Physics

13.8 Decay rate of living carbon-containing matter, R = 15 decay / min

Half life of ${}_6{}^{14}C$ , $T_{1/2}$ = 5730 years

Decay rate of the specimen obtained from the Mohenjo-Daro site, R' = 9 decays/min

Let N be the number of radioactive atoms present in a normal carbon-containing matter.

Let N' be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period.

We can relate the decay constant, $\lambda$ and time t as:

$\frac{N}{N\text{'}}$ = $\frac{R\text{'}}{R}$ = $e^{-\lambda t}$

$e^{-\lambda t}=$ $\frac{R\text{'}}{R}$ = $\frac{9}{15}$ = $\frac{3}{5}$

By taking log (ln) on both sides,

$-\lambda t=$ ${\log }_e?3$ - ${\log }_e?5$

t = $\frac{0.5108}{\lambda }$

Since $\lambda$ = $\frac{0.693}{T_{1/2}}$ = $\frac{0.693}{5730}$

t = $\frac{5730*0.5108}{0.693}$ = 4223.5 years

Hence, the approximate age of the Indus-valley is 4223.5 years.

13.7 Half life of the radioactive isotope = T years

Original amount of the radioactive isotope = $N_o$

After decay, the amount of radioactive isotope = N

It is given that only 3.125% of $N_o$ remains after decay. Hence, we can write,

$\frac{N}{N_o}$ = 3.125% = $\frac{3.125}{100}$ = $\frac{1}{32}$

But $\frac{N}{N_o}$ = $e^{-\lambda t}$ , where $\lambda$ = decay constant, t = time

Therefore,

$e^{-\lambda t}=\frac{1}{32}$

By taking log on both sides

${\log }_e?e^{-\lambda t}$ = ${\log }_e?\frac{1}{32}$

$-\lambda t=$ ${\log }_e?1$ - ${\log }_e?32$

$-\lambda t$ = 0 – 3.465

$t$ = $\frac{3.465}{\lambda }$

Since $\lambda$ = $\frac{0.693}{T}$

t = $\frac{3.465}{\frac{0.693}{T}}$ = 5T years

Hence, all the isotopes will take about 5T years to reduce 3.125% of its original value.

After decay, the amount of radioactive isotope = N

It is given that only 1.0% of $N_o$ remains after decay. Hence, we can write,

$\frac{N}{N_o}$ = 1% = $\frac{1}{100}$

But $\frac{N}{N_o}$ = $e^{-\lambda t}$ , where $\lambda$ = decay constant, t = time

Therefore,

$e^{-\lambda t}=\frac{1}{100}$

By taking log on both sides

${\log }_e?e^{-\lambda t}$ = ${\log }_e?\frac{1}{100}$

$-\lambda t=$ ${\log }_e?1$ - ${\log }_e?100$

$-\lambda t$ = 0 – 4.605

$t$ = $\frac{4.605}{\lambda }$

Since $\lambda$ = $\frac{0.693}{T}$

t = $\frac{4.605}{\frac{0.693}{T}}$ = 6.645T years

Hence, all the isotopes will take about 6.645T years to reduce 1.0% of its original value.

13.5 Mass of the copper coin, m' = 3.0 g

Atomic mass of ${}_{29}{}^{63}Cu$ , m = 62.92960 u

The total number of ${}_{29}{}^{63}Cu$ atoms in the coin, N = $\frac{N_A*m\text{'}}{Massnumber}$ , where

$N_A$ = Avogadro's number = 6.023 $*{10}^{23}$ atoms / g

Mass number = 63 g

Therefore, N = $\frac{6.023\mathrm{}*{10}^{23}*3}{63}$ = 2.868 $*{10}^{22}$ atoms

${}_{29}{}^{63}Cu$ has 29 protons and (63 – 29) 34 neutrons

Hence the mass defect of the nucleus Δm = 29 $*m_p$ + 34 $*m_n$ - $m$

Mass of a proton, $m_p$ = 1.007825 u

Mass of a neutron, $m_n$ = 1.008665 u

Δm = 29 $*1.007825\mathrm{}$ + 34 $*1.008665\mathrm{}$ - 62.92960

Δm = 0.591935 u

Mass defect of all the atoms present in the coin, Δm = 0.591935 $*N$

= 0.591935 $*$ 2.868 $*{10}^{22}$ = 1.69766958 $*{10}^{22}$ u

But 1 u = 931.5 MeV/ $c^2$

Δm = 1.581 $*{10}^{25}$ MeV/ $c^2$

The binding energy of the nucleus, $E_b$ = Δm $c^2$ , where c = speed of light

$E_b=$ (1.581 $*{10}^{25}$ / $c^2$ ) $*$ $c^2$ = 1.581 $*{10}^{25}$ MeV

1 MeV = 1.6 $*{10}^{-13}$ J

Hence, $E_b=$ 2.530 $*{10}^{12}$ J

This much of energy required to separate all the neutrons and protons from the given coin.

14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is $\stackrel{?}{A+A}$ = $\stackrel{?}{A}$

The truth table for the same is given as:

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are $\stackrel{?}{A}$ and $\stackrel{?}{B}$  , as shown in the following figure

Above is given the inputs for the last NOR gate.

Hence, the output for the circuit can be written as:

Y = $\stackrel{?}{A+B}$ = $\stackrel{?}{\stackrel{?}{A}.\stackrel{?}{B}}$   = A.B

The truth table for the same can be written as:

This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.