Part A Calculate the mass defect of the fluorine nucleus 19 9F. The mass of neut

Question

Part A

Calculate the mass defect of the fluorine nucleus 19  9F. The mass of neutral 19  9F is equal to 18.998403 atomic mass units.

Express your answer in atomic mass units to four significant figures.

Part B

Calculate the binding energy E of the fluorine nucleus 19  9F (1eV=1.602×1019J).

Express your answer in millions of electron volts to four significant figures.

Part C

Calculate the binding energy per nucleon of the fluorine nucleus 19  9F.

Express your answer in millions of electron volts to four significant figures.

Part D

Calculate the mass defect of the helium nucleus 42He. The mass of neutral 42He is given by MHe=4.002603amu.

Express your answer in atomic mass units to four significant figures.

Part E

Calculate the binding energy E of the helium nucleus 42He (1eV=1.602×1019J).

Express your answer in millions of electron volts to four significant figures.

Part F

Calculate the binding energy per nucleon of the helium nucleus 42He.

Express your answer in millions of electron volts to four significant figures.

Explanation / Answer

Part A

The given isotope is 199F . Its atomic number is 9 and mass number is 19.

Atomic number is equal to number of protons. Therefore

p = 9.

Mass number = total protons + total neutrons.

19 = 9 + neutrons

n = 10.

Let us find calculated mass of the given isotope.

Calculated mass = Mass of total protons + Mass of total neutrons.

From standard reference table, we get mass of proton = 1.007276 amu and mass of neutron = 1.008665 amu

Let's plug in these values to find calculated mass

Calculated mass = 9 * 1.007276 + 10* 1.008665

Calculated mass = 9.065484 + 10.08665

Calculated mass = 19.152134 amu

Mass defect = Calculated mass - Actual mass

Mass defect = 19.152134 amu - 18.998403 amu

Mass defect = 0.1537 amu

_____________________________________________________________________________________________

Part B

binding energy can be calculated using following equation.

E = m* c2

Where E is the binding energy

m is mass defect

and c is velocity of light , 3.0 x 10^8 m/s

Let's convert mass defect from amu to kg unit.

0.1537 amu * 1.6605 x 10^-27 kg / 1 amu = 2.5522 x 10^-28 kg

E = 2.5522 x 10^-28 kg * ( 3.0 x 10^8 m/s)^2

E = 2.5522 x 10^-28 kg * 9 x 10^16 m^2/s^2

E = 2.29698 x 10^-11 kg m^2/s^2 .................... but kg m^2/s^2 = J

E = 2.297 x 10^-11 J

Let's convert this to eV using the given conversion factor

E = 2.297 x 10^-11 J * 1 eV / 1.602 x 10^-19 * 1 MeV / 10^6 eV = 1.434 x 10^2 MeV

Binding energy of the Fluorine nucleus is 143.4 MeV

_________________________________________________________________________________________

c)

Total nucleons for the given atom are 9 protons + 10 neutrons = 19 nucleons.

Binding energy per nucleon = Binding energy of the nucleus / total nucleons

Binding energy per nucleon = 143.4 MeV / 19 nucleons

Binding energy per nucleon = 7.547 MeV

____________________________________________________________________________________________

Part D

Calculated mass for H nucleus is = 2 * 1.007276 amu + 2 * 1.008665 amu

Calculated mass = 2.014552 + 2.01733

Calculated mass = 4.031882 amu

Mass defect = 4.031882 amu - 4.002603 amu

Mass defect = 0.02923 amu

___________________________________________________________________________________________

Part E

Binding energy of the nucleus = 0.02923 amu * 1.6605 x 10^-27 kg/ 1 amu * ( 3.0 X 10^8 m/s)^2

Binding energy of the nucleus = 4.368 x 10^-12 kg m^2 / s^2

Binding energy of the nucleus = 4.368 x 10^-12 J

Binding energy of the nucleus = 4.368 x 10^-12 J * 1 eV / 1.602 x 10^-19 J * 1 MeV / 10^6 eV

Binding energy of the nucleus = 27.27 MeV

__________________________________________________________________________________________

Part F

Total nucleons for the given atom = 2 protons + 2 neutrons = 4

Binding energy per nucleon = 27.27 MeV / 4

Binding energy per nucleon = 6.816 MeV

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.