1 Marks Question

Question 11 Mark

Calculate the energy equivalent of $1 g$ of substance.

Answer

Energy, $E=10^{-3} \times\left(3 \times 10^8\right)^2 J$
$
E=10^{-3} \times 9 \times 10^{16}=9 \times 10^{13} J
$
Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.

View full question & answer→

Question 21 Mark

Given the mass of iron nucleus as $55.85 u$ and $A =56$, find the nuclear density?

Answer

$
m_{ Fe }=55.85, \quad u =9.27 \times 10^{-26} kg
$
Nuclear density $=\frac{\text { mass }}{\text { volume }}=\frac{9.27 \times 10^{-26}}{(4 \pi / 3)\left(1.2 \times 10^{-15}\right)^3} \times \frac{1}{56}$
$
=2.29 \times 10^{17} kg m ^{-3}
$
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

View full question & answer→

Question 31 Mark

Why is it found experimentally difficult to detect neutrinos in nuclear $\beta$-decay?

Answer

Neutrinos are neutral (chargeless), (almost) massless particles that hardly interact with matter.

Alternate Answer

The neutrinos can penetrate large quantity of matter without any interaction

Alternate Answer

Neutrinos are chargeless and (almost) massless particles.

View full question & answer→

Question 41 Mark

Define the activity of a given radioactive substance. Write its S.I. unit.

Answer

The rate of decay of a radioactive substance is called activity of that substance. It is negative of the rate of decay of the radioactive substance.

Activity, $\text{R} = \frac{\text{N}' - \text{N}}{\text{t}' - \text{t}}$

$ = - \frac{\Delta\text{N}}{\Delta\text{t}}$

$ \DeclareMathOperator*{\median}{\text{lim}} \median_{\Delta\text{t}\rightarrow0}\bigg(-\frac{\Delta\text{N}}{\Delta\text{t}}\bigg) = -\frac{\text{dN}}{\text{dt}}$ S.I. unit of activity

becquerel (Bq).
decay per second.

View full question & answer→

Question 51 Mark

Write any two characteristic properties of nuclear force.

Answer

Characteristic properties of nuclear force are short ranged, strong, attractive, charge independent, spin dependent, does not obey inverse square law, saturated, non central.

View full question & answer→

Question 61 Mark

Two nuclei have mass numbers in the ratio 1 : 8. What is the ratio of their nuclear radii?

Answer

$\text{R}= \text{R}_{0}\text{A}^{1/3}$

$\frac{\text{R}_{1}}{\text{R}_{2}} =\frac{1}{2}.$

View full question & answer→

Question 71 Mark

State two characteristic properties of nuclear force.

Answer

Properties of Nuclear Force:

Short range/saturation.
Strongest.
Charge independent.

View full question & answer→

Question 81 Mark

Define the term 'activity' of radionuclide. Write its SI unit.

Answer

The total decay rate of a sample of one or more radionuclide is called activity.

Alternate Answer

Radioactive disintegration taking place per second.

Alternate Answer

$\text{R} = - \frac{\text{dN}}{\text{dt}} , $where N is the total number of radionuclides at any time ‘t’. SI unit-becquerel(Bq) .

View full question & answer→

Question 91 Mark

Two nuclei have mass numbers in the ratio 1:2. What is the ratio of their nuclear densities?

Answer

Ratio of nuclear density equals 1:1.

View full question & answer→

Question 101 Mark

State the reason, why heavy water is generally used as a moderator in a nuclear reactor.

Answer

(HOTS) Lesser absorption probability of neutrons/more effective in slowing down neutrons to thermal energies.

View full question & answer→

Question 111 Mark

An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other?

Answer

$\text{K.E.} = \frac{p^2}{2m}$

$\therefore \text{K.E.} \propto \frac{1}{m} $ ( For same P)

(as $\lambda = h/ p $ is the same)

Alternate Answer

$\therefore \frac{E_ke}{E_ka} = \frac{m_a}{m}_e$

View full question & answer→

Question 121 Mark

The radioactive isotope D decays according to the sequence $\text{D}\xrightarrow{\beta}^{- }$ $\text{D}_{1}\xrightarrow{\alpha - particle}$ $\text{D}_{2}$

If the mass number and atomic number of $\text{D}_{1}$ are 176 and 71 respectively, what is (i) the mass number (ii) atomic number of D?

Answer

Mass number 180
Atomic number 72

Alternate Answer

$\text{D}^{180}_{72}$

View full question & answer→

Question 131 Mark

Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two - the parent or the daughter nucleus - would have higher binding energy per nucleon?

Answer

Daughter nucleus have higher binding energy per nucleon because they are more stable and higher stabilitymeans higher binding energy per nucleon.

View full question & answer→

Question 141 Mark

The nuclear radius of $^{27}_{13}\text{Al}$ is 3.6 fermi. Find the nuclear radius of $^{64}_{29}\text{Cu}.$

Answer

$\frac{\text{R}_2}{\text{R}_1}=\Big(\frac{\text{A}_2}{\text{A}_1}\Big)^\frac{1}{3}$
$\Rightarrow\text{R}_2=3.6\Big(\frac{64}{27}\Big)^\frac{1}{3}=3.6\times\frac{4}{3}$
$=4.3\text{ Fermi}.$

View full question & answer→

Question 151 Mark

Among alpha, beta and gamma radiations, which get affected by electric field?

Answer

Alpha and beta radiations are charged, so they are affected by electric field.

View full question & answer→

Question 161 Mark

Calculate the energy released if ²³⁸U emits an $\alpha$-particle.
Calculate the energy to be supplied to ²³⁸U it two protons and two neutrons are to be emitted one by one. The atomic masses of ²³⁸U, ²³⁴Th and ⁴He are 238.0508u, 234.04363u and 4.00260u respectively.

Answer

U²³⁸ ₂He⁴ + Th²³⁴

E = [Mu - (N_HC + M_Th)]u = 238.0508 - (234.04363 + 4.00260)]u = 4.25487Mev = 4.255Mev.

E = U²³⁸ - [Th²³⁴ + 2n'₀ + 2p'₁]

= {238.0508 - [234.64363 + 2(1.008665) + 2(1.007276)]}u

= 0.024712u = 23.0068 = 23.007MeV.

View full question & answer→

Question 171 Mark

What will be the ratio of radii of two nuclei of mass numbers A₁ and A₂?

Answer

Ra dius of nuceus R $=\text{R}_0\text{A}^{\frac{1}{3}}$
$\Rightarrow \frac{\text{R}_1}{\text{R}_2}=\Big(\frac{\text{A}_1}{\text{A}_2}\Big)^{\frac{1}{3}}$

View full question & answer→

Question 181 Mark

³²P beta-decays to ³²S. Find the sum of the energy of the antineutrino and the kinetic energy of the $\beta$-particle. Neglect the recoil of the daughter nucleus. Atomic mass of ³²P = 31.974u and that of ³²S = 31.972u.

Answer

$\text{P}^{32}\rightarrow\text{S}^{32}+ \ _0\bar{\text{v}}^0+ \ _1\beta^0$
Energy of antineutrino and $\beta$-particle
= (31.974 - 31.972)u = 0.002u = 0.002 × 931 = 1.862MeV = 1.86.

View full question & answer→

Question 191 Mark

A nucleus of mass number A, has a mass defect $\triangle \text{m}.$ Give the formula, for the binding energy per nucleon, of this nucleus.

Answer

BE per nucleon, B_n $=\frac{\text{Total binding energy}}{\text{Number of nucleons}}$
$=\frac{\triangle \text{mc}^2}{\text{A}}$
Where c is the speed of light in vacuum.

View full question & answer→

Question 201 Mark

When a p-type impurity is doped in a semiconductor, a large number of holes are created. This does not make the semiconductor charged. But when holes diffluse from the p-side to the n-side in a p-n junction, the n-side gets positively charged. Explain.

Answer

A p-type semiconductor is formed by doping a group 13 element with group 14 element (Si or Ge). As the group 13 element has only 3 electrons in its valence shell and the group 14 element has 4 electrons in its valence shell, when the group 13 element, say, Al, replaces one Si in the silicon crystal, only 3 covalent bonds are formed by it. And the fourth covalent bond is left in need of one electron. So, it creates a hole. Since the atom as a whole is electriclly neutral, the p-type semiconductor is also neutral.
In a p‒n junction, when the diffusion of holes takes place across the junction because of the difference in the concentration of charge carriers from p to n sides, these holes neutralise some of the electrons on the n side. So, the atom attached with that electron becomes one electron deficient and hence positively charged. This makes the n side of the p‒n junction positively charged and the p side of the p‒n junction negatively charged.

View full question & answer→

Question 211 Mark

What percentage of a given mass of a radioactive substance will be left undecayed after four half periods?

Answer

Percentage of mass of radioactive substance undecayed after n = 4 half-lives.
$=\Big(\frac{1}{2}\Big)^4\times100\%=\frac{100}{16}$
$=6.25\%$

View full question & answer→

Question 221 Mark

Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons it terms of the masses M_Z.N, M_Z_,N-1 and the mass of the neutron.

Answer

$\text{E}_2\text{N}=\text{E}_{\text{Z,N}-1}+\text{ }^1_0\text{n}.$
Energy released = (Initial Mass of nucleus - Final mass of nucleus)c² = (M_Z.N-1 + M₀ - M_ZN)c².

View full question & answer→

Question 231 Mark

Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is:
$\Delta\text{E}=(\text{M}_{\text{Z}-1,\text{N}}+\text{M}_{\text{H}}-\text{M}_{\text{Z,N}})\text{c}^2$
where M_Z,N = mass of an atom with Z protons and N neutrons in the nucleus and M_H = mass of a hydrogen atom. This energy is known as proton-separation energy.

Answer

E_Z.N. → E_Z-1, N + P₁ ⇒ E_Z.N. → E_Z-1, N + ₁H¹ [As hydrogen has no neutrons but protons only]
$\Delta$E = (M_Z-1,N + N_H - M_Z,N)c²

View full question & answer→

Question 241 Mark

Write any one equation representing nuclear fusion reaction.

Answer

Equation of fusion reaction.
$^2_1\text{H}\ \ +\ \ ^2_1\text{H}\ \ \rightarrow\ \ ^3_1\text{H}\ \ +\ \ ^1_1\text{H}\ \ +\ \ 4.03\text{MeV}$

View full question & answer→

Question 251 Mark

The binding energy per nucleon of the two nuclei A and B are 4 MeV and 8.2 MeV. Which of the two nuclei is more stable?

Answer

The nucleus (B) having larger binding energy is more stable.

View full question & answer→

Question 261 Mark

If the nuclei of masses X and Y are fused together to form a nucleus of mass m and some energy is released, then:

X + Y = m
X + Y < m
X + Y > m
X − Y = m

Answer

X + Y > m

Explanation:

Since some energy is released, so one part of total mass of reactants is converted into energy
Therefore, total mass of reactants

X + Y

is more than the mass of nucleus (m).

View full question & answer→

Physics 1 Marks Question

Test yourself on this topic