parts (a), (b), or (c)?
6---Select---(b) matches (e) only.(a) matches (d) only.No. The answers do not match.Yes. (a) matches (d), and (b) matches(e).
Section 42-3 4He 4.002 60 u 2H 2.014 10 u 3H 3.016 05 u 1H 1.007 83 u n 1.008 67 u - -
Section 42-3 4He 4.002 60 u 2H 2.014 10 u 3H 3.016 05 u 1H 1.007 83 u n 1.008 67 u - -

Explanation / Answer

(a) Energy required to remove a proton is :       E1 = ( m3H + m1H -m4He ) C2              = (3.01605 + 1.00783 - 4.00260) 931.5 MeV / u              = 19.8 M eV (b)     E2  =   ( m2H + mn - m3H ) C2             = ( 2.01410 + 1.00867 - 3.01605) *931.5 M eV / u             = 6.26 MeV (c)     E3 =   ( mH + mn - m2H ) C2            = 2.23 MeV (d)         E  =   E1 +   E2  +   E3                = 19.8 M eV + 6.26  +    2.23 MeV                = 28.3 M eV (e)        binding energy pernucleon is :                Eb =   E  /A   = 28.3 / 4   = 7.07 M eV (f)       No, the answers do not match. Hope this helps u!                              = (3.01605 + 1.00783 - 4.00260) 931.5 MeV / u              = 19.8 M eV (b)     E2  =   ( m2H + mn - m3H ) C2             = ( 2.01410 + 1.00867 - 3.01605) *931.5 M eV / u             = 6.26 MeV (c)     E3 =   ( mH + mn - m2H ) C2            = 2.23 MeV (d)         E  =   E1 +   E2  +   E3                = 19.8 M eV + 6.26  +    2.23 MeV                = 28.3 M eV (e)        binding energy pernucleon is :                Eb =   E  /A   = 28.3 / 4   = 7.07 M eV (f)       No, the answers do not match. Hope this helps u!                       

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