A source contains two phosphorous radio nuclides ^{32}_{15}P\, (T_{1/2}\, =\, 14.3d) and ^{33}_{15}P\, (T_{1/2}\, =\, 25.3d) . Initially, 10% of the decays come from ^{32}_{15}P . How long one must wait until 90% to do so?

Let the number of atoms of ^{32}_{15} P be N_1 and atoms of ^{33}_{15} P be N_2 .The decay constants of two atoms be \lambda_1 and \lambda_2 respectively.The initial activity of ^{33}_{15} P is \dfrac19 times that of ^{32}_{15} P .Therefore, we can write the decay constants as: N_1 \lambda_1 = \dfrac{N_2 \lambda_2}{9} ___(i)Let after time t the activity of ^{33}_{15} P be 9 times that of ^{32}_{15} P N_1 \lambda_1 e^{-\lambda_1 t} = 9 N_2 \lambda_2 e^{-\lambda_2 t} ___(ii)Now. Divide equation (ii) by (i) and taking the natural log of both sides we get -\lambda_1 t = ln\ 81 - \lambda_2 t t = \dfrac{ln\ 81}{\lambda_2 - \lambda_1} where \lambda_2 = 0.048/day and \lambda_1 = 0.027 /day t=\dfrac{4.394}{0.048-0.027}=209.2\ days t comes out to be 209.2 days

Nuclei - Difficult Conceptual Numericals

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

The half life of the fuel is 5\ years we know that the energy produced in fission of Uranium is 200\ MeV .So, energy produced by 1kg of Uranium: E=\dfrac{6.022\times10^{23}}{235}\times1000\times200 \approx8.17\times10^{13}\ J Since the reactor operate 80% of time, so time of operation is 4\ years .The energy produced is: E_r={1000\times1066\times4\times365\times24\times60\times60} The mass required for producing energy is: m=\dfrac{8.17\times10^{13}\ J}{1000\times1066\times4\times365\times24\times60\times60} =1544\ kg Since this is the amount consumed and initial amount should have been double of this. So, the initial amount of Uranium would have been 3088\ kg

Q: (a)Calculate the energy released if ^{238}U emits an \alpha -particle.(b) Calculate the energy supplied to ^{238}U if two protons and two neutrons are to be emitted one by one. The atomic masses of ^{238}U , ^{234}Th and ^4He are 238.0508 u ,234.04363 u and 4.00260 u respectively.

Given, Atomic mass of ^{238}U = 238.0508 u Atomic mass of ^{234} Th = 234.04363 u Atomic mass of ^4 He = 4.00260 u To find, a) Energy released if ^{238}u emits an \alpha -particle.b) Energy to be supplied to ^{238} u if two protons and two neutrons are emitted one by one. a) We have, when ^{238} u emits an \alpha -particle, ^{238}u \rightarrow \, ^{234}Th + \, ^4He \therefore Mass defect, \Delta m = [m. \, ^{238} u - (m. \, ^{234} Th + m \, ^4He)] \Rightarrow \Delta m = [238.0508 - (234.0436 + 4.0026)] \Rightarrow \Delta m = 0.00457 u So, energy released \Rightarrow E = \Delta mc^2 \Rightarrow E = (0.00457 u) \times 931.5 MeV \Rightarrow E = 4.25 MeV b) When two protons and two neutrons are emitted one by one, ^{233}u \rightarrow \, ^{234} Th + 2n + 2p \therefore Mass defect, \Delta m = m ^{238}u - [m \, ^{234}Th + 2m_n + 2m_p] \Rightarrow \Delta m = 238.0508 u - [234.04363 + 2 \times 1.008665 + 2 (1.007276) u] \Rightarrow \Delta m =- 0.024712 u \therefore Energy supplied, E = \Delta mc^2 \Rightarrow E = (0.024712) \times 931.5 MeV \Rightarrow E = 23.019 MeV \Rightarrow E = 23.02 MeV Therefore, a) 4.25 MeV ; b) 23.02 MeV

Q: Suppose India had a target of producing by 2020 \mathrm{AD}, 200000 MW of electric power, 10 \% of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilisation (i.e.conversion to electric energy) of thermal energy produced in a reactor was 25 \% How much amount of fissionable uranium would our country need per year by 2020 ? Take 235 \mathrm{U} to be about 200 \mathrm{MeV} .

Total electric power which is aimed to be produced by 2020 = 2 \times 10^5 \ MW Power obtained from the nuclear power plant = 10 \text{% of } 2 \times 10^5 \ MW = 2 \times 10^4 \ MW= 2 \times 10^{10} \ W Energy required from the nuclear plant in one year= \text{Power} \times \text{time} = 2 \times 10^{10} \times 365 \times 24 \times 60 \times 60 = 6.3 \times 10^{17} \ J Available electric energy per fission= 25 \text{% of } 200 \ MeV = (0.25 \times 200 \times 1.602 \times 10^{-13}) \ J = 8 \times 10^{-12} \ J Required no. of fission per year = \dfrac{6.3 \times 10^{17} }{8 \times 10^{-12}} = 0.7875 \times 10^{29} Now, 6.023 \times 10^{23} nuclei of _{92}^{235} U have mass = 235 g \therefore Mass required to produce 7.9 \times 10^{28} nuclei = \dfrac{ 235}{ 6.023 \times 10^{23} } \times 3.95 \times 10^{28} g \approx 3.084 \times 10^4 \ kg

Q: A radioactive sample can decay by two different processes. The half-life for the first process is T_{1} and that for the second process is T_{2} . The effective half-life T of the radioactive sample is

Radioactive decay rate is directly proportional to number of nucleus present at any time \dfrac{dN}{dt} = -{\lambda}N ....(i)where \lambda is a constant related to half life as \lambda =\dfrac{\ln 2}{T_\frac{1}{2}} where T_\frac{1}{2} is half life time of radioactive decayThe total rate of decay is \dfrac{dN}{dt} =- {(\lambda_1 +\lambda_2)}N =\lambda_{effective}N ...(ii) \lambda_1 =\dfrac{\ln 2}{T_1} \lambda_2 =\dfrac{\ln 2}{T_2} \lambda_{effective} =\dfrac{\ln 2}{T_{effective}} substituting in equation (ii) (\dfrac{\ln 2}{T_1} +\dfrac{\ln 2}{T_2})N =\dfrac{\ln 2}{T_{effective}}N \dfrac{1}{T} =\dfrac{1}{T_1} +\dfrac{1}{T_2}

Q: When an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose a 5.00 MeV alpha particle has a head on elastic collision with a gold nucleus that is initially at rest. What is the kinetic energy of (a) the recoiling nucleus and (b) the rebounding alpha particle?

The conservation laws of (classical kinetic) energy and (linear) momentum determine the outcome of the collision. The final speed of the \alpha particle is v_{\alpha f} = \dfrac {m_\alpha - m_{Au}} {m_\alpha + m_{Au}} v_{\alpha i}. and that of the recoiling gold nucleus is v_{Au, f} = \dfrac {2m_\alpha}{m_\alpha + m_{Au}} v_{\alpha i} (a) Therefore, the kinetic energy of the recoiling nucleus is K_{Au, f} = \dfrac 12 m_{Au} v^2_{Au,f} = \dfrac 12 m_{Au} \left(\dfrac {2m_\alpha}{m_\alpha + m_{Au}} \right)^2 v^2_{\alpha i} = K_{\alpha i} \dfrac{4m_{Au} m_{\alpha}} {(m_{\alpha} + m_{Au})^2} = (5.00 MeV)\dfrac{4(197 u)(4.00 u)} {(4.00u + 197 u)^2} = 0.390 MeV (b) The final kinetic energy of the alpha particle is K_{\alpha f} = \dfrac 12 m_{\alpha} v^2_{\alpha f} = \dfrac 12 m_{\alpha} \left(\dfrac {m_{\alpha} - m_{Au}} {m_{\alpha} + m_{Au}} \right)^2 v^2_{\alpha i} = K_{\alpha i} \left(\dfrac {m_{\alpha} - m_{Au}} {m_{\alpha} + m_{Au}} \right) ^2 = (5.00 MeV) \left (\dfrac {4.00u - 197u} {4.00u + 197u} \right)^2 = 4.61 MeV. We note that K_{af} + K_{Au, f} = K_{\alpha i} is indeed satisfied.

Q: The element curium _{96}^{218}Cm has a mean-life of 10^{13}s . Its primary decay modes are spontaneous fission and \alpha-decay , the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV energy. The masses involved in \alpha-decay are as follows: _{98}^{218}Cm=248.07220 u, _{94}^{244}Pu=244.064100 u , and _2^4He=4.002603 u .Calculate the power output from a sample of 10^{20} Cm atoms. (1 u=931 MeV c^{-2})

Activity (or rate) of fission: A=\dfrac {dN}{dt}=\lambda N=\dfrac {1}{\tau}N=\dfrac {10^{20}}{10^{13}}=10^7 As probability of fission is 8% only, therefore, actual rate of fission is: \dfrac {8}{100}\times 10^7=8\times 10^5 s^{-1} Energy released per fission =200 MeV Power output of fission =8\times 10^5\times 200 MeV s^{-1} =16\times 10^7 MeV s^{-1} Probability of \alpha-particle decay is 92%. Therefore, rate of decay of \alpha-particle is: \dfrac {92}{100}\times 10^7=92\times 10^5 s^{-1} For \alpha-decay , the equation is: \underset {(248.072220)}{_{96}^{248}Cm}\rightarrow \underset {(244.064100)}{_{94}^{244}Pu}+\underset {(4.002603)}{_2^4He} Mass of decay products is: 244.064100+4.002603=248.06673 .Mass defect, \Delta m=24.072220-248.066703=0.005517 u Energy released per \alpha-decay is: 92\times 10^5\times 5/1363 MeV s^{-1}=4.725\times 10^7 MeV s^{-1} Total power output =16\times 10^7+4.725\times 10^7 =20.725\times 10^7 MeV s^{-1} =20.725\times 10^7\times 1.6\times 10^{-13} Js^{-1} =33.16\times 10^{-6}W\\ =33.16\mu W

Solve Study Textbooks Guides

Use app

Difficult Conceptual Numericals

1 min read

Nuclei

- Solve the toughest problems

1000 M W

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.

View solution

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

View solution

A source contains two phosphorous radio nuclides

_{15}^{32} P (T_{1 / 2} = 14.3 d)

and

_{15}^{33} P (T_{1 / 2} = 25.3 d)

. Initially, 10% of the decays come from

_{15}^{32} P

. How long one must wait until 90% to do so?

View solution

(a)Calculate the energy released if

^{238} U

emits an

α

-particle.

(b) Calculate the energy supplied to

^{238} U

if two protons and two neutrons are to be emitted one by one. The atomic masses of

^{238} U

^{234} T h

and

^{4} H e

are

238.0508 u, 234.04363 u

and

4.00260 u

respectively.

View solution

Suppose India had a target of producing by

2020 A D, 200000

M W

of electric power,

10

%

of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilisation (i.e.conversion to electric energy) of thermal energy produced in a reactor was

25

%

How much amount of fissionable uranium would our country need per year by

2020

?

Take

235

U

to be about

200

M e V

View solution

A radioactive sample can decay by two different processes. The half-life for the first process is

T_{1}

and that for the second process is

T_{2}

. The effective half-life

T

of the radioactive sample is

T = T_{1} + T_{2}

\frac{1}{T} = \frac{1}{T _{1}} + \frac{1}{T _{2}}

T = \frac{T _{1} + T _{2}}{T _{1} T _{2}}

T = \frac{T _{1} - T _{2}}{T _{1} T _{2}}

View solution

When an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose a 5.00 MeV alpha particle has a head on elastic collision with a gold nucleus that is initially at rest. What is the kinetic energy of (a) the recoiling nucleus and (b) the rebounding alpha particle?

View solution

The element curium

_{96}^{218} C m

has a mean-life of

1 0^{13} s

. Its primary decay modes are spontaneous fission and

α - d e c a y

, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV energy. The masses involved in

α - d e c a y

are as follows:

_{98}^{218} C m = 248.07220 u,_{94}^{244} P u = 244.064100 u

, and

_{2}^{4} H e = 4.002603 u

.
Calculate the power output from a sample of

1 0^{20} C m

atoms.

(1 u = 931 M e V c^{- 2})

View solution