JEE Main & JEE Advanced & CBSE Board Numerical Examples

JEE Main & JEE Advanced Numerical Examples -

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

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

3. The most stable isobar has N/Z ≈1. Therefore, neutrons to be added or removed: $$N_{stable}-N_{initial}=140-100\times \frac{1}{1.4}$$ $$=140-71.43=68.57$$ 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: $$\begin{array}{}\text{(1)}& Mv=mv\Rightarrow \frac{M}{m}=\frac{v}{V}\end{array}$$

Kinetic energy conservation: $$\frac{1}{2}M{v}^2=\frac{1}{2}m{v}^2+5MeV$$ $$\begin{array}{}\text{(2)}& (M-m)v^2=10MeV\end{array}$$

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

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

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 $$=\mathrm{\Delta }m=1.0078u-1.0087u=0.0009u$$ $$\mathrm{\Delta }E=(\mathrm{\Delta }m)(931.5MeV/u)=0.84MeV$$

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

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) $$TotalQ=Q_1+Q_2+Q_3+Q_4+Q_5=51.7MeV$$

CBSE Board Exam Numerical Examples -

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

2. Nuclear density $$=\frac{Massofthenucleus}{Volumeofthenucleus}$$ 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) $$Nucleardensity=\frac{Massofthenucleus}{\frac{4}{3}\pi R^{3}A}$$ $$\therefore \frac{{\rho }_{He}}{{\rho }_C}=\frac{A_{He}}{A_C}\frac{(r_C{)}^3}{(r_{He}{)}^3}$$ $$=\frac{4}{12}\frac{(R\cdot {12}^{1/3}{)}^3}{(R\cdot 4^{1/3}{)}^3}$$ $$=\frac{4}{12}\times \frac{({12}^{1/3}{)}^3}{(4^{1/3}{)}^3}$$ $$=3$$

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

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

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

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