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Physics Nuclei

13.9 Obtain the amount of ${}_{27}^{60}{C o}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{C o}$ is 5.3 years.

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Payal Gupta

Contributor-Level 10

13.9 The strength of the radioactive source is given as:

$\frac{d N}{d t}$ = 8.0 mCi = 8 $* 10^{- 3}$ $* 3.7 * 10^{10}$ decay/s = 296 $* 10^{6}$ decay/s, where

N = Required number of atoms

Given, half life of ${}_{27}^{60}{C o}$ , $T_{1 / 2}$ = 5.3 years = 5.3 $* 365 * 24 * 60 * 60$ secs = 167 $* 10^{6}$ s

For decay constant $λ$ , we have rate of decay as:

$\frac{d N}{d t}$ = $λ N$ or

N = $\frac{1}{λ} \frac{d N}{d t}$ , where $λ$ = $\frac{0.693}{T_{1 / 2}}$ = $\frac{0.693}{167 * 10^{6}}$ /s = 4.1497 $* 10^{- 9}$ $s^{- 1}$

N = $\frac{296 * 10^{6}}{4.1497 * 10^{- 9}}$ = 7.133 $* 10^{16}$ atoms

For ${}_{27}^{60}{C o}$ , mass of 6.023 $* 10^{23}$ atoms = 60 gms

Therefore, the mass of 7.133 $* 10^{16}$ atoms = $\frac{60}{6.023 * 10^{23}} *$ 7.133 $* 10^{16}$ gms = 7.106 $* 10^{- 6}$ g

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Physics Nuclei

13.8 The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive ${}_{6}^{14}C$ present with the stable carbon isotope ${}_{6}^{12}C$ . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of ${}_{6}^{14}C$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ${}_{6}^{14}C$ dating used in archaeology. Suppose a specimen from Mohenjo-Daro

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Payal Gupta

Contributor-Level 10

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, $λ$ and time t as:

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

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

By taking log (ln) on both sides,

$- λ t =$ $\log_{e} ? 3$ - $\log_{e} ? 5$

t = $\frac{0.5108}{λ}$

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

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

Hence, the appro

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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, $λ$ and time t as:

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

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

By taking log (ln) on both sides,

$- λ t =$ $\log_{e} ? 3$ - $\log_{e} ? 5$

t = $\frac{0.5108}{λ}$

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

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

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

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8 months ago

Physics Nuclei

13.7 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

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Payal Gupta

Contributor-Level 10

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 $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{32}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{32}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 32$

$- λ t$ = 0 – 3.465

$t$ = $\frac{3.465}{λ}$

Since $λ$ = $\frac{0.693}{T}$

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

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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 $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{32}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{32}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 32$

$- λ t$ = 0 – 3.465

$t$ = $\frac{3.465}{λ}$

Since $λ$ = $\frac{0.693}{T}$

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 $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{100}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{100}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 100$

$- λ t$ = 0 – 4.605

$t$ = $\frac{4.605}{λ}$

Since $λ$ = $\frac{0.693}{T}$

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.

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8 months ago

Physics Nuclei

13.5 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of ${}_{29}^{63}{C u}$ atoms (of mass 62.92960 u).

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Payal Gupta

Contributor-Level 10

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

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

The total number of ${}_{29}^{63}{C u}$ atoms in the coin, N = $\frac{N_{A} * m'}{M a s s n u m b e r}$ , where

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

Mass number = 63 g

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

${}_{29}^{63}{C u}$ 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 $* 1.007825$ + 34 $* 1.008665$ - 62.92960

Δm = 0.591935 u

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

= 0.591935 $*$ 2.

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

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

The total number of ${}_{29}^{63}{C u}$ atoms in the coin, N = $\frac{N_{A} * m'}{M a s s n u m b e r}$ , where

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

Mass number = 63 g

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

${}_{29}^{63}{C u}$ 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 $* 1.007825$ + 34 $* 1.008665$ - 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.

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8 months ago

Physics Nuclei

13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus ( ${}_{7}^{14}{N)}$ , given

m( ${}_{7}^{14}{N)}$ =14.00307 u

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Payal Gupta

Contributor-Level 10

13.3 Atomic mass of ${}_{7}^{14}N$ nitrogen , m = 14.00307 u

A nucleus of ${}_{7}^{14}N$ nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7 $m_{p}$ + 7 $m_{n}$ - m, where

Mass of a proton, $m_{p}$ = 1.007825 u

Mass of a neutron, $m_{n}$ = 1.008665 u

Therefore, Δm = 7 $*$ 1.007825+ 7 $*$ 1.008665 – 14.00307 = 0.11236 u

But 1 u = 931.5 MeV/ $c^{2}$

Δm = 104.66334 MeV/ $c^{2}$

The binding energy of the nucleus, $E_{b}$ = Δm $c^{2}$ , where c = speed of light

$E_{b} =$ (104.66334/ $c^{2}$ ) $*$ $c^{2}$ = 104.66334 MeV

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8 months ago

Physics Nuclei

13.2 The three stable isotopes of neon: ${}_{10}^{20}{N e},$ ${}_{10}^{21}{N e}$ and ${}_{10}^{22}{N e}$ have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

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Payal Gupta

Contributor-Level 10

13.2 Atomic mass of ${}_{10}^{20}{N e}$ neon isotope, $m_{1}$ = 19.99 u ad the abundance $η_{1}$ = 90.51 %

Atomic mass of ${}_{10}^{21}{N e}$ neon isotope, $m_{2}$ = 20.99 u ad the abundance $η_{2}$ = 0.27 %

Atomic mass of ${}_{10}^{22}{N e}$ neon isotope, $m_{3}$ = 21.99 u ad the abundance $η_{3}$ = 9.22 %

The average atomic mass of neon is given as:

m = $\frac{m_{1} η_{1} + m_{2 η_{2}} + m_{3} η_{3}}{η_{1} + η_{2} + η_{3}}$ = $\frac{19.99 * 90.51 + 20.99 * 0.27 + 21.99 * 9.22}{90.51 + 0.27 + 9.22}$ = $\frac{2017.71}{100}$ = 20.1771 u

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8 months ago

Physics Semiconductor Electronics: Materials, Devices and

14.15 Write the truth table for the circuits given in Fig. 14.40 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

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Payal Gupta

Contributor-Level 10

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 $\overset{?}{A + A}$ = $\overset{?}{A}$

The truth table for the same is given as:

A	Y = ( $\overset{?}{A}$ )
0	1
1	0

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 $\overset{?}{A}$ and $\overset{?}{B}$ , 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 = $\overset{?}{A + B}$ = $\overset{?}{\overset{?}{A} . \overset{?}{B}}$ = A.B

The truth table for the same can b

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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 $\overset{?}{A + A}$ = $\overset{?}{A}$

The truth table for the same is given as:

A	Y = ( $\overset{?}{A}$ )
0	1
1	0

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

Above is given the inputs for the last NOR gate.

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

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

The truth table for the same can be written as:

A	B	Y(=A!B)
0	0	0
0	1	0
1	0	0
1	1	1

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

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Physics Semiconductor Electronics: Materials, Devices and

14.14 Write the truth table for circuit given in Fig. 14.39 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)

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Payal Gupta

Contributor-Level 10

14.14 A and B are the inputs of the given circuit. The output of the first NOR gate is $\overset{?}{A}$ + $\overset{?}{B}$ . It can be observed from the following figure that the inputs of the second NOR gate become the output of the first one.

Hence, the output of the combination is given as:

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

The truth table for this operation is given as:

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

A	B	Y ( = A + B)
0	0	0
0	1	1
1	0	1
1	1	1

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

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Physics Semiconductor Electronics: Materials, Devices and

14.13 You are given two circuits as shown in Fig. 14.38, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

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Payal Gupta

Contributor-Level 10

14.13 The output of the left NAND gate will be $\overset{?}{A . B}$ , as shown in the following figure:

Hence, the output of the combination of two NAND gates is given as:

Y = ( $\overset{?}{A . B}$ ).( $\overset{?}{A . B}$ ) = $\overset{?}{A . B}$ + $\overset{?}{A . B}$ = AB

Hence the circuit functions as an AND gate.

$\overset{?}{A}$ is the output of the upper left of the NAND gate and $\overset{?}{B}$ is the output of the lower half of the NAND gate, as shown in the following figure.