Physics 5 Marks Questions

1. Activity $=\frac{\text{dN}}{\text{dt}}=\lambda\text{n}=\frac{0.693}{\text{t}_{\frac{1}{2}}}\text{N}=\frac{0.693}{12.3}\times4\times10^{23}\text{dis/year}.$

$=7.146\times10^{14}\text{dis/sec}.$

2. $\frac{\text{dN}}{\text{dt}}=7.146\times10^{14}$

No. of decays in next 10 hours $=7.146\times10^{14}\times10\times36$

$=257.256\times10^{17}=2.57\times10^{19}$

3. $\text{N = N}_0\text{e}^{-\lambda\text{t}}=4\times10^{23}\times\text{e}^{\frac{-0.693}{20.3}\times6.16}\\=2.82\times10^{23}$ = No. of atoms remained

No. of atoms disintegrated $=(4-2.82)\times10^{23}=1.18\times10^{23}$

$\Rightarrow0.05042=0.06789\text{e}^{\frac{-0.693}{4.47\times10^9}}$

$\Rightarrow\log\Big(\frac{0.05042}{0.06789}\Big)=\frac{-0.693\text{t}}{4.47\times10^9}$

$\Rightarrow\text{t}=1.92\times10^9\text{ years}.$

1. Energy radiated per fission $=2\times10^8\text{ev}$

Usable energy $=2\times10^8\times\frac{25}{100}=5\times10^7\text{ev}=5\times1.6\times10^{-12}$

Total energy needed $300\times10^8 =3\times10^8 \text{J/s}$

No. of fission per second $=\frac{3\times10^8}{5\times1.6\times10^{-12}}=0.375\times10^{20}$

No. of fission per day $=0.375\times10^{20}\times3600\times24 = 3.24\times10^{24}\text{ fissions}$

2. From ‘a’ No. of atoms disintegrated per day $= 3.24 \times 10^{24}$

We have, $6.023 \times 10^{23}$ atoms for 235g

for $3.24 × 10^{24} \text{ atom}=\frac{235}{6.023\times10^{23}}\times3.24\times10^{24}\text{g}$

$=1264\text{g/day}=1.264\text{kg/day}$

3. Total uranium needed per month = 1.264 × 30kg

Let x kg of uranium enriched to 3% in ²³⁵U be used.

$\Rightarrow\text{x}\times\frac{3}{100}=1.264\times30$

$\Rightarrow\text{x}=1264\text{kg}$

1. Here we should take R₀ at time is t₀ = 30 × 10⁹s^-1

1. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_1}\Big)=\text{ln}\Big(\frac{30\times10^9}{30\times10^9}\Big)=0$

2. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_2}\Big)=\text{ln}\Big(\frac{30\times10^9}{16\times10^9}\Big)=0.63$

3. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_3}\Big)=\text{ln}\Big(\frac{30\times10^9}{8\times10^9}\Big)=1.35$

4. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_4}\Big)=\text{ln}\Big(\frac{30\times10^9}{3.8\times10^9}\Big)=2.06$

5. $\text{ln}\Big(\frac{\text{R}_0}{\text{R}_5}\Big)=\text{ln}\Big(\frac{30\times10^9}{2\times10^9}\Big)=2.7$

2. $\therefore$ The decay constant $\lambda=0.028\text{min}^{-1}$
3. $\therefore$ The half life period $\text{t}_{\frac{1}{2}}$

$\text{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}=\frac{0.693}{0.028}=25\text{min}.$

1. Decay of potassium-40 by $\beta^-$ emission is given by

$\text{ }_{19}\text{K}^{40}\rightarrow\text{ }_{20}\text{Ca}^{40}+\beta^-+\bar{\text{v}}$

Decay of potassium-40 by $\beta^+$ emission is given by

$\text{ }_{19}\text{K}^{40}\rightarrow\text{ }_{18}\text{Ar}^{40}+\beta^++\text{v}$

Decay of potassium-40 by electron capture is given by

$\text{ }_{19}\text{K}^{40}+\text{e}^-\rightarrow\text{ }_{18}\text{Ar}^{40}+\text{v}$

Q_value in the $\beta^-$ decay is given by

Q_value = [m(₁₉K⁴⁰) - m(₂₀Ca⁴⁰)]c²

= [39.9640u - 39.9626u]c²

= 0.0014 × 931MeV

= 1.3034MeV

Q_value in the $\beta^+$ decay is given by

Q_value = [m(₁₉K⁴⁰) - m(₂₀Ar⁴⁰) - 2m_e]c²

= [39.9640u - 39.9624u - 0.0021944u]c²

= (39.9640 - 39.9624)931MeV - 1022keV

= 1489.96keV - 1022keV

= 0.4679MeV

Q_value in the electron capture is given by

Q_value = [m(₁₉K⁴⁰) - m(₂₀Ar⁴⁰)]c²

= (39.9640 - 39.9624)uc²

= 1.4890 = 1.49MeV

1. 198 grams of Ag contains → N₀ atoms.

$1\mu\text{g of Ag contains}\rightarrow\frac{\text{N}_0}{198}\times1\mu\text{g}=\frac{6\times10^{23}\times1\times10^{-6}}{198}$ atoms

Activity $=\lambda\text{N}=\frac{0.963}{\text{t}_{\frac{1}{2}}}\times\text{N}=\frac{0.693\times6\times10^{17}}{198\times2.7}$ disintegrations/day.

$=\frac{0.693\times6\times10^{17}}{198\times2.7\times3600\times24}$ disintegration/sec

$=\frac{0.693\times6\times10^{17}}{198\times2.7\times36\times24\times3.7\times10^{10}}$ curie = 0.244 Curie.

2. $\text{A}=\frac{\text{A}_0}{2\text{t}_{\frac{1}{2}}}=\frac{0.244}{2\times\frac{7}{2.7}}=0.0405=0.040$ Curie.

1. $\text{ }^2_1\text{H}+\text{ }^2_1\text{H}\rightarrow\text{ }^3_1\text{H}+\text{ }^1_1\text{H}$

Q value $=2\text{M}\big(\text{ }^2_1\text{H}\big)=\big[\text{M}\big(\text{ }^3_1\text{H}\big)+\text{M}\big(\text{ }^{3}_1\text{H}\big)\big]$

$=[2\times2.014102-(3.016049 + 1.007825)]\text{u}$

$= 4.0275\text{Mev} = 4.05\text{Mev.}$

2. $\text{ }^2_1\text{H}+\text{ }^2_1\text{H}\rightarrow\text{ }^3_2\text{H}+\text{n}$

Q value $=2\big[\text{M}\big(\text{ }^2_1\text{H}\big)-\text{M}\big(\text{ }^3_2\text{He}\big)+\text{M}_{\text{n}}\big]$

$=[2\times2.014102 -(3.016049 + 1.008665)]\text{u}$

$=3.26\text{Mev}= 3.25\text{Mev.}$

3. $\text{ }^2_1\text{H}+\text{ }^3_1\text{H}\rightarrow\text{ }^4_1\text{H}+\text{n}$

Q value $=\big[\text{M}\big(\text{ }^2_1\text{H}\big)+\text{M}\big(\text{ }^3_1\text{He}\big)-\text{M}\big(\text{ }^4_2\text{He}\big)+\text{M}_{\text{n}}\big]$

$=(2.014102 + 3.016049)-(4.002603 + 1.008665)]\text{u}$

$=17.58\text{Mev} = 17.57\text{Mev.}$