JEE Main & JEE Advanced & CBSE Board Numerical Examples

JEE Main & JEE Advanced Numerical Examples -

1. Mass defect $$=(16.000\ u -15.995\ u)= 0.005\ u$$ Then, Energy equivalent of mass defect $$\Delta E = (0.005 u)(931.5 MeV/u)=4.66\text{ MeV}$$

2. Binding energy per nucleon $$= \frac{Binding\ energy}{Number\ of\ nucleon}$$ $$=(92.16 MeV/12 nucleons)=7.68\ MeV$$

3. The most stable isobar has N/Z ≈1. Therefore, neutrons to be added or removed: $$N_{stable} - N_{initial} = 140 - 100 × \frac{1}{1.4}$$ $$ = 140 - 71.43 = 68.57$$ Thus, 69 neutrons must be removed.

4. Let M and V be the mass and speed of the daughter nucleus, and m and v be those of the alpha particle. Then, momentum conservation: $$Mv = mv \Rightarrow \frac{M}{m} = \frac{v}{V} \tag1$$

Kinetic energy conservation: $$ \frac{1}{2}Mv^2 = \frac{1}{2}mv^2 + 5 MeV$$ $$(M-m)v^2 = 10 MeV \tag2$$

Substituting (1) in (2): $$ (M-m)(\frac{M}{m}) = \frac{10 MeV}{v} \Rightarrow M^2 - M = \frac{10 MeV}{v}$$

$$M=\frac{1}{2}\pm\sqrt{\frac{1}{4}+\frac{10 MeV}{v}}$$ $$V=\frac{v}{M}=\frac{1}{2\pm \sqrt{\frac{1}{4}+\frac{10 MeV}{v}}}-\frac{4}{v}$$

With v=2×10^7ms−1, we find V=4.2×10^6 ms−1 and 1/V = 0.24×10^{−6} m=2.4×10^{−15} m.

5. Difference in mass $$ =\Delta m=1.0078 u - 1.0087u=0.0009u$$ $$\Delta E=(\Delta m) (931.5\ MeV/u) = 0.84 \ MeV$$

6. Decay constant $$\lambda = \frac{0.693}{t_{1/2}}=\frac{0.693}{12 h}=5.78 \times 10^{-3} h^{-1}$$ $$Activity (A) = \lambda N = 5.78 \times 10^{-3} h^{-1} \times \frac{N_A}{gmol^{-1}} \times \frac{1g}{gmol} \times 6.022 \times 10^{23} atom \ g^{mol}$$ $$A= 3.49 \times 10^{21} Bq$$

7.

1. ( ^{238}U \rightarrow ^{234}Th+ ^4He + Q_1)
2. ( ^{234}Th \rightarrow ^{234}Pa + \beta^-+Q_2)
3. ( ^{234}Pa \rightarrow ^{234}U + \beta^-+Q_3)
4. ( ^{234}U \rightarrow ^{230}Th+ ^4He + Q_4)
5. ( ^{230}Th \rightarrow ^{226}Ra + \alpha + Q_5) $$Total \ Q = Q_1 +Q_2 +Q_3 +Q_4 +Q_5 = 51.7\ MeV$$

CBSE Board Exam Numerical Examples -

1. Mass of ( ^4He ) nucleus in amu Binding energy per nucleon $$= 7.07\ MeV = \frac{Binding\ energy}{4nucleon}$$ Binding energy of ( ^4He) nucleus $$= 7.07 MeV \times 4 =28.28 MeV $$ $$ \therefore Mass \ defect = \frac{Binding \ energy}{931.5 MeV} = \frac{28.28 MeV}{931.5} = 0.0304 amu$$ Therefore, mass of ( ^4He) nucleus $$= 4 -0.0304 = 3.9696 amu$$

2. Nuclear density $$= \frac{Mass \ of \ the \ nucleus}{Volume \ of \ the \ nucleus}$$ Assuming nuclei are spherical their volume $$V=\frac{4}{3}\pi r^3$$ Also $$r = R A^{1/3}$$ (where A is the mass of the nucleus and R is a constant) $$Nuclear \ density = \frac{Mass \ of \ the \ nucleus}{\frac{4}{3}\pi R^3 A}$$ $$\therefore \frac{\rho_{He}}{\rho_{C}}=\frac{A_{He}}{A_{C}}\frac{(r_C)^3}{(r_{He})^3}$$ $$=\frac{4}{12} \frac{(R \cdot 12^{1/3})^3}{(R\cdot 4^{1/3})^3}$$ $$= \frac{4}{12}\times\frac{(12^{1/3})^3}{(4^{1/3})^3}$$ $$= 3$$

similarly $$\frac{\rho_{He}}{\rho_{U}}=\frac{4}{238} \frac{(R \cdot 238^{1/3})^3}{(R\cdot 4^{1/3})^3}$$ $$=\frac{4}{238} \times \frac{(238^{1/3})^3}{(4^{1/3})^3}$$ $$= 7.8$$

3. Radioactive decay constant (\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730 \ y} $$ $$= 1.21\times 10^{-4} y^{-1}$$ $$Activity (A) = \lambda N = 1.21\times 10^{-4} y^{-1} \times \frac{N_A}{gmol^{-1}} \times \frac{1g}{gmol} \times 6.022 \times 10^{23} atom \ g^{mol}$$ $$= 7.3\times 10^{18} Bq$$

4. Remaining activity after 12 hours,
$$A_t = A_0 e^{-\lambda t}$$ $$A_{12} = (10.0\ mCi)(e^{-(6.0h^{-1}) \cdot (12h)})$$ $$= 7.38 mCi$$ Remaining activity after 24 hours, $$A_{24} = (10.0\ mCi)(e^{-(6.0h^{-1}) \cdot (24h)})$$ $$= 5.54 mCi$$

5. Number of ( ^{235}U ) atoms $$N = (\frac{0.72g}{1000g})\times (1g) \times (\frac{1 \ mole }{235g }) \times (6.022 \times 10^{23} atoms \ mole^{−1})$$ $$= 1.86 \times 10^{20}$$

$$t_{1/2}=7.04 \times 10^8 \ years = 7.04 \times 10^{8} \times 365 \times 24 \times 60 \times 60 s = 2.21\times 10^{17} s$$

Total Activity $$A=\lambda N$$ $$= \frac{0.693}{2.21 \times 10^{17}s} \times (1.86 \times 10^{20})$$ $$= 5.53 \times 10^2Bq= 5.53 \times 10^{−2}Ci$$