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Class 12th Nuclei

13.22 For the $β^{+}$ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K–shell, is captured by the nucleus and a neutrino is emitted).

$e^{+}$ + ${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + v}$

Show that if $β^{+ +}$ emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.

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Payal Gupta

Contributor-Level 10

13.22 Let the amount on energy released during the electron capture process be $Q_{1}$ . The nuclear reaction can be written as:

$e^{+}$ + ${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + v} + Q_{1}$ …………………….(1)

Let the amount of energy released during the positron capture process be $Q_{2}$ . The nuclear reaction can be written as:

${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + e^{+} + v}$ + $Q_{2}$ …………………….(2)

Let us assume

$m_{N} ({}_{Z}^{A}X)$ = Nuclear mass of ${}_{Z}^{A}X$

$m_{N} ({}_{Z}^{A}X)$ = Atomic mass of ${}_{Z}^{A}X$

$m_{N} ({}_{Z - 1}^{A}Y)$ = Nuclear mass of ${}_{Z - 1}^{A}Y$

$m_{N} ({}_{Z - 1}^{A}Y)$ = Atomic mass of ${}_{Z - 1}^{A}Y$

$m_{e}$ = mass of the electron

c = speed of light

Q-value of the electron capture reaction is given as:

$Q_{1}$ = [ $m_{N} ({}_{Z}^{A}X) + m_{e} - m_{N} ({}_{Z - 1}^{A}Y)] c^{2}$

= [ $m ({}_{Z}^{A}X) - Z m_{e} + m_{e} - m ({}_{Z - 1}^{A}Y) + (Z - 1) m_{e}] c^{2}$

= [ $m ({}_{Z}^{A}X) - m ({}_{Z - 1}^{A}Y)] c^{2}$ ………………….(3)

Q-value of the positron capture reaction is given as:

$Q_{2}$ = [ $m_{N} ({}_{Z}^{A}X) - m_{N} ({}_{Z - 1}^{A}Y) - m_{e}] c^{2}$

= [

...more

13.22 Let the amount on energy released during the electron capture process be $Q_{1}$ . The nuclear reaction can be written as:

$e^{+}$ + ${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + v} + Q_{1}$ …………………….(1)

Let the amount of energy released during the positron capture process be $Q_{2}$ . The nuclear reaction can be written as:

${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + e^{+} + v}$ + $Q_{2}$ …………………….(2)

Let us assume

$m_{N} ({}_{Z}^{A}X)$ = Nuclear mass of ${}_{Z}^{A}X$

$m_{N} ({}_{Z}^{A}X)$ = Atomic mass of ${}_{Z}^{A}X$

$m_{N} ({}_{Z - 1}^{A}Y)$ = Nuclear mass of ${}_{Z - 1}^{A}Y$

$m_{N} ({}_{Z - 1}^{A}Y)$ = Atomic mass of ${}_{Z - 1}^{A}Y$

$m_{e}$ = mass of the electron

c = speed of light

Q-value of the electron capture reaction is given as:

$Q_{1}$ = [ $m_{N} ({}_{Z}^{A}X) + m_{e} - m_{N} ({}_{Z - 1}^{A}Y)] c^{2}$

= [ $m ({}_{Z}^{A}X) - Z m_{e} + m_{e} - m ({}_{Z - 1}^{A}Y) + (Z - 1) m_{e}] c^{2}$

= [ $m ({}_{Z}^{A}X) - m ({}_{Z - 1}^{A}Y)] c^{2}$ ………………….(3)

Q-value of the positron capture reaction is given as:

$Q_{2}$ = [ $m_{N} ({}_{Z}^{A}X) - m_{N} ({}_{Z - 1}^{A}Y) - m_{e}] c^{2}$

= [ $m ({}_{Z}^{A}X) - Z m_{e} - m ({}_{Z - 1}^{A}Y) + (Z - 1) m_{e} - m_{e}] c^{2}$

= [ $m ({}_{Z}^{A}X) - m ({}_{Z - 1}^{A}Y) - 2 m_{e}] c^{2}$ ……………(4)

It can be inferred that if $Q_{2} > 0$ , then $Q_{1} > 0$ , also if $Q_{1} > 0, i t d o e s n o t n e c e s s a r i l y$ means that $Q_{2} > 0$ .

In other words, this means that if $β^{+}$ emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because Q-value must be positive for an energetically-allowed nuclear reaction.

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Class 12th Electric Charges and Fields

(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 $\times 10^{- 7}$ C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

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Class 12th Nuclei

13.21 From the relation R = $R_{0} A^{1 / 3}$
, where $R_{0}$ is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

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Payal Gupta

Contributor-Level 10

13.21 We have the expression for nuclear radius as:

R = $R_{0} A^{1 / 3}$

Where $R_{0}$ = constant

A = mass number of nucleus

Let m be the average mass of the nucleus, hence mass of the nucleus = mA

Nuclear matter density $ρ$ can be written as

$ρ = \frac{M a s s o f t h e n u c l e u s}{V o l u m e o f t h e n u c l e u s}$ = $\frac{m A}{\frac{4}{3} π R^{3}}$ = $\frac{3 m A}{4 π {(R_{0} A^{\frac{1}{3}})}^{3}}$ = $\frac{3 m A}{4 π R_{0}^{3} A}$ = $\frac{3 m}{4 π R_{0}^{3}}$

Hence, the nuclear mass density is independent of A. It is nearly constant

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Class 12th Nuclei

13.20 Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

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Payal Gupta

Contributor-Level 10

13.20 When two deuterons collide head-on, the distance between their centers, d is given as

d = Radius of 1st deuteron + Radius of 2nd deuteron

Radius of the deuteron nucleus = 2 fm = 2 $* 10^{- 15}$ m

Hence d = 2 $* 10^{- 15} + 2 * 10^{- 15}$ = 4 $* 10^{- 15}$ m

Charge on a deuteron nucleus = Charge on an electron = 1.6 $* 10^{- 19}$ C

Potential energy of two-deuteron system: $V$ = $\frac{e^{2}}{4 π ?_{0} d}$

Where $?_{0}$ = permittivity of free space

$\frac{1}{4 π ?_{0}}$ = 9 $* 10^{9}$ N $m^{2} C^{- 1}$

V = $\frac{{(1.6 * 10^{- 19})}^{2} * 9 * 10^{9}}{4 * 10^{- 15}}$ J = $\frac{{(1.6 * 10^{- 19})}^{2} * 9 * 10^{9}}{4 * 10^{- 15} * 1.6 * 10^{- 19}}$ eV = 360 $* 10^{3}$
eV = 360 keV

Hence the height of the potential barrier of the two-deuteron system is 360 keV.

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Class 12th Physics

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Class 12th Electric Charges and Fields

1.11 A polythene piece rubbed with wool is found to have a negative charge of 3 $10 -7$ C.

(a) Estimate the number of electrons transferred (from which to which?)

(b) Is there a transfer of mass from wool to polythene?

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alok kumar singh

Contributor-Level 10

1.11

(a) When polythene rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene negatively charged.

Amount of charge on the polythene piece, q = -3 $* 10^{- 7}$ C.

Amount of charge of 1 electron e = -1.6 $* 10 -19$

So number of electron transferred from wool to polythene

=
- $\frac{3 * 10^{- 7}}{- 1.6 * 10^{- 19}}$ 1.875 $* 10^{12}$

(b) Since electron has a mass, so there will be transfer of mass also.

Mass of single electron, $m_{e}$ = 9.1 $* 10^{- 31}$ kg

Total mass transferred from wool to polythene = 1.875 $* 10^{12} *$ 9.1 $* 10^{- 31}$ kg

= 1.706 $* 10^{- 18}$ kg $?$ negligible

Hence a negligible amount of mass is transferred from wool to polythene.

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Class 12th Nuclei

13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

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Payal Gupta

Contributor-Level 10

13.19 The given fusion reaction is

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023 $* 10^{23}$
atoms

Hence 2 kg of deuterium contains =
$\frac{6.023 * 10^{23}}{2}$ $* 2 * 10^{3}$ atoms = 6.023 $* 10^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released.

Hence total energy released from 2 kg of deuterium, E =
$\frac{3.27}{2}$ $*$ 6.023 $* 10^{26}$ MeV

= $\frac{3.27}{2}$ $*$ 6.023 $* 10^{26} * 10^{6} * 1.6 * 10^{- 19}$ J = 1.5756 $* 10^{14}$ J

Power of the electric bulb, P = 100 W = 100 J/s

Hence energy consumed by the bulb per second = 100 J

Therefore, total time the electric bulb will glow = $\frac{1.5756 * 10^{14}}{100}$ seconds = 1.5756 $* 10^{12}$ secs

= 49.96 $* 10^{3}$ years

...more

13.19 The given fusion reaction is

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023 $* 10^{23}$
atoms

Hence 2 kg of deuterium contains =
$\frac{6.023 * 10^{23}}{2}$ $* 2 * 10^{3}$ atoms = 6.023 $* 10^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released.

Hence total energy released from 2 kg of deuterium, E =
$\frac{3.27}{2}$ $*$ 6.023 $* 10^{26}$ MeV

= $\frac{3.27}{2}$ $*$ 6.023 $* 10^{26} * 10^{6} * 1.6 * 10^{- 19}$ J = 1.5756 $* 10^{14}$ J

Power of the electric bulb, P = 100 W = 100 J/s

Hence energy consumed by the bulb per second = 100 J

Therefore, total time the electric bulb will glow = $\frac{1.5756 * 10^{14}}{100}$ seconds = 1.5756 $* 10^{12}$ secs

= 49.96 $* 10^{3}$ years

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Class 12th Nuclei

13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ${}_{92}^{235}U$
did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ${}_{92}^{235}U$ and that this nuclide is consumed only by the fission process.

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Payal Gupta

Contributor-Level 10

13.18 Half life of the fuel in the fission reactor, $T_{1 / 2}$
= 5 years = 5 $* 365 * 24 * 60 * 60 s$

= 157.68 $* 10^{6} s$

We know that in the fission of 1 g of ${}_{92}^{235}U$
, the energy released = 200 MeV

1 mole i.e. 235 gm of ${}_{92}^{235}U$ contains 6.023 $* 10^{23}$ atoms

Therefore 1 gm of ${}_{92}^{235}U$
contains = $\frac{6.023 * 10^{23}}{235}$ = 2.563 $* 10^{21}$ atoms

The total energy Q generated per gm of ${}_{92}^{235}U$ = 200 $*$ 2.563 $* 10^{21}$ MeV/g = 5.126 $* 10^{23}$
MeV/g = 5.126 $* 10^{23} * 1.6 * 10^{- 19} * 10^{6}$ J/g = 8.20 $* 10^{10}$
J/g

Since the reactor operates only 80% of the time, hence the amount of ${}_{92}^{235}U$
in 5 years is given by $\frac{0.8 * 157.68 * 10^{6} * 1000 * 10^{6}}{8.20 * 10^{10}}$ = 1538 kg

Hence, initial amount of fuel = 2 $* 1538 k g$ = 3076 kg

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Class 12th Nuclei

13.17 The fission properties of ${}_{94}^{239}{P u}$ are very similar to those of ${}_{92}^{235}U$ . The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ${}_{94}^{239}{P u}$ undergo fission?

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Payal Gupta

Contributor-Level 10

13.17 The average energy released per fission of ${}_{94}^{239}{P u}$ $E_{a v g}$ = 180 MeV

Amount of pure ${}_{94}^{239}u$ , m = 1 kg = 1000 g

Avogadro's number,
$N_{A}$ = 6.023 $* 10^{23}$

Mass number of ${}_{94}^{239}{P u}$ = 239 gm

Hence, number of atoms in 1000 g ${}_{94}^{239}{P u}$ , N = $\frac{6.023 * 10^{23}}{239}$ $* 1000$ = 2.52 $* 10^{24}$

Total energy released during the fission of 1 kg of ${}_{94}^{239}{P u}$ , E = $E_{a v g}$ $*$ N

= 180 $*$ 2.52 $* 10^{24}$ MeV = 4.536 $* 10^{26}$ MeV

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Class 12th Nuclei

13.16 Suppose, we think of fission of a ${}_{26}^{56}{F e}$ nucleus into two equal fragments, ${}_{13}^{28}{A l}$ . Is the fission energetically possible? Argue by working out Q of the process. Given

m ( ${}_{26}^{56}{F e})$ = 55.93494 u and m ( ${}_{13}^{28}{A l})$ = 27.98191 u.

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Payal Gupta

Contributor-Level 10

13.16 The fission can be shown as:

${}_{26}^{56}{F e} \to 2 {}_{13}^{28}{A l}$

It is given that atomic mass

m ( ${}_{26}^{56}{F e})$ = 55.93494 u

m ( ${}_{13}^{28}{A l})$ = 27.98191 u

The Q-value of this reaction is given as:

Q = $[m ({}_{26}^{56}{F e}) - 2 m ({}_{13}^{28}{A l})] c^{2}$

= $[55.93494 - 2 * 27.98191] c^{2}$

= -0.02888 $c^{2}$

= -0.02888 $* 931.5$ MeV

= - 26.902 MeV

The Q value of the fission is negative, therefore the fission is not possible energetically. Q value needs to be positive for a fission.

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