Q: The value of 1 amu is equal to:

One an amu is \$\dfrac{1}{12}\$ of the mass of one carbon-12 atom. It is equal to \$ \dfrac{1}{N_A} \$\$= \dfrac{1}{6.022\times10^{23}} = 1.66 \times 10^{-24} g\$.

Q: What is the mass of one atom of \$C-12\$ in grams?

Mass of \$1\$ mole = \$12\,gm\$Mass of \$6.022 \times 10^{23}\,atom = 12\,gm\$mass of \$1\$ atom = \$\dfrac{12}{6.023 \times 10^{23}} = 1.993 \times 10^{-23}\,gm\$

Q: The binding energy per nucleon of deuteron \$(_{1}H^{2})\$ and helium \$(_{2}He^{4})\$ are \$1.1\ MeV\$ and \$7.0\ MeV\$, respectively. The energy released when two deuterons fuse to form a helium nucleus is

The nuclear reaction is,\$_{1}H^{2} + _{1}H^{2} \rightarrow _{2}He^{4} + Q\$Total binding energy of helium nucleus\$= 4\times 7 = 28\ MeV\$Total binding energy of each deuteron\$= 2\times 1.1 = 2.2\ MeV\$Hence, energy released when two deuteron fuse to form helium,\$= 28 - 2 \times 2.2 = 28 - 44 = 23.6\ MeV\$.

Q: Which of the following isotopes is used for treatment of cancer?

\$Co^{60}\$Cobalt therapy or cobalt -60 therapy is the medical use of gamma rays from the radioisotope cobalt 60 to treat conditions such as cancer.

Q: The total energy of an electron in the first excited state of the hydrogen atom is \$-3.4\$eV. Find the out its (i) kinetic energy and (ii) potential energy in this state.

We know that \$\dfrac{|P.E|}{2} = |T.E| = K.E\$So, \$\dfrac{|P.E|}{2} =3.4 = K.E\$So, kinetic energy \$K.E = 3.4 \ eV\$Potential energy \$P.E =- 2\times |T.E| = -2\times (3.4) = -6.8 \ eV\$

Question 1

The value of 1 amu is equal to:

Accepted Answer

One an amu is $\dfrac{1}{12}$ of the mass of one carbon-12 atom.  It is equal to $ \dfrac{1}{N_A} $$= \dfrac{1}{6.022\times10^{23}} = 1.66 \times 10^{-24} g$.

Question 2

Draw a graph showing the variation of potential energy between a pair of nucleons as a function of their separation. Indicate the regions in which the nuclear force is(i) Attractive(ii) Repulsive.Write two important conclusions which you can draw regarding the nature of the nuclear forces.

Accepted Answer

Conclusions:(i) The potential energy is minimum at a distance $r_0$ of about $0.8\,fm.$(ii) Nuclear force is attractive for distance larger than $r_0.$(iii) Nuclear force is repulsive if two are separated by distance less than $r_0.$(iv) Nuclear force decreases very rapidly at $r_0 / $equilibirium position.

Question 3

What is the mass of one atom of $C-12$ in grams?

Accepted Answer

Mass of $1$ mole = $12\,gm$Mass of $6.022 \times 10^{23}\,atom = 12\,gm$mass of $1$ atom = $\dfrac{12}{6.023 \times 10^{23}} = 1.993 \times 10^{-23}\,gm$

Question 4

Draw a plot potential energy of a pair of nucleons as a functions of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces.

Accepted Answer

The potential energy (PE) vs separation graph is shown in figure. From the graph- (i) the nuclear force is repulsive when the separation is less 1 fm $(1 fermi =10^{-15} m)$. (ii) It is attractive when the separation is greater than 1 fm. The characteristic properties of nuclear force:i) The nuclear force is short range force.iii) The nuclear force is independent of electric charge.

Question 5

Composition of the nuclei of two atomic species $X$ and $Y$ are given as under                      $X$        $Y$Protons $=$     $6$         $6$Neutrons$=$    $6$         $8$Given the mass numbers of $X$ and $Y$. What is the relation between the two species?

Accepted Answer

Atomic mass=number of protons+number of neutronsAtomic mass of X = 6+6=12Atomic mass of Y =6+8=14Both X and Y have same atomic number but different atomic mass number so they are isotopes.

Question 6

State the law of radioactive decay. Hence derive the expression $N={ N }_{ 0 }{ e }^{ -\lambda t }$ where symbols have their usual meanings.

Accepted Answer

Law of radioactive decay : At any instant, the rate of radioactive disintegration is directly proportional to the number of nuclei of the radioactive element present at that instant.We know by radioactive decay law$\dfrac { dN }{ dt } \propto N$or $\dfrac { dN }{ dt } =-\lambda N$where $\lambda$ is the constant of proportionality called radioactive decay constant. Let $ { N }_{ 0 }$ be the number of nuclei present at time $t=0$, and $N$ the number of nuclei present at time $t$.$\int _{ { N }_{ 0 } }^{ N }{ \dfrac { dN }{ N }  } =-\int _{ 0 }^{ t }{ \lambda dt } =-\lambda \int _{ 0 }^{ t }{ dt }$$ \log _{ e }{ N } -\log _{ e }{ { N }_{ 0 } } =-\lambda t$$\log _{ e }{ \left( \dfrac { N }{ { N }_{ 0 } }  \right)  } =-\lambda t$$\dfrac { N }{ { N }_{ 0 } } ={ e }^{ -\lambda t }$$N={ N }_{ 0 }{ e }^{ -\lambda t }$

Question 7

The binding energy per nucleon of deuteron $(_{1}H^{2})$ and helium $(_{2}He^{4})$ are $1.1\ MeV$ and $7.0\ MeV$, respectively. The energy released when two deuterons fuse to form a helium nucleus is

Accepted Answer

The nuclear reaction is,$_{1}H^{2}  + _{1}H^{2} \rightarrow _{2}He^{4} + Q$Total binding energy of helium nucleus$= 4\times 7 = 28\ MeV$Total binding energy of each deuteron$= 2\times 1.1 = 2.2\ MeV$Hence, energy released when two deuteron fuse to form helium,$= 28 - 2 \times 2.2 = 28 - 44 = 23.6\ MeV$.

Question 8

An atom of an element contains $13$ electrons. Its nucleus has $14$ neutrons. Find out its atomic number and approximate atomic mass. An isotope has atomic mass $2$ units higher. What will be the number of protons, neutrons and electrons in the isotope?

Accepted Answer

Atomic number=No. of protons=No. of electrons in ground state(unchanged atom)Atomic number=$13$=No. of protonsAtomic mass=no. of protons+no. of neutrons=$13+14=27$For isotope no. of proton=$13$(same atomic number but different mass number are isotopes)no. of electrons=$13$no. of neutrons=$14+2=16$

Question 9

Which of the following isotopes is used for treatment of cancer?

Accepted Answer

$Co^{60}$Cobalt therapy or cobalt -60 therapy is the medical use of gamma rays from the radioisotope cobalt 60 to treat conditions such as cancer.

Question 10

The total energy of an electron in the first excited state of the hydrogen atom is $-3.4$eV. Find the out its (i) kinetic energy and (ii) potential energy in this state.

Accepted Answer

We know that     $\dfrac{|P.E|}{2} = |T.E| = K.E$So,   $\dfrac{|P.E|}{2} =3.4 = K.E$So, kinetic energy    $K.E = 3.4 \ eV$Potential energy    $P.E =- 2\times |T.E| = -2\times (3.4) = -6.8 \ eV$