Q: Question 16A nucleus ZAX_{Z}^{A}\text{X}ZAX emits 2 α particles and 1 β particle to form a 85222R_{85}^{222}\text{R}85222R nucleus. Find the atomic number and mass number of X.

85222R_{85}^{222}\text{R}85222R before the emission of β particle, would have atomic number 84, and before 2 α emissions, atomic number would have been 88. Hence, atomic number (Z) = 88.85222R_{85}^{222}\text{R}85222R before the emission of β particle, would have mass number 222, and before 2 α emissions, mass number would have been 230. Hence, mass number (A) = 230.

Q: Question 17What are radio isotopes? Give one example of a radio isotope. State one use of radio isotopes.

One example of radio isotopes is Cobalt 2760Co_{27}^{60}\text{Co}2760Co (Z = 27 and A = 60).Radio isotopes are used in the Medical field. Many diseases such as leukemia, cancer, etc. are cured by radio therapy with the help of radio isotopes.

Question 1

Question 3A radioactive source emits three type of radiations.(a) Name the radiation of zero mass.(b) Name the radiation which has the lowest ionising power.(c) Name the radiation which has the lowest penetrating power.(d) Give the charge and mass of particles composing the radiation in part (c).(e) When the particle referred to in part (c) becomes neutral, it is found to be the atom of a rare gas. Name this rare gas and draw a model of it's neutral atom.(f) From which part of the atom do these radiations come?

Accepted Answer

(a) The radiation of zero mass are γ radiation.(b) The radiation which has the lowest ionising power is the γ radiation.(c) The radiation which has the lowest penetrating power is the α radiation.(d) α particles have a charge of +3.2 × 10-19 C (twice the charge of a proton) and rest mass of 6.68 × 10-27 kg (four times the mass of proton)(e) When α particles become neutral, it is found to be the atom of Helium gas. Below is the model of its neutral atom:(f) The radiations come from the nucleus of the atom

Question 2

Question 7State following four properties each of α, β and γ radiations — (a) nature, (b) charge, (c) mass, and (d) effect of electric field.

Accepted Answer

α RadiationNature — Stream of positively charged particles i.e., helium nucleusCharge — Positive charge (two times that of a proton) = +3.2 × 10-19 C (or +2e)Mass — Four times the mass of proton i.e., 6.68 × 10-27 kgEffect of Electric Field — Less deflectedβ RadiationNature — Stream of negatively charged particles, i.e., energetic electronsCharge — Negative charge = -1.6 × 10-19 C (or –e)Mass — Equal to the mass of electron i.e., 9.1 × 10-31 kgEffect of Electric Field — More deflected than α radiations, but in a direction opposite to those of α radiationsγ RadiationNature — Highly energetic electromagnetic radiationCharge — No charge (Neutral)Mass — No mass (rest mass is zero)Effect of Electric Field — Undeviated

Question 3

Question 12A nucleus 1124Na_{11}^{24}	ext{Na}1124​Na is β-radioactive.(a) What are the numbers 24 and 11 called?(b) Write the equation representing β-decay.(c) What general name is given to the product nucleus with respect to 1124Na_{11}^{24}	ext{Na}1124​Na.

Accepted Answer

(a) The number 24 is called the mass number and the number 11 is called the atomic number.(b) The equation representing β-decay is as follows —1124Na⟶ 1224Mg+−1   0e_{11}^{24}	ext{Na} \longrightarrow \space _{12}^{24}	ext{Mg} + _{-1}^{\space \space\space 0}	ext{e}1124​Na⟶ 1224​Mg+−1   0​e(c) The product nucleus is called the isobar because they have same mass number and different atomic number.

Question 4

Question 5What are isobars? Give one example.

Accepted Answer

The atoms of different elements which have the same mass number A, but different atomic number Z are called isobars.Example — 1123Na_{11}^{23}	ext{Na}1123​Na and 1223Mg_{12}^{23}	ext{Mg}1223​Mg are isobars.

Question 5

Question 6Name the atoms of a substance having same atomic number, but different mass numbers. Give one example of such a substance. How do the structures of such atoms differ?

Accepted Answer

The atoms of a substance having same atomic number, but different mass number are known as isotopes.Example — Hydrogen has three isotopes namely protium 11H_{1}^{1}	ext{H}11​H, deuterium 12H_{1}^{2}	ext{H}12​H and tritium 13H_{1}^{3}	ext{H}13​H. Each isotope in it's nucleus has one proton (Z = 1) but protium (11H_{1}^{1}	ext{H}11​H) has no neutron, deuterium (12H_{1}^{2}	ext{H}12​H) has one neutron and tritium (13H_{1}^{3}	ext{H}13​H) has two neutrons. There is one electron outside the nucleus in the atom of each isotope. Thus, the structures of these isotopes differ by the number of neutrons in their nuclei.

Question 6

Question 14What happens to the position of an element in the periodic table when it's nucleus emits (a) an α particle, (b) a β particle and (c) γ radiation? Give reason for your answer.

Accepted Answer

(a) When the nucleus of an element emits an α particle, the position of the resultant element changes to two places earlier (i.e., to the left of the parent element in the periodic table).This happens because the atomic number decreases by 2, hence the new element shifts two places to the left on emission of an α particle.(b) When the nucleus of an element emits a β particle, the position of the resultant element changes to one place after (i.e., one place to the right ).This happens because the atomic number increases by 1, hence the new element shifts one place to the right on emission of a β particle.(c) When the nucleus of an element emits a γ radiation, there is no change in the position of the element in the periodic table.This happens because there is no change in the atomic number, hence there is no change in the position of the element on emission of γ radiations.

Question 7

Question 15What changes occur in a nucleus of a radioactive element when it emits (a) an alpha particle, (b) a beta particle, (c) gamma radiation? Give one example, in each case (a) and (b) in support of your answer.

Accepted Answer

(a) If the nucleus X of a radioactive element of mass number A and atomic number Z emits an α particle, the daughter nucleus Y of a new element has mass number equal to (A - 4) and atomic number equal to (Z - 2).Thus, atomic number decreases by 2 units and mass number decreases by 4 units.ZAX⟶Z−2A−4Y+ 24HeParent nucleusDaughter nucleus(α-particle)\begin{matrix} _{Z}^{A}	ext{X} & \longrightarrow & _{Z-2}^{A-4}	ext{Y} & + & \space _{2}^{4}	ext{He} \ \footnotesize{	ext{Parent nucleus}} & & \footnotesize{	ext{Daughter nucleus}} & & \footnotesize{	ext{(α-particle)}} \end{matrix}ZA​XParent nucleus​⟶​Z−2A−4​YDaughter nucleus​+​ 24​He(α-particle)​Example — When a radioactive uranium nucleus 92238U_{92}^{238}	ext{U}92238​U emits an α particle, a new nucleus thorium 90234Th_{90}^{234}	ext{Th}90234​Th is formed and the change is represented as follows —92238U⟶ 90234Th+ 24He_{92}^{238}	ext{U} \longrightarrow \space _{90}^{234}	ext{Th} + \space _{2}^{4}	ext{He}92238​U⟶ 90234​Th+ 24​He(b) In emitting a β particle, the number of nucleons in the nucleus remains same, but the number of neutrons is decreased by one and the number of protons is increased by one.If a radioactive nucleus P with mass number A and atomic number Z emits a β particle to form a daughter nucleus Q with mass number A and atomic number Z + 1, the change can be represented as follows:ZAP⟶Z+1     AQ+ −1   0eParent nucleusDaughter nucleus(β-particle)\begin{matrix} _{Z}^{A}	ext{P} & \longrightarrow & _{Z+1}^{\space \space \space \space \space A}	ext{Q} & + & \space _{-1}^{\space \space \space 0}	ext{e} \ \footnotesize{	ext{Parent nucleus}} & & \footnotesize{	ext{Daughter nucleus}} & & \footnotesize{	ext{(β-particle)}} \end{matrix}ZA​PParent nucleus​⟶​Z+1     A​QDaughter nucleus​+​ −1   0​e(β-particle)​Example — A radioactive carbon nucleus 614C_{6}^{14}	ext{C}614​C, having 14 nucleons (6 protons and 8 neutrons) emits a β particle and changes to a new nucleus nitrogen 714N_{7}^{14}	ext{N}714​N having 14 nucleons (7 protons and 7 neutrons).The change is represented as follows —614C⟶   714N+−1   0e_{6}^{14}	ext{C} \longrightarrow \space _{\space \space 7}^{14}	ext{N} + _{-1}^{\space \space \space 0}	ext{e}614​C⟶   714​N+−1   0​e(c) There is no change in the mass number A and atomic number Z of the nucleus in gamma emission. It is represented as follows:ZAX*⟶ZAX+ γParent nucleusDaughter nucleus(gamma radiation)\begin{matrix} _{Z}^{A}	ext{X*} & \longrightarrow & _{Z}^{A}	ext{X} & + & \space 	ext{γ} \ \footnotesize{	ext{Parent nucleus}} & & \footnotesize{	ext{Daughter nucleus}} & & \footnotesize{	ext{(gamma radiation)}} \end{matrix}ZA​X*Parent nucleus​⟶​ZA​XDaughter nucleus​+​ γ(gamma radiation)​Here the star indicates the excited state of the nucleus. Thus in gamma emission, the excited nucleus comes to its ground state.

Question 8

Question 16A nucleus ZAX_{Z}^{A}	ext{X}ZA​X emits 2 α particles and 1 β particle to form a 85222R_{85}^{222}	ext{R}85222​R nucleus. Find the atomic number and mass number of X.

Accepted Answer

85222R_{85}^{222}	ext{R}85222​R before the emission of β particle, would have atomic number 84, and before 2 α emissions, atomic number would have been 88. Hence, atomic number (Z) = 88.85222R_{85}^{222}	ext{R}85222​R before the emission of β particle, would have mass number 222, and before 2 α emissions, mass number would have been 230. Hence, mass number (A) = 230.

Question 9

Question 17What are radio isotopes? Give one example of a radio isotope. State one use of radio isotopes.

Accepted Answer

One example of radio isotopes is Cobalt 2760Co_{27}^{60}	ext{Co}2760​Co (Z = 27 and A = 60).Radio isotopes are used in the Medical field. Many diseases such as leukemia, cancer, etc. are cured by radio therapy with the help of radio isotopes.

Question 10

Question 18Why are the alpha particles not used in radio therapy? [Hint : alpha particles cannot penetrate the human skin]

Accepted Answer

The alpha particles are not used in radio therapy because they cannot penetrate the human skin.

Question 11

Question 19When does the nucleus of an atom tend to become radioactive?

Accepted Answer

The nucleus of an atom becomes radioactive when number of neutrons in the nucleus exceeds the number of protons in it.

Question 12

Question 20State the medical use of radioactivity.

Accepted Answer

Many diseases such as leukemia, cancer, etc., are cured by radiation therapy. γ radiations from cobalt-60 (2760Co_{27}^{60}	ext{Co}2760​Co) are used to treat cancer by killing the cells in the malignant tumor of the patient.The salt of weak radioactive isotopes such as radio-sodium chloride, radio-iron and radio-iodine are used for diagnosis. Such radio isotopes are called the tracers. These tracers are used to detect the suspected brain tumors and blood clots before they become dangerous.γ rays emitted by the radio isotopes are used to sterilize bandages, dressings, syringes and other equipments to make them free from germs.

Question 13

Question 21Arrange the α, β, and γ radiation in ascending order of their biological damage. Give reason.

Accepted Answer

α < β < γAn α-particle rapidly dissipates it's energy as it moves through a medium and therefore it's penetrating power is quite small. It can penetrate only through 3-8 cm in air. It can easily be stopped by a thin card sheet or a thick paper. As it is least penetrating so it causes least biological damage.The penetrating power of β-particles is more than that of the α-particles but less than that of γ-particles. They can travel through nearly 5 m in air and pass through thin card sheet, and even through thin aluminium foil, but a 5 mm thick aluminium sheet can stop the beta particles. As the penetrating power is in between α and γ, hence the biological damage cause by β radiation is also in between α and γ.The penetrating power of γ-rays is high. It is about (100,00) times that of α-particles and (100) times that of β-particles. They can pass through 500 m in air or through 30 cm thick sheet of iron. A thick sheet of lead is required to stop them. As penetrating power is most hence biological damage caused by γ-rays is most.

Question 14

Question 22Name two main sources of nuclear radiations. How are the nuclear radiations harmful?

Accepted Answer

Two main sources of nuclear radiations are:Radioactive fall out from nuclear plants — The nuclear power plants are now a major source of electricity in the world. If somehow, there is an accident in the reactor of power plant, a large amount of radioactive material and radiations will escape out into the atmosphere. This will not only affect the population around the plant, but will also affect the life at far off places where they will reach due to air currents.Nuclear waste — The nuclear waste are still very radioactive and are a source of harmful radiations therefore, they should not be dumped in the open garbage. They can contaminate water and soil and affect the human and living organisms.The harmful biological effects of nuclear radiations are of three types:Short term recoverable effects like diarrhea, sore throat, loss of hair, nausea, etc.Long term irrecoverable effects like leukemia and cancer.Genetic effects that appear in the later generations of the person exposed to the radiation as the genes in the living cell get modified.

Question 15

Question 23State two safety measures to be taken while establishing a nuclear power plant?

Accepted Answer

Two safety measures to be taken while establishing a nuclear power plant are:The nuclear reactor of the power plant must be shielded with lead and steel walls so as to stop radiations from escaping out to the environment during it's normal operation.The nuclear reactor must be housed in an air tight building of strong concrete structure which can withstand earthquakes, fires and explosion.

Question 16

Question 24What is meant by nuclear waste? State one way for the safe disposal of nuclear waste?

Accepted Answer

The fuel rods used in the nuclear power reactors are rejected when their activity decreases below a certain level. They then become a nuclear waste.These rods are still quite radioactive and are the source of harmful radiations therefore they should not be dumped in open garbage. They can contaminate water and soil and affect the human and living organism.

Question 17

Question 25State three safety precautions that you would take while handling the radioactive substances.

Accepted Answer

Safety precautions that people should take while handling the radioactive substances are —They should put on special lead lined aprons and lead gloves.They should handle the radioactive materials with long lead tongs.The radioactive substances must be kept in a thick lead container with a very narrow opening, so that the walls of the container absorb the radiations which strike on them from the inside and thus radiations only come out through the narrow opening.

Question 18

Question 26Why should a radioactive substance not be touched by hands?

Accepted Answer

Radioactive substance should not be touched by hands because these radiations are harmful and can damage the living tissues and cause radiation burns.

Question 19

Question 27What do you mean by background radiations? Name it's two sources. Is it possible for us to keep ourselves away from it?

Accepted Answer

Background radiations are the radioactive radiations (such as α, β and γ) to which we all are exposed, even in the absence of an actual visible radioactive source.The sources of background radiation are:Internal source — the radioactive substances such as potassium (K-40), carbon (C-14) and radium are present inside our body.External source — cosmic rays, naturally occurring radioactive elements such as radon-222 and solar radiation.It is not possible for us to keep ourselves away from the background radiations.

Question 20

Question 8Complete the following nuclear changes —(a) xaP⟶Q+−10β_{x}^{a}	ext{P} \longrightarrow Q + _{-1}^{0}	ext{β}xa​P⟶Q+−10​β(b) 92238U⟶ 90234Th+......+energy_{92}^{238}	ext{U} \longrightarrow \space _{90}^{234}	ext{Th} + ...... + 	ext{energy}92238​U⟶ 90234​Th+......+energy(c) 92238P⟶αQ⟶βR⟶βS_{92}^{238}	ext{P} \overset{\alpha} \longrightarrow 	ext{Q} \overset{\beta} \longrightarrow 	ext{R} \overset{\beta} \longrightarrow 	ext{S}92238​P⟶α​Q⟶β​R⟶β​S(d) ZAX⟶αX1⟶γX2⟶2βX3_{Z}^{A}	ext{X} \overset{\alpha} \longrightarrow 	ext{X}_{1} \overset{\gamma} \longrightarrow 	ext{X}_{2} \overset{2\beta} \longrightarrow 	ext{X}_{3}ZA​X⟶α​X1​⟶γ​X2​⟶2β​X3​(e) X⟶βX1⟶αX2⟶α 69172X3	ext{X} \overset{\beta} \longrightarrow 	ext{X}_{1} \overset{\alpha} \longrightarrow 	ext{X}_{2} \overset{\alpha} \longrightarrow \space _{69}^{172}	ext{X}_{3}X⟶β​X1​⟶α​X2​⟶α​ 69172​X3​

Accepted Answer

(a) Due to emission of a β particle, the atomic number increases by 1 and the mass number is unchanged. Hence, the nuclear change is —xaP⟶x+1aQ+−10β_{x}^{a}	ext{P} \longrightarrow _{x + 1}^{a}	ext{Q} + _{-1}^{0}	ext{β}xa​P⟶x+1a​Q+−10​β(b) Due to α emission, atomic number decreases by 2 and mass number decreases by 4. Also, an alpha particle is same as a Helium nucleus with 2 protons and 2 neutrons. Hence, we get the nuclear change as —92238U⟶90234Th+24He+energy_{92}^{238}	ext{U} \longrightarrow _{90}^{234}	ext{Th} + _{2}^{4}	ext{He} + 	ext{energy}92238​U⟶90234​Th+24​He+energy(c) When there is an α emission from 92238P_{92}^{238}	ext{P}92238​P , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get 90234Q_{90}^{234}	ext{Q}90234​Q .Now, when 90234Q_{90}^{234}	ext{Q}90234​Q undergoes β emission, the atomic number increases by 1 and the mass number is unchanged. Hence, we get 91234R_{91}^{234}	ext{R}91234​R.In the final step when 91234R_{91}^{234}	ext{R}91234​R undergoes β emission again the atomic number increases by 1 and the mass number is unchanged. Hence we get 92234S_{92}^{234}	ext{S}92234​S.So the complete nuclear change is as follows —92238P⟶α 90234Q⟶β 91234R⟶β 92234S_{92}^{238}	ext{P} \overset{\alpha} \longrightarrow \space _{90}^{234}	ext{Q} \overset{\beta} \longrightarrow \space _{91}^{234}	ext{R} \overset{\beta} \longrightarrow \space _{92}^{234}	ext{S}92238​P⟶α​ 90234​Q⟶β​ 91234​R⟶β​ 92234​S(d) When there is an α emission from ZAX_{Z}^{A}	ext{X}ZA​X , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get Z−2A−4X1_{Z - 2}^{A - 4}	ext{X}_{1}Z−2A−4​X1​After the γ emission, there is no change in atomic number and mass number. Hence, we get Z−2A−4X2_{Z - 2}^{A - 4}	ext{X}_{2}Z−2A−4​X2​In the final step when there are two β emissions, the atomic number increases by 2 and the mass number is unchanged and we get ZA−4X3_{Z}^{A - 4}	ext{X}_{3}ZA−4​X3​Hence, the nuclear change is as follows —ZAX⟶αZ−2A−4X1⟶γZ−2A−4X2⟶2β     ZA−4X3_{Z}^{A}	ext{X} \overset{\alpha} \longrightarrow _{Z - 2}^{A - 4}	ext{X}_{1} \overset{\gamma} \longrightarrow _{Z - 2}^{A - 4}	ext{X}_{2} \overset{2\beta} \longrightarrow _{\space \space \space \space \space Z}^{A - 4}	ext{X}_{3}ZA​X⟶α​Z−2A−4​X1​⟶γ​Z−2A−4​X2​⟶2β​     ZA−4​X3​(e) The daughter nucleus   69172X3_{\space \space 69}^{172}	ext{X}_{3}  69172​X3​ is formed after α decay of X2	ext{X}_{2}X2​, so atomic number of X2	ext{X}_{2}X2​ will be 2 units more and it's mass number will be 4 units more i.e.,   71176X2_{\space \space 71}^{176}	ext{X}_{2}  71176​X2​The nucleus   71176X2_{\space \space 71}^{176}	ext{X}_{2}  71176​X2​ is formed after α decay of X1	ext{X}_{1}X1​ so atomic number of X1	ext{X}_{1}X1​ will be 2 units more and it's mass number will be 4 units more i.e.,   73180X2_{\space \space 73}^{180}	ext{X}_{2}  73180​X2​The nucleus   73180X2_{\space \space 73}^{180}	ext{X}_{2}  73180​X2​ is formed after β decay of X, the atomic number will be one unit less and the mass number is unchanged i.e.,   72180X_{\space \space 72}^{180}	ext{X}  72180​XHence, the nuclear change is as follows —  72180X⟶β   73180X1⟶α   71176X2⟶α   69172X3_{\space \space 72}^{180}	ext{X} \overset{\beta} \longrightarrow \space _{\space \space 73}^{180}	ext{X}_{1} \overset{\alpha} \longrightarrow \space _{\space \space 71}^{176}	ext{X}_{2} \overset{\alpha} \longrightarrow \space _{\space \space 69}^{172}	ext{X}_{3}  72180​X⟶β​   73180​X1​⟶α​   71176​X2​⟶α​   69172​X3​

Radioactivity

Exercise 12A Long Questions

Exercise 12A Short Questions

Exercise 12A Very Short Questions

Exercise 12B Short Questions

Exercise 12B Very Short Questions