1. (a) Calculate BE/ A , the binding energy per nucleon, for 18 O. This is the r

Question

1. (a) Calculate BE/A, the binding energy per nucleon, for 18O. This is the rarer of the two most common oxygen isotopes.
1 MeV/nucleon
(b) Repeat for 16O. Most of oxygen is 16O. Note that 16O has an equal number of  protons and neutrons.
2 MeV/nucleon
Note: This difference in BE/A values is sufficiently large to explain the different isotope abundances. Question Part Points Submissions Used 1 1. (a) Calculate BE/A, the binding energy per nucleon, for 18O. This is the rarer of the two most common oxygen isotopes.
1 MeV/nucleon
(b) Repeat for 16O. Most of oxygen is 16O. Note that 16O has an equal number of  protons and neutrons.
2 MeV/nucleon
Note: This difference in BE/A values is sufficiently large to explain the different isotope abundances.
1. (a) Calculate BE/A, the binding energy per nucleon, for 18O. This is the rarer of the two most common oxygen isotopes.
1 MeV/nucleon
(b) Repeat for 16O. Most of oxygen is 16O. Note that 16O has an equal number of  protons and neutrons.
2 MeV/nucleon
Note: This difference in BE/A values is sufficiently large to explain the different isotope abundances.
1. (a) Calculate BE/A, the binding energy per nucleon, for 18O. This is the rarer of the two most common oxygen isotopes.
1 MeV/nucleon
(b) Repeat for 16O. Most of oxygen is 16O. Note that 16O has an equal number of  protons and neutrons.
2 MeV/nucleon
Note: This difference in BE/A values is sufficiently large to explain the different isotope abundances.
1. (a) Calculate BE/A, the binding energy per nucleon, for 18O. This is the rarer of the two most common oxygen isotopes.
1 MeV/nucleon
(b) Repeat for 16O. Most of oxygen is 16O. Note that 16O has an equal number of  protons and neutrons.
2 MeV/nucleon
Note: This difference in BE/A values is sufficiently large to explain the different isotope abundances.

Question Part Points Submissions Used 1 Question Part Points Submissions Used 1 Question Part Points Submissions Used 1 Question Part Points Submissions Used 1 Question Part Points Submissions Used 1 Question Part Points Submissions Used 1 Question Part Points Submissions Used 1 1 Question Part Points Submissions Used

Explanation / Answer

1) 180 has 10 neutrons and 8 protons

dm = 10 x mass of neutron + 8 x mass of proton - mass on nucleus

dm = 10 x 1.008665 + 8 x 1.007276 - 17.999161

dm = 0.145697

BE = 0.145697 x 931.5

BE = 135.7167 MeV

BE/N = 135.7167 /10

BE/N = 13.57167

160 has 8 neutrons and 8 protons

dm = 8 x mass of neutron + 8 x mass of proton - mass on nucleus

dm = 8 x 1.008665 + 8 x 1.007276 - 15.9949146

dm = 0.1326134

BE = 0.1326134 x 931.5

BE = 123.5293821 MeV

BE/N = 135.7167 /8

BE/N = 15.44117

2) Given 9.2 MeV proton

Energy = 9.2 MeV

=9.2 x 1.6 x 10-13 J

= 14.72 x 10-13 J

so 1/2 mV2 = 14.72 x 10-13 J

V2 = 29.44 x 10-13 / 1.67 x 10-27

V=4.12 x 107 m/s

velocity of proton is 4.12 x 107 m/s

radius of curvature = mV/bq

R= 1.67 x 10-27 x 4.12 x 107 / 4 x 1.6 x 10-19

R= 10.75 cm

radius of curvature of proton is 10.75 cm

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