The radionuclide ^{56}Mn has a half-life of 2.58 h and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope ^{55}Mn , and the manganesedeuteron reaction that produces ^{56}Mn is ^{55}Mn + d \rightarrow ^{56}Mn +p If the bombardment lasts much longer than the half-life of ^{56}Mn , the activity of the ^{56}Mn produced in the target reaches a final value of 8.88 \times 10^10 Bq . (a) At what rate is ^{56}Mn being produced? (b) How many ^{56} Mn nuclei are then in the target? (c) What is their total mass?

If M_He is the mass of an atom of helium and M_C is the mass of an atom of carbon, then the energy released in a single fusion event is Q= (3M_He - Mc)c^2 = 3 [3(4.0026 u) (12.0000 u)](931.5 MeV/u)= 7.27 MeV . Note that 3M_He contains the mass of six electrons and so does M_C . The electron masses cancel and the mass difference calculated is the same as the mass difference of the nuclei.

In a radioactive decay chain, _{90}^{232}Th nucleus decays to _{82}^{212} Pb nucleus. Let N_{\alpha} and N_{\beta} be the number of \alpha and \beta^{-} particles respectively, emitted in this decay process. Which of the following statements is (are) true?

_{90}^{232} Th has converted into _{82}^{212}Pb Change in mass number (A) = 20 \therefore no of \alpha particle = \dfrac{20}{4} = 5 Due to 5 \alpha particle, z will change by 10 unit. Since given change is 8 , therefore no. of the beta particle is 10-8=2

Nuclei - Most Asked Questions

Q: Consider the nuclear fission, Ne^{20} \rightarrow 2He^4 + C^{12} Given that the binding energy/nucleon of Ne^{20} , He^4 and C^{12} are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:

Energy absorbed; \Delta Q = \text{Binding energy of products - Binding energy of reactants} \text{Binding energy of products} = 2 \times (\text{B.E of } He^4 ) + (\text{B.E of } C^{12} ) = 2 (4 \times 7.07 ) + (12 \times 7.86) =150.88 \ MeV \text{Binding energy of Reactants} = 20 \times 8.03 = 160.6 \ MeV \Delta Q = \text{Binding energy of products - Binding energy of reactants} = 150.88 - 160.6 = -9.72 \ MeV \text{ Hence, 9.72 MeV energy will be released}

Q: The activity of a nuclide is 15 millicurie. If its decay constant is 0.005 sec ^{-1} , the number of atoms present in it is

Decay constant = 0.005 sec^{-1} = \lambda Activity of nuclei = 15 \times3.7\times10^{7} dps [\because 1 millicurie = 3.7\times10^{7} dps] = 5.55\times10^{8} dps Now, from differential rate law: \dfrac{dN}{dt} = \lambda N [ where N is no of atoms present] \Rightarrow \dfrac{5.55\times10^{8}}{0.005} = N \Rightarrow N = 11.1\times 10^{10} atoms

Q: A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is 50 kg of the material present and after two hours it is observed that the material has lost 10%of its original mass, then the

\dfrac{dN}{dt} \propto N \dfrac{dN}{dt} = -\lambda N \dfrac{dN}{N} = -\lambda dt \ell nN = -\lambda t+c t = 0 , N = 50 \ell nN = -\lambda t + \ell n50 N = 50e^{-\lambda t} ......(1)At t = 2, N=45 45=50e^{-\lambda .2} \dfrac{9}{10} = e^{-2\lambda} \sqrt{\frac{9}{10}} = e^{-\lambda} \therefore N = 50(\dfrac{9}{10})^{t/2} ......(2)At t = 4 N = 50(0.9)^2 When N = 25kg 25 = 50(0.9)^{t/2} \Rightarrow \dfrac{1}{2}=(0.9)^{t/2} \Rightarrow t = \dfrac{2\ell n\frac{1}{2}}{\ell n0.9}

Q: The activity of a radioactive sample is measured as N _{0} counts per minute at t = 0 and N _{0} /e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

According to question, at t=5 min N=\cfrac{N_o}{e} Therefore, equation of nuclei left undecayed after time, t is given by, N=N_o\left(\cfrac{1}{2}\right)^{t/T_{1/2}} when t=5 min , N=\cfrac{N_o}{e} \implies \cfrac{N_o}{e}=N_o\left(\cfrac{1}{2}\right)^{5/T_{1/2}} \implies e^{-1}=\left(\cfrac{1}{2}\right)^{5/T_{1/2}} Calling log both sides, \implies -\ln(e)=\cfrac{5}{T_{1/2}}\ln\left(\cfrac{1}{2}\right) And as we know ln(e)=1 \implies T_{1/2}=-5\ln (1/2) \implies T_{1/2}=-5(-\ln 2) \implies T_{1/2}=5\ln 2 \implies T_{1/2}=5\log_{e}2

Q: The initial number of atoms of a radioactive element with half life 100 days, is 9.6\times10^{26} . The number of atoms remaining undecayed after 500 days, will be

Fraction undecayed after 500 days is =\dfrac{9.6\times10^{26}}{2^{(500/100)}}=\dfrac{9.6\times10^{26}}{32}=3\times10^{25}

Q: A piece of bone from an archaeological site is found to give a count rate of 15 counts per minute. A similar sample of fresh bone gives a count rate of 19 counts per minute. Calculate the age of the specimen. Given T_{1/2} = 5570\ years

Count rate of fresh sample N_{0} = 19 counts/minuteCount rate of bone N = 15 counts/minute T_{1/2} = 5570 years, Age of the sample, t = ? N = N_{0}e^{-\lambda}t \lambda = \dfrac {0.6931}{5570} 15 = 19e^{-\lambda t} (or) e^{\lambda t} = \dfrac {19}{15} \log e \dfrac {19}{15} = \lambda t (or) t = \log e \dfrac {19}{15} \times \dfrac {1}{\lambda} t = \dfrac {5570}{0.6931}\times 2.3026 \times \log_{10} \dfrac {19}{15} = \dfrac {5570}{0.6931}\times 2.3026 \times 0.1026 t = 1899\ years .

Q: In an experiment the initial number of radioactive nuclei is 3000 . It is found that 1000\pm 40 nuclei decayed in the 1.0\ s . For |x| < < 1, \ln(1+x) = x up to first power in x . The error \Delta \lambda , in the determination of the decay constant \lambda , is s^{-1} , is

N = N_{o} e ^{-\lambda t} ln N = lnN_{o} - \lambda t \dfrac{dN}{N} = - \lambda t Converting to error \dfrac{\Delta N}{N} = \Delta \lambda t \Delta \lambda = \dfrac{40}{2000 \times L} =0.02

Q: _{92}U^{238} decays to _{90}Th^{234} with half-life 4.5\times 10^9 year. The resulting _{90}Th^{234} is in excited state and hence, emits a further a gamma ray to come to the ground state, with half-life 10^{-8}\ s . A sample of _{92}U^{238} emits 20 gamma rays per second. In what time, the emission rate will drop to 5 gamma ray per second ?

\dfrac{dN}{dt}=-\lambda N [ \dfrac{dN}{dt} rate of decay N = no. of nucleus] \Rightarrow \dfrac{dN}{N}=-\lambda dt \Rightarrow \log N/N_0=-\lambda t ............(1)Now, \lambda N_0=20 and \lambda N=5 N/N_0=1/4 Now from (1) we get \Rightarrow \log 1/4=-\dfrac{\log 2}{t_{1/2}}t \Rightarrow t=-t_{1/2}\dfrac{\log 2^{-2}}{\log 2} \Rightarrow t=2t_{1/2}=2\times 4.5\times 10^9 =9\times 10^9 Years.

Q: A small amount of solution containing Na^{24} radionuclide with activity A= 2.0.10^3 disintegration per second was injected into the bloodstream of a man. The activity of 1 cm^3 of the blood sample taken t= 5.0 hours later turned out to be A'= 16 disintegration per minute per cm^3 . The half-life of the radionuclide is T= 15 hours. Find the volume of the man's blood.

Let V = volume of blood in the body of the human being. Then the total activity of the blood is A'V . Assuming all this activity is due to the injected Na^{24} and taking account of the decay of this radionuclide, we get V A' = A\ e^{-\lambda\ t} Now \lambda = \dfrac{ln\ 2}{15} per hour , t = 15 hourThus V = \dfrac{A}{A'} e^{-ln\ 2/3} = \dfrac{2.0 \times 10^{3}}{(16 / 60)} e^{-ln\ 2/3} cc = 5.95 litre

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Most Asked Questions

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Nuclei

- Understand the topics that are frequently asked in JEE exam

JEE Mains

Binding energy and binding energy per nucleon

If a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass, an energy

E_{b}

will be released in the process. The energy

E_{b}

is called the binding energy of the nucleus.

E_{b} = Δ M c^{2}

The ratio of the binding energy

E_{b}

of a nucleus to the number of the nucleons, A, in that nucleus is called the binding energy per nucleon,

E_{b n}

E_{b n} = \frac{E _{b}}{A}

Consider the nuclear fission,

N e^{20} \to 2 H e^{4} + C^{12}

Given that the binding energy/nucleon of

N e^{20}

H e^{4}

and

C^{12}

are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:

8.3 MeV energy will be released

energy of 12.4 MeV will be supplied

energy of 3.6 MeV has to be supplied

none of the above

View solution

Decay constant

N

is the number of nuclei in the sample and

Δ N

undergo decay in time

Δ t

then

\frac{Δ N}{Δ t} \propto N

\frac{Δ N}{Δ t} = λ N

where

λ

is called the radioactive decay constant or disintegration constant.

The activity of a nuclide is

15

millicurie. If its decay constant is

0.005

sec

^{- 1}

, the number of atoms present in it is

11.1 x 10

^{10}

1 x 10

^{9}

111 x 10

^{10}

1.1 x 10

^{10}

View solution

Activity of a radioactive substance

The total decay rate R of a sample of one or more radionuclides is called the activity of that sample.

R = - \frac{d N}{d t} = λ N e^{- λ t} = R_{0} e^{- λ t} = λ N

Here

R_{0}

is the radioactive decay rate at time t = 0, and R is the rate at any subsequent time t.

A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is

50 k g

of the material present and after two hours it is observed that the material has lost 10%of its original mass, then the

mass of the material after four hours is

50 (\frac{9}{1 0})^{2}

mass of the material after four hours is

50 . e^{- 0.5 I n 9}

time at which the material has decayed to half of its initial mass (in hours) is

\frac{( ℓ n \frac{1}{2} )}{( - \frac{1}{2} ℓ n 0 . 9 )}

time at which the material has decayed to half of its initial mass (in hours) is

\frac{2 l n \frac{1}{2}}{l n 0 . 9}

View solution

The activity of a radioactive sample is measured as N

_{0}

counts per minute at t = 0 and N

_{0}

/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

log

_{e}

2/5

\frac{5}{l o g _{e} 2}

5 log

_{10}

5 log

_{e}

View solution

Half life of a radioactive substance

The half-life

T_{1 / 2}

of a radio nuclide is the time at which both N and R have been reduced to one-half their initial values.
Putting

R = (1 / 2) R_{0}

and

t = T_{1 / 2}

in the law of radioactive decay, we get:

T_{1 / 2} = \frac{l n 2}{λ} = \frac{0 . 6 9 3}{λ}

The initial number of atoms of a radioactive element with half life

100

days, is

9.6 \times 1 0^{26}

. The number of atoms remaining undecayed after

500

days, will be

9.6 \times 1 0^{25}

3.84 \times 1 0^{25}

3 \times 1 0^{25}

19.2 \times 1 0^{25}

View solution

Fraction of substance decayed in a period of time

Tritium has a half-life of 12.5 y undergoing beta decay.What fraction of a sample of pure tritium will remain undecayed after 25 y.
Solution
By definition of half-life, half of the initial sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure tritium will remain undecayed.

A piece of bone from an archaeological site is found to give a count rate of

15

counts per minute. A similar sample of fresh bone gives a count rate of

19

counts per minute. Calculate the age of the specimen. Given

T_{1 / 2} = 5570 y e a r s

View solution

JEE Advanced

Decay constant

N

is the number of nuclei in the sample and

Δ N

undergo decay in time

Δ t

then

\frac{Δ N}{Δ t} \propto N

\frac{Δ N}{Δ t} = λ N

where

λ

is called the radioactive decay constant or disintegration constant.

In an experiment the initial number of radioactive nuclei is

3000

. It is found that

1000 \pm 40

nuclei decayed in the

1.0 s

. For

∣ x ∣ < < 1, ln (1 + x) = x

up to first power in

x

. The error

Δ λ

, in the determination of the decay constant

λ

, is

s^{- 1}

, is

0.04

0.03

0.02

0.01

View solution

_{92} U^{238}

decays to

_{90} T h^{234}

with half-life

4.5 \times 1 0^{9}

year. The resulting

_{90} T h^{234}

is in excited state and hence, emits a further a gamma ray to come to the ground state, with half-life

1 0^{- 8} s

. A sample of

_{92} U^{238}

emits

20

gamma rays per second. In what time, the emission rate will drop to

5

gamma ray per second ?

2 \times 1 0^{- 8} s

0.25 \times 1 0^{- 8} s

9 \times 1 0^{9}

year

1.125 \times 1 0^{9}

year

View solution

Activity of a radioactive substance

The total decay rate R of a sample of one or more radionuclides is called the activity of that sample.

R = - \frac{d N}{d t} = λ N e^{- λ t} = R_{0} e^{- λ t} = λ N

Here

R_{0}

is the radioactive decay rate at time t = 0, and R is the rate at any subsequent time t.

A small amount of solution containing

N a^{24}

radionuclide with activity

A = 2.0.1 0^{3}

disintegration per second was injected into the bloodstream of a man. The activity of 1

c m^{3}

of the blood sample taken t= 5.0 hours later turned out to be A'= 16 disintegration per minute per

c m^{3}

. The half-life of the radionuclide is T= 15 hours. Find the volume of the man's blood.

View solution

A human body excretes ( removes by waste discharge sweating etc.) certain materials by a law similar to radioactivity. if technitium is injected in some form in a human body.the body exerts half - the amount in 24hours . A patient is given an injection containing

^{99} t c

this isotope is radioactive with a half-life of 6 hours.the activity frim the body just after the injection is

6 μ C i

How much time will elapse before the activity falls to

3 μ C i

View solution

Half life of a radioactive substance

The half-life

T_{1 / 2}

of a radio nuclide is the time at which both N and R have been reduced to one-half their initial values.
Putting

R = (1 / 2) R_{0}

and

t = T_{1 / 2}

in the law of radioactive decay, we get:

T_{1 / 2} = \frac{l n 2}{λ} = \frac{0 . 6 9 3}{λ}

The radionuclide

^{56} M n

has a half-life of 2.58 h and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope

^{55} M n

, and the manganesedeuteron reaction that produces

^{56} M n

^{55} M n + d \to^{56} M n + p

If the bombardment lasts much longer than the half-life of

^{56} M n

, the activity of the

^{56} M n

produced in the target reaches a final value of

8.88 \times 1 0^{1} 0 B q

. (a) At what rate is

^{56} M n

being produced? (b) How many

^{56}

Mn nuclei are then in the target? (c) What is their total mass?

View solution

A source contains two phosphorous radio nuclides

_{15}^{33} P (T_{1 / 2} = 14.3 d)

and

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

. Initially,

10

% of the decays come from

T_{1 / 2}

. How long one must wait until

90

% do so?

View solution

Changes within the nucleus in alpha, beta and gamma emission

Alpha emission: If the nucleus of a radioactive element X of mass number A and atomic number Z emits an

α

particle, a new element Y (daughter nucleus) is formed which has mass number equal to (A-4) and atomic number equal to (Z-2). Thus due to emission of an alpha particle, atomic number Z decreases by two units and mass number decreases by 4 units.
Beta emission: In emitting a beta particle the number of nucleons in the nucleus (i.e. protons and neutrons) remain same, but the number of nuetrons is decreased by one and the number of protons is increased by one.
Gamma emission: In emitting gamma particle there is no change in mass number A and atomic number Z of the nucleus.

In a radioactive decay chain,

_{90}^{232} T h

nucleus decays to

_{82}^{212} P b

nucleus. Let

N_{α}

and

N_{β}

be the number of

α

and

β^{-}

particles respectively, emitted in this decay process. Which of the following statements is (are) true?

This question has multiple correct options

N_{α} = 5

N_{α} = 6

N_{β} = 2

N_{β} = 4

View solution