Physics

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

0 Follower 5 Views

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

so from

...more

This is a long answer type question as classified in NCERT Exemplar

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

so from above eqn

change in PE +change in KE = A $ρ g {(l}^{2} + y^{2})$ +A $ρ l v^{2}$

system being conservative

so change in total energy =0

A $ρ g {(l}^{2} + y^{2}) + A ρ l v^{2}$ =0

Differentiating both with respect to time t

A $ρ g (2 y v) + A ρ l (2 v) a = 0$ =0

Dy/dt =v and dv/dt=a

(gy+la)2A $ρ v$ =0

2A $ρ v = c o n s t a n t$ and 2A $ρ v \neq 0$

La+gy=0

A+(gy/l)=0

d²y/dt²+gy/l=0

d²y/dt²+w²y=0

w= $\sqrt[]{\frac{g}{l}}$

T=2 $π \sqrt[]{\frac{l}{g}}$

less

0 0

New answer posted

a year ago

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. 2 m T A g π ρ = where m is mass of the body and ρ is density of the liquid.

0 Follower 2 Views

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let the log be passed and the vertical displacement at the vertical displacement at the equilibrium position

So mg= buoyant forces = $ρ A x_{o} g$

When it is displaced by further displacement x, the buoyant force is A (x_o+x) $ρ g$

Net restoring force = buoyant forces -weight

=A (x_o+x) $ρ g$ -mg

=A $ρ g x$

As displacement x is downward and restoring force is upward

F_restoring₌-A $ρ g x$ =-kx

So motion is SHM

Acceleration a=F_restoring/m=-kx/m

a=-w²x

w²=k/m

w= $\sqrt[]{\frac{k}{m}}$

T= 2 $π \sqrt[]{\frac{m}{k}} = 2 π \sqrt[]{\frac{m}{A ρ g}}$

0 0

New answer posted

a year ago

A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand. (a) What is the amplitude of oscillation? (b) Find the frequency of oscillation?

0 Follower 7 Views

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x' is the extension in that case

F= kx'

F=mg

Mg=kx'

X'=mg/k

By dividing x by x'

x/x'= $\frac{2 m g / k}{m g / k} = 2$

so x=2x'

x'=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural l

...more

This is a long answer type question as classified in NCERT Exemplar

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x' is the extension in that case

F= kx'

F=mg

Mg=kx'

X'=mg/k

By dividing x by x'

x/x'= $\frac{2 m g / k}{m g / k} = 2$

so x=2x'

x'=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural length is equal to amplitude of the oscillation

A=x'=2cm

(b) Time period of the oscillating system depends on mass spring constant given by

T= $2 π \sqrt[]{\frac{m}{k}}$

2mg/k=x

2mg/k=4cm =4 $* 10^{- 2} m$

m/k =4 $* 10^{- 2}$ /2g

k/m=g/2 $* 10^{- 2}$

$v = \frac{1}{T} \sqrt[]{\frac{k}{m}}$

$v = \frac{1}{2 * 3.14} \sqrt[]{\frac{g}{2 * 10^{- 2}}}$

= $\frac{10}{628} * 2.21$ =3.51Hz

less

0 0

New answer posted

a year ago

A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.

(a) Will there be any change in weight of the body, during the oscillation?

(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

0 Follower 3 Views

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) The weight of the body changes during oscillations.

(b) Considering the situations in two extreme positions

we can say mg-N= ma

so at the highest point the platform is accelerating downward.

N=mg-ma

a=w²A

N=mg-mw²A

A= amplitude of motion m=50kg v=2m/s

w=2 $π v = 4 π r a d / s$

A= 5cm = 5 $* 10^{- 2} m$

N= 50 $* 9.8 - 50 * {(4 π)}^{2} * 5 * 10^{- 2} = 95.5 N$

When it is accelerating towards mean position that is vertically upwards

N-mg=ma=Mw²A

N=mg+mw²A

N=m (g+w²A)

N= 50 [9.8+ ${(4 π)}^{2} * 5 * 10^{- 2}$ ]

N= 884N

Machine reads the normal reaction

Maximum weight =884N

Minimum weight=95.5N

0 0

New answer posted

a year ago

13.28 Consider the D–T reaction (deuterium–tritium fusion)

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

(a) Calculate the energy released in MeV in this reaction from the data:

m( ${}_{1}^{2}{H)}$ =2.014102 u

m( ${}_{1}^{3}{H)}$ =3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?

(Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzmann's constant, T = absolute temperature.)

0 Follower 8 Views

Contributor-Level 10

13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $* 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

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

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

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

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

Hence, V = $\frac{{(1.6 * 10^{- 19})}^{2}}{4 * 10^{- 15}}$ $*$ 9 $* 10^{9}$ = 5.76 $* 10^{- 14}$ J =

...more

13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $* 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

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

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

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

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

Hence, V = $\frac{{(1.6 * 10^{- 19})}^{2}}{4 * 10^{- 15}}$ $*$ 9 $* 10^{9}$ = 5.76 $* 10^{- 14}$ J = $\frac{5.76 * 10^{- 14}}{1.6 * 10^{- 19}}$ eV = 360 keV

Hence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.

It is given that

KE = 2 $* \frac{3}{2} k T$

Where

k = Boltzmann constant = 1.38 $* 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

T = Temperature required to trigger the reaction

Therefore T = $\frac{K E}{3 k}$ = $\frac{5.76 * 10^{- 14}}{3 * 1.38 * 10^{- 23}}$ = 1.39 $* 10^{9}$ K

Hence the gas must be heated to 1.39 $* 10^{9}$ K to initiate the reaction.

less

0 0

New answer posted

a year ago

13.27 Consider the fission of ${}_{92}^{238}U$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ${}_{58}^{140}{C e}$ and ${}_{44}^{99}{R u})$ . Calculate Q for this fission process. The relevant atomic and particle masses are

m( ${}_{92}^{238}U)$ =238.05079 u

m( ${}_{58}^{140}{C e}$ ) =139.90543 u

m( ${}_{44}^{99}{R u})$ = 98.90594 u

0 Follower 4 Views

Contributor-Level 10

13.27 In the fission of ${}_{92}^{238}U$ , $β^{-}$ 10 particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}^{238}U + {}_{0}^{1}{n \to {}_{58}^{140}{C e +}} {}_{44}^{99}{R u + 10 {}_{- 1}^{0}e}$

It is given that:

Mass of a ${}_{92}^{238}U n u c l e u s, m_{1}$ = 238.05079 u

Mass of a ${}_{58}^{140}{C e} n u c l e u s, m_{2}$ =139.90543 u

Mass of a ${}_{44}^{99}{R u} n u c l e u s, m_{3}$ = 98.90594 u

Mass of a neutron ${}_{0}^{1}n$ , $m_{4}$ = 1.00865 u

Q value of the above equation,

Q = $[m^{'} ({}_{92}^{238}U) + m' ({}_{0}^{1}{n) - m^{'} (} {}_{58}^{140}{C e) - m^{'} ({}_{44}^{99}{R u) - 10 m_{e}}}] c^{2}$

Where

m' = represents the corresponding atomic masses of the nuclei.

$m^{'} ({}_{92}^{238}U)$ = $m_{1} - 92 m_{e}$

m'( ${}_{58}^{140}{C e)}$ = $m_{2} - 58 m_{e}$

m'( ${}_{44}^{99}{R u)}$ = $m_{3} - 44 m_{e}$

m'( ${}_{0}^{1}{n)} = m_{4}$

Substituting these values, we get

Q = $[m_{1} - 92 m_{e} + m_{4} - m_{2} + 58 m_{e} - m_{3} + 44 m_{e} - 10 m_{e}] c^{2}$

= $[m_{1} + m_{4} - m_{3} - m_{2}] c^{2}$

= $[238.05079 + 1.00865 - 98.90594 - 139.90543] c^{2}$ u

=0.24807 $c^{2}$ u

= 231.077 MeV

0 0

New answer posted

a year ago

13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $α -$ particle. Consider the following decay processes:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

0 Follower 4 Views

Contributor-Level 10

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ &n

...more

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 219.00948 - 4.00260) $c^{2}$ u

= 6.42 $* 10^{- 3} c^{2}$ u

= 6.42 $* 10^{- 3} * 931.5$ MeV = 5.980 MeV

Hence, the Q-value of the nuclear reaction is 5.98 MeV, since the value is positive, the reaction is energetically allowed.

less

0 0

New answer posted

a year ago

13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ${}_{20}^{41}{C a}$ and from the following data:

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

0 Follower 8 Views

Contributor-Level 10

13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $* 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m( ${}_{0}^{1}n) -$ m( ${}_{13}^{27}{A l)}$

= 25.

...more

13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $* 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m( ${}_{0}^{1}n) -$ m( ${}_{13}^{27}{A l)}$

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

=13.0586985 MeV

less

0 0

New answer posted

a year ago

13.15 The Q value of a nuclear reaction A + b $\to$ C + d is defined by

Q = [ $m_{A} + m_{b}$ – $m_{C}$ – $m_{d}$ ] $c^{2}$

where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

(i) ${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

(ii) ${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic masses are given to be

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

0 Follower 12 Views

Contributor-Level 10

13.15 The given nuclear reaction is

${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

Q = $[m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}$

= $[1.007825 + 3.016049 - 2 * 2.014102] c^{2}$

= (-4.33 $* 10^{- 3}$ ) $c^{2}$

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

Q = -4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

The given nuclear reaction is

${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic mass

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 * 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $* 10^{- 3} c^{2}$ u

=4.958 $* 10^{- 3} * 931.5$

=4.6183 MeV

...more

13.15 The given nuclear reaction is

${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

Q = $[m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}$

= $[1.007825 + 3.016049 - 2 * 2.014102] c^{2}$

= (-4.33 $* 10^{- 3}$ ) $c^{2}$

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

Q = -4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

The given nuclear reaction is

${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic mass

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 * 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $* 10^{- 3} c^{2}$ u

=4.958 $* 10^{- 3} * 931.5$

=4.6183 MeV

The positive sign shows that the reaction is exothermic.

less

0 0

New answer posted

a year ago

13.13 The radionuclide ${}_{6}^{11}C$ decays according to ${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v :}} T_{1 / 2}$ =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

0 Follower 4 Views