Q: 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}

Q: The Binding energy per nucleon of ^7_3Li and _2^4He nucleon are 5.60 MeV and 7.06 MeV , respectively. In the nuclear reaction _3^7Li+_1^1H\rightarrow _2^4He+^4_2He+Q , the value of energy Q released is

_4Li^7 + _1H^1 \rightarrow _2(_2He^4) BE of products = (5.6 MeV) \times 7 + 0 = 39.2 MeV E = -39.2\ MeV BE of reactant = (7.06) \times 4 \times 2 = 56.48 MeV E_f =- 56.48\ MeV As nuclear energy decreases, so some energy will be released Q_{release}=E_i-E_f=(-39.2)-(-56.48)=17.28\ MeV

Question 1

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

Accepted Answer

_{92}U^{235} + \ _on^1 	o \ _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 	imes 0 	herefore 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 	imes 1 	herefore LHS = RHS P + 92= 236 P = 144 So, the element is  _{56}Ba^{144}

Question 2

The energy equivalent of  0.5 g  of a substance is :

Accepted Answer

The energy equivalent of the particular mass is given as: E=mc^2 mass is given as  =0.5g=0.5	imes 10^{-3}\ kg and the speed of light  (c)=3	imes 10^8\ m/s ;.So,  E=0.5	imes10^{-3}	imes(3	imes 10^{8})^2 E=4.5	imes 10^{!3}\ J

Question 3

\alpha -particle consists of :

Accepted Answer

\alpha   - particle is nucleus of Helium which has two protons and two neutrons.Hence option A is correct answer.

Question 4

Radioactive material ' A ' has decay constant ' 8\lambda ' and material ' B ' has decay constant ' \lambda '. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be  {e} ?

Accepted Answer

N_A=N_{A_o}e^{(-8\lambda)t} N_B = N_{B_o}e^{-\lambda t} N_{A_o} = N_{B_0} \Rightarrow \dfrac{N_B}{N_A}  = e \Rightarrow \dfrac{e^{-\lambda t}}{e^{-8\lambda t}} = e^{1} \Rightarrow e^{7\lambda t } = e^{1} 7\lambda t = 1 t = \dfrac{1}{7\lambda}

Question 5

For a radioactive material, half-life is  10  minutes. If initially there are  600  number of nuclei, the time taken (in minutes) for the disintegration of  450  nuclei is?

Accepted Answer

Half life   = 10min = \dfrac {ln2}{\lambda} Nuclei disintegrate = 450Nuclei left = 600-450 =150 N=N_{o}e^{-\lambda}{t} 150 = 600 e^{\dfrac{-ln2}{10} t} so, t = 20 minutes

Question 6

In the reaction  _{1}^{2}	extrm{H}+_{1}^{3}	extrm{H}ightarrow  _{2}^{4}	extrm{He}+_{0}^{1}	extrm{n} . If the binding energies of  _{1}^{2}	extrm{H}, _{1}^{3}	extrm{H}  and  _{2}^{4}	extrm{He}   are respectively  a, b  and  c  (in MeV), then the energy(in MeV) released in this reaction is:

Accepted Answer

_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\longrightarrow _{ 2 }^{ 4 }{ He }+_{ 0 }^{ 1 }{ n } Energy released,  E=\left( 	riangle m ight) 	imes 931MeV (Note:- 1  atomic mass unit =931MeV ) 	riangle m= mass of product -  mass of reactant 	riangle m=c-a-b E=\left( 	riangle m ight) 	imes 931  or  E=\left( c-a-b ight)  (The binding energies of  _{ 1 }^{ 2 }{ H },_{ 1 }^{ 3 }{ H }  and  _{ 2 }^{ 4 }{ He }  are respectively a,b and c)

Question 7

The Binding energy per nucleon of  ^7_3Li  and  _2^4He  nucleon are  5.60 MeV  and  7.06 MeV , respectively. In the nuclear reaction  _3^7Li+_1^1H\rightarrow _2^4He+^4_2He+Q , the value of energy  Q  released is

Accepted Answer

_4Li^7 + _1H^1 ightarrow _2(_2He^4) BE of products  = (5.6 MeV) 	imes 7 + 0 = 39.2 MeV E = -39.2\ MeV BE of reactant  = (7.06) 	imes 4 	imes 2 = 56.48 MeV E_f =- 56.48\ MeV As nuclear energy decreases, so some energy will be released Q_{release}=E_i-E_f=(-39.2)-(-56.48)=17.28\ MeV

Question 8

The half life of a radioactive isotope 'X' is 20 years. It decays to another element 'Y' which is stable. The two elements 'X' and 'Y' were found to be in the ratio 1:7 in a sample of a given rock. The age of the rock is estimated to be

Accepted Answer

\displaystyle \frac{N}{N_{0}}=\frac{1}{8}=\frac{1}{2^{3}} 3  half lives,  T=3	imes 20=60  years

Question 9