When a uranium isotope ^{235}_{92}U is bombarded with a neutron, it generates ^{89}_{36}Kr , three neutrons and :

_{92}U^{235} + \ _on^1 \to \ _QX^P + _{36}Kr^{89} + 3\ _on^1+ Q(energy) \sum Atomic number on LHS = 89 + 0 = 89 + 3 \sum Atomic number on RHS = Q + 36 + 3 \times 0 \therefore LHS = RHS Q+36 = 92 Q = 56 also, \sum atomic mass number on LHS = 235 + 1 = 236 \sum Atomic mass number on RHS=P + 89 + 3 \times 1 \therefore LHS = RHS P + 92= 236 P = 144 So, the element is _{56}Ba^{144}

Nuclei - Most Asked Questions

Q: A slow neutron strikes a nucleus of ^{235}_{92}U splitting it into lighter nuclei of ^{141}_{56}Ba and ^{92}_{36}Kr along with three neutrons. The energy released in this reaction is :(The masses of Uranium, Barium and Krypton in this reaction are 235.043933, 140.917700 and 91.895400\ amu respectively. The mass of a neutron is 1.008665 \ amu )

^{235}_{92}U + ^1n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3 ^1n \Delta m = [235.043933 - 140.9177 - 91.8954 - 2(1.008665)] = 0.213503 \ amu E = 0.21503\times 931.5 MeV = 198.87 \ MeV

Q: 20 % of a radioactive element disintegrates in 1hr .The percentage of the radioactive element disintegrated in 2hrs will be

Let initially the number of radioactive particles be 100 .Using the formula: N = N_0 e^{-\lambda t} where \lambda = \dfrac{2.303}{t} log(\dfrac{N_0}{N}) \lambda = \dfrac{2.303}{1} log(\dfrac{100}{80}) \Rightarrow \lambda = 0.223 hrs^{-1} Radioactive material left in 2\ hrs = N_0 e^{-\lambda t} = 100 e^{-0.223\times2} =63.99 \approx 64% Radioactive material decayed in 2 hours = (100-64)% = 36 %

Q: For a radioactive element, the correct representation of number of nuclei (N) decayed after time (t) is represented by

Number of nuclei decayed after time t is =N_0(1-e^{-\lambda t}) Hence option (b) represents the correct variation.

Q: A radioactive sample has 5000 disintegration per minute initially and 1250 disintegration per minute after 5 seconds. Then the decay constant is

Using formulae, N=N_o e^{-\lambda t} \Rightarrow 1250 =5000 e^{-\lambda 5} \Rightarrow ln(\dfrac{1}{4})= -\lambda 5 \Rightarrow 2 ln 2=5 \lambda \Rightarrow \lambda=0.4 ln 2.

Q: A radioactive substance after 48 days remains 25% of initial. Find disintegration constant.

t = 48 days, a = 100, a - x = 25 K = \displaystyle \frac{2.303}{t} log \frac{a}{a - x} = \frac{2.303}{48} log \frac{100}{25} = 0.048 log 4 = 0.048 \times 0.060 = 0.0289 time^{-1} = 2.89 \times 10^{-2} time^{ -1}

Q: The rate of disintegration of a radioactive substance falls from 800 decay/min to 100 decay/min in 6 hours. The half-life of the radioactive substance is:

A=A_0e^{-\lambda t} \Rightarrow 100=800e^{-\lambda (6\times60)} \Rightarrow e^{-360\lambda }=\dfrac{1}{8} \Rightarrow-360\lambda =ln\dfrac{1}{8} =-ln8 \Rightarrow \lambda =\dfrac{ln2^{3}}{360}=\dfrac{ln2}{120} \therefore T_{1/2}=\dfrac{ln2}{\lambda }=\dfrac{ln2}{(ln2/120)}=120\ min or T_{1/2}=2\ hrs

Q: How many half-lives does Carbon-10 undergo in 2 minutes and 32 seconds? Given the half-life of Carbon-10 is about 19 seconds.

Given : T_{1/2} = 19 sec Time of observation, t =2 min and 32 sec = 152 secThus number of half lives, n = \dfrac{t}{T_{1/2}} = \dfrac{152}{19} = 8 half-lives

Q: The energy equivalent of 0.5 g of a substance is :

The energy equivalent of the particular mass is given as: E=mc^2 mass is given as =0.5g=0.5\times 10^{-3}\ kg and the speed of light (c)=3\times 10^8\ m/s ;.So, E=0.5\times10^{-3}\times(3\times 10^{8})^2 E=4.5\times 10^{!3}\ J

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}

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Nuclei

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Examples of nuclear fission reactions

$_{0}^{1} n +_{92}^{235} U \to_{92}^{236} U \to_{56}^{144} B a +_{36}^{89} K r + 3_{0}^{1} n$
Nuclear fission of uranium to produce Barium and Krypton.
$_{0}^{1} n +_{92}^{235} U \to_{54}^{140} X e +_{38}^{94} S r + 2_{0}^{1} n$

A slow neutron strikes a nucleus of

_{92}^{235} U

splitting it into lighter nuclei of

_{56}^{141} B a

and

_{36}^{92} K r

along with three neutrons. The energy released in this reaction is :
(The masses of Uranium, Barium and Krypton in this reaction are

235.043933, 140.917700

and

91.895400 a m u

respectively. The mass of a neutron is

1.008665 a m u

)

198.87 M e V

156.9 M e V

186.9 M e V

196.9 M e V

View solution

When a uranium isotope

_{92}^{235} U

is bombarded with a neutron, it generates

_{36}^{89} K r

, three neutrons and :

_{40}^{91} Z r

_{36}^{101} K r

_{36}^{103} K r

_{56}^{144} B a

View solution

Law of radioactive decay

We have

\frac{d N}{d t} = - λ N

\frac{d N}{N} = - λ d t

Integrating both sides,

\int_{N_{0}}^{N} \frac{d N}{N} = - λ \int_{t_{0}}^{t} d t

l n (\frac{N}{N _{0}}) = - λ t

N = N_{0} e^{- λ t}

This is the law of radioactive decay

20

% of a radioactive element disintegrates in

1 h r

.The percentage of the radioactive element disintegrated in

2 h r s

will be

36

64

60

40

View solution

For a radioactive element, the correct representation of number of nuclei (N) decayed after time (t) is represented by

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.

A radioactive sample has

5000

disintegration per minute initially and

1250

disintegration per minute after

5

seconds. Then the decay constant is

0.4 l n 2

0.2 l n 2

0.1 l n 2

0.8 l n 2

View solution

A radioactive substance after 48 days remains 25% of initial. Find disintegration constant.

2.89 \times 1 0^{- 2} T^{- 1}

3.89 \times 1 0^{- 3} T^{- 1}

4.89 \times 1 0^{- 2} T^{- 1}

none of these

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 rate of disintegration of a radioactive substance falls from 800 decay/min to 100 decay/min in 6 hours. The half-life of the radioactive substance is:

6/7 hr

2 hr

3 hr

1 hr

View solution

How many half-lives does Carbon-10 undergo in 2 minutes and 32 seconds? Given the half-life of Carbon-10 is about 19 seconds.

8 half-lives

7 half-lives

6 half-lives

5 half-lives

4 half-lives

View solution

Relationship between mass and energy

By the theory of special relativity Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa.Einstein gave the famous mass-energy equivalence relation

E = m c^{2}

where

c = 3 \times 1 0^{8} m / s

is the speed of light.

The energy equivalent of

0.5 g

of a substance is :

4.5 \times 1 0^{13} J

1.5 \times 1 0^{13} J

0.5 \times 1 0^{13} J

4.5 \times 1 0^{16} J

View solution

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

Consider the nuclear reaction: X

^{200}

\to

^{110}

^{90}

+ energy .
The binding energy per nucleon for X

^{200}

, A

^{110}

, B

^{90}

is respectively 7.4MeV, 8.2MeV and 8.2Mev. The energy released is

200MeV

160MeV

110MeV

90MeV

View solution