Energy-Mass Equivalence : Nuclear Binding Energy

Question
ICSEENIPH12030069

What is meant by the mass defect ?

Calculate the disintegration energy Q when a heavy nucleus of mass number A = 240 with binding energy/nucleon ~7.6 MeV is split into two equal fragments of mass number A' = 120, each with BE/nucleon ~8.5 MeV.   

Solution

Mass defect : The difference between the sum of the mass of the nucleons constituting a nucleus and the rest mass of the nucleus is known as mass defect.
It is denoted by incrementm.

Δm = sum of mass of nucleons - mass of nucleus.

Given,
Mass No. of heavy nucleus is A = 240 , and
binding energy/nucleon = 7.5 MeV

Therefore,
Total binding energy of nucleus = 240 x 7.6 Mev

                                         = 1824 MeV

It splits into two nuclei of Mass No. A1 = 120.

Therefore,
Binding energy per nucleon = 8.5 MeV

∴ Total binding energy of two nuclei = 2 x 120 x 8.5 MeV

                                               = 2040 MeV.

∴ Disintegration energy Q = 2040 - 1824 MeV

                              Q = 216 MeV

Question
ICSEENIPH12030070

Calculate the equivalent energy in MeV of a unified atomic mass unit. 

Solution

We know that,
1 amu = 1.660565 x 10-27kg

Energy equivalent to this mass, E = mc2

E = (1.660565 x 10-27) x (2.998 x 108)2

E = 1.4925 x 10-10J
straight E space equals space fraction numerator 1.4925 space straight x space 10 to the power of negative 10 end exponent over denominator 1.602 space straight x space 10 to the power of negative 19 space straight x space 10 to the power of 6 end exponent end fraction space MeV

straight i. straight e. comma space straight E space equals space 931. space 64 space MeV

1 space amu space equals space 931 space MeV

Question
ICSEENIPH12030118

The following equation represents a fusion reaction :

straight H presubscript 1 presuperscript 2 space plus space straight H presubscript 1 presuperscript 3 space rightwards arrow space He presubscript 2 presuperscript 4 space plus space straight n presubscript 0 presuperscript 1 space plus space straight Q space semicolon

space where space straight Q space is space the space enrgy space released

Mass space of space straight H presubscript 1 presuperscript 2 space space atom space equals space 2.014102 space straight u

Mass space of space straight H presubscript 1 presuperscript 3 space atom space equals space 3.016050 space straight u

Mass space of space He presubscript 2 presuperscript 4 space atom space equals space 4.002603 space straight u

Mass space of space space neutron space space straight n presubscript 0 presuperscript 1 space equals space 1.008665 space straight u

Calculate the value of Q.

Solution

straight H presubscript 1 presuperscript 2 space plus space straight H presubscript 1 presuperscript 3 space rightwards arrow space He presubscript 2 presuperscript 4 space plus space straight n presubscript 0 presuperscript 1 space plus space straight Q space
Mass defect Δ m = Mass of reactant - Mass of product

equals space straight m space left parenthesis straight H presubscript 1 presuperscript 2 right parenthesis space plus space straight m left parenthesis straight H presubscript 1 presuperscript 3 right parenthesis space minus space open square brackets straight m open parentheses He presubscript 2 presuperscript 4 close parentheses space plus space straight n presubscript 0 presuperscript 1 space plus space straight Q close square brackets space

equals space left parenthesis 2.014102 space plus space 3.016050 right parenthesis space minus space left parenthesis 4.002603 space plus space 1.008665 right parenthesis

equals space 5.030152 minus 5.011268 space equals space 0.018884 space straight u

straight Q space equals space increment mc squared

space space space space equals space 0.018884 space straight x space 931 space MeV space
space
space space space space equals space 17.581004 space MeV space

space space space space equals space 17.581004 space straight x space 10 to the power of 6 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent space straight J space space

space space space space equals space 28.13 space straight x space 10 to the power of negative 13 space end exponent

space space space equals space 2.81 space straight x space 10 to the power of negative 12 end exponent space straight J space

Question
ICSEENIPH12030123

If the temperature of the cold junction of a thermocouple is lowered, what will be the effect on its neutral temperature and temperature of inversion ?

Solution

Neutral temperature is independent of temp. of cold junction, hence neutral temp, will not change, if temp, of cold junction is lowered.

T, = 2 Tn T0

If temp. of cold junction decreases, temperature of inversion increases.