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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 951. |
Nitrogen atom has atomic number ? And oxygen has atomic number `6` calculate the total number of electrons in nitrate ion |
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Answer» Formula of nitrate ion `= NO_(3)^(2-)` Number of electron in `NO_(3)^(2) =` Electrons in` N = 3 xx ` electron in O `+ 1` (due to negative charge) `= 7 + 3+ 8 + 1 = 32` |
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| 952. |
Nitrogen atom has an atomic number of 7 and oxygen has an atomic number 8.The total number of electrons in a nitrate ion will beA. 8B. 16C. 32D. 64 |
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Answer» Correct Answer - C Numbers of electrons in nitrogen =7 and number of electron is oxygen =8 we know that formula of nitrate ion is `NO_3^-` we also know that number of electron =(1 x Number of electrons in nitrogen ) + ( 3 x number of electrons in oxygen ) +1 =(1 x 7) + (3 x 8) + 1 =32 |
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| 953. |
Which of the following is not isoelectronic ?A. `Na^+`B. `Mg^(2+)`C. `O^(2-)`D. `Cl^-` |
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Answer» Correct Answer - D `Na^+=1s^2 2s^2 2p^6=10e^-` `Mg^(++)=1s^2 2s^2 2p^6 =10e^-` `O^(2-)=1s^2 2s^2 2p^6 =10e^-` `Cl^(-)=1s^2 2s^2 2p^6 3s^2 3p^6 =18 e^-` |
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| 954. |
Which one of the following has unit positive charge and 1 amu massA. ElectronB. NeutronC. protonD. None of these |
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Answer» Correct Answer - C The proton has unit positive charge (`+1.602xx10^(-19)` coulombs ) and its mass is 1.007 amu `(1.677 xx 10^(-27) kg)` |
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| 955. |
Which of the following statements is incorrect for anode rays?A. They are deflected by electric and magnetic fields.B. Their e/m ratio depends on the gas in the discharge tube used to produce the anode rays.C. The e/m ratio of anode rays is constant.D. They are produced by the ionization of the gas in the discharge tube. |
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Answer» Correct Answer - C |
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| 956. |
An Electro magnetic radiation of wavelength 484 nm is just sufficient of ionise a sodium atom. Calculate the ionisation energy of sodium in kJ/mo, approximately?A. `494.5`B. `246.9`C. `989.0`D. `794.5` |
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Answer» Correct Answer - B `E = (hc)/(lambda) = E xx N_(A)` |
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| 957. |
An Electro magnetic radiation of wavelength 484 nm is just sufficient of ionise a sodium atom. Calculate the ionisation energy of sodium in kJ/mo, approximately?A. `246.9`B. `594.5`C. `694.5`D. `794.5` |
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Answer» Correct Answer - A `E = (hc)/(lambda)` Energy of one mole `=E xx N_(A)` |
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| 958. |
The nucleus of the atom consists of:A. protons and neutronsB. protons and electronsC. neutrons and electronsD. protons, neutrons and electrons. |
| Answer» Correct Answer - A | |
| 959. |
A neutral atom of an element has `2K,KL ,9M,`and `2N` electon .Find and the followinga. Atomic numberb. Total number of s electron c Total number of p electrond.Total number of d electron e.Velocity of the element f.Number of unpaired electrons |
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Answer» Electronic configuration of the neutral atom : `(1s^(2))/(K) ,(2s^(2)2p^(6))/(L),(3s^(2)3p^(6)3d^(1))/(M) ,(4s^(2))/(N)` a. Atomic number = Total number of electrons in neutrual atom `= 21` b.Total number of s electrons `= 8` c.Total number of p electrons `= 12` d. Total number of d electron `= 1` e. VAlency of element `= + 2 and + 3` (due to number of electrons in outer shell and penadimate d sub-shell) f. Number of unpaired electrons `= 1 (of 3d)` |
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| 960. |
In neutral atom, which particles are equivalentA. `p^+, e^+`B. `e^- , e^+`C. `e^- , p^+`D. `p^+ , n^0` |
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Answer» Correct Answer - C Electrons and Protons are same in neutral atom. |
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| 961. |
A neutral atom of an element has ` 2 k , 8L, 8L, 9M` and ` 2N` electrons . Find out the following : (a) Atomic no. (b ) total no. of s-electrons , (c ) total no. of s-electrons , (d) total no. of s-electrons , (e ) Valency of element . (f) No, of unpaired elecrons) . |
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Answer» Deltaronic configuraion fo neutral atom : ` 9 1s^2 /K , ( 2s^2 2P^6)/(L) , ( 3s^2 3P^6 3d^1)/M , ( 4s^1 )/N` (a) At. No . = Total no. fo electron in nertral atom ` =- 21` ( b) Total no .of s-electrons `= 8` ( c) Total no. of d-electrons =`21` (d) Total no. of d-electron `= 1` ( e) Valency fo element ` = + 2` and ` +3` ( due to of electron in outer shell and penultimate d-subshell ) (f) No. of upaired electrons ` = 1 ( "of " 3 d)` . |
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| 962. |
One of the fundamental particles is missing in one of the isotopes of hydrogen atom. The particle and isotope are respectively.A. Neutron, protiumB. Neutron, tritiumC. Proton, protiumD. Electron, tritium |
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Answer» Correct Answer - A `H_(1)^(1)` (Pr otium) number of neutrons `= 1-1 = 0` |
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| 963. |
The fundamental particles present in equal numbers in neutral atoms (atomic number 71) areA. protons and electronsB. neutrons and electronsC. Protons and neutronsD. protons and positrons |
| Answer» Correct Answer - A | |
| 964. |
Identify the electronic configuration of manganese (Z=25)A. 2,8,13,2B. 2,8,8,7C. 2,8,15D. 2,8,10,5 |
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Answer» Correct Answer - A Electronic configuration of maganese is 2,8,13,2 . |
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| 965. |
The mass of which of the following fundamental particles is negligible ?A. ElectronsB. ProtonsC. NeutronsD. Both (1) and (3) |
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Answer» Correct Answer - A Mass of electrons is negligible |
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| 966. |
What is the ratio of distance between successive orbits of 1 and 2 to 2 and 3 of hydrogen atom ? |
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Answer» (i) effects of high voltage and low pressure on the gas molecles in the discharge tube. (ii) the effect of velocity of electrons on ionisation (iii) relation between the voltage and the velocity of electrons. (iv) relation between the pressure and the number of gas molecules. (v) the effect of the number of gas molecules on the impact of colision. |
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| 967. |
To draw the geometrical representation for the structure of the oxygen atom the following steps are given . Identify the correct sequence of steps. (a) The eight electrons present in the extra-nuclear part would be distributed in the first two orbits that is K and L. As per the rules , two electrons would occupy K orbit and the remaining six electrons occupy the L orbit. (b) The atomic number of oxygen is 8. ltbr. (c) In the nucleus , 8 protons and 8 neutrons and present and in the extra-nuclear part, that is in the orbits, 8 electrons are present. (d) Oxygen atom has 8 electrons and 8 protrons . The mass number is 16 , hence number of neutrons is equal to `8[""_(8)O^(16)]`A. b a c dB. c a d bC. b d c aD. c d b a |
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Answer» Correct Answer - C (i) The atomic number of oxygen is 8. (ii) Oxygen atom has 8 electrons and 8 protons . The mass number is 16, hence hte number of neutrons is equal to `8[""_(8)O^(16)]`. (iii) In the nucleus, 8 protons and 8 neutrons are present and in the extra-nuclear part that is in the orbits , 8 electrons are present .(iv) The eight electrons present in extra nuclear part, would be distributed in first two orbits that is K and L . As per the rules ,two electrons would occupy K orbits and remaining six electrons in L-orbit. |
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| 968. |
Is the velocity of an electron in all orbits the same for an atom of a particular element ? How does it very for different single electron specites ? Give reasons in support of your answer. |
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Answer» (i) forces acting on moveing electron. (ii) position of the electron in the orbit. (iii) relationship between position, forces and velocity. (iv) comparison of nuclear charge in different single electron species. |
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| 969. |
The outermost electron configuration of Cr is ………. |
| Answer» Correct Answer - `4s^(1)` | |
| 970. |
The wavelength of a charged particle `…………..` the square root of the potential difference through which it is accelerated .A. is inversely proportional toB. is directly proportional toC. is independent ofD. is unrelated with |
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Answer» Correct Answer - A |
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| 971. |
Identify the atomic number corresponding to least number of valence electronsA. 19B. 15C. 35D. 34 |
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Answer» Correct Answer - A The electronic confriguration of element with atomic number 19 is 2, 8 , 8 , 1, number of valence electrons are 1 similarly the electronic configuration of elements with atomic number 15 , 35 , and 34 , are 2, 8 , 5 , 2 , 8 , 18 ,7 , 2 , 8 ,18 , 6 respectively and the valence electrons are 5 , 7 and 6 respectively .So the atomic number corresponding to least number of valence electrons is 19. |
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| 972. |
Assertion (A): Electrons present in ground states of different single electron species `(H,He^(+), Li^(++))` possess different amount of energy. Reason (R) : Distances of electrons from the nuclei of different single electron are equal.A. Both A and R are correct and R is the correct explanation of A.B. Both A and R are correct and R is not the correct explanation of A.C. A is correct and R is wrongD. A is wrong and R is correct |
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Answer» Correct Answer - C The electrons present in different single electon species posses different amounts of energy due to the difference in the atomic numbers. The distance of the electrons from the nuclei of different single electron species is different because of difference in atomic number. |
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| 973. |
The atoms of the same element may differ in the number of ________.A. electrons onlyB. protons onlyC. neutrons onlyD. both electrons and protons. |
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Answer» Correct Answer - C The atoms of the same element which have the same atomic number but different mass numbers re called isotopes. Hence these atoms posses the same number of electrons and protons but differ in the number of neutons . |
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| 974. |
Which amoung the following set of elements contain same number of valence electrons ?A. `""_(11)^(23)X,""_(20)^(40)Y`B. `""_(10)^(20)X,""_(19)^(39)Y`C. `""_(11)^(23)X,""_(19)^(39)Y`D. `""_(18)^(40)X,""_(26)^(40)Y` |
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Answer» Correct Answer - C `""_(11)^(23)X,""_(19)^(39)Y` have one valence electron as they are sodium and potassium with electronic arrangements 2,8,1 and 2,8,8,1 respectively. |
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| 975. |
using s, p, d notations describe the orbital with the following quantum numbers . (a) n = 2 l=1, (b) n = 4 , l = 0, (c) n= 5 , l=3 , (d) n = 3 , l = 2 |
| Answer» `{:(,n,l,"orbital"),((a),2,1,2p),((b),4,0,4s),((c),5,3,5f),((d),3,2,3d):}` | |
| 976. |
The correct set of quantum number for the unpaired electron of chlirine atom is A. `n = 2,l = 1,m = 0`B. `n = 2,l = 1,m = 0`C. `n = 3,l = 1,m = 1`D. `n = 3,l = 0,m = 0` |
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Answer» Correct Answer - C `Cr(Z = 18)` `[Ne]3s^(2)3p^(5)` For the last electron `n = 3,l = 1,m = -1 or + 1` |
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| 977. |
Calculate the binding energy of the oxygen isotope `._(8)^(16)O`. The mass of the isotope is 16.0 amu. (Given e=0.0005486 amu, p=1.00757 amu and n=1.00893 amu). |
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Answer» The isotope `._(8)^(16)O` contains 8 protons, 8 neutrons and 8 electrons. Actual mass of the nucleus of `._(8)^(16)O` `=16-`mass of 8 electrons `=16-8xx0.0005486=15.9956`amu Mass of the nucleus of `._(8)^(16)O` =Mass of 8 protons + Mass of 8 neutrons `=8xx1.00757+8xx1.00893=16.132`amu Mass defect `=(16.132-15.9956)=0.1364` amu Binding energy `=0.1364xx931=127`MeV |
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| 978. |
The binding energy of `._(2)^(4)He` is 28.57 MeV. What shall be the binding energy per nucleon of this element? |
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Answer» The nucleus of `._(2)^(4)He` consists of 4 nucleons. So, Binding energy per nucleon `=("Total binding energy")/("No. of nucleons")` `=(28.57)/(4)=7.14MeV` |
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| 979. |
A certain dye absorbs `4530A^(@)` and fluoresence at `5080A^(@)` these being wavelength of maximum absorption that under given condition `47%` of the absorbed energy is emitted. Calculate the ratio of the no of quanta emitted to the number absorbed.A. `0.527`B. `1.5`C. `52.7`D. 3 |
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Answer» Correct Answer - A `E = (Nhc)/(lambda), (E_(1))/(E_(2)) = (lambda_(2))/(lambda_(1))` |
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| 980. |
The outermost electric configuration of the most electron of chlorine atom isA. `ns^(2)np^(3)`B. `ns^(2)np^(4)`C. `ns^(2)np^(5)`D. `ns^(2)np^(6)` |
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Answer» Correct Answer - C `ns^(2),np^(2)` is the electronic configuration of a halogen and halogens are most electronegative |
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| 981. |
Which of the following lines will have a wave no equal in magnitude to the value of R in the H-Spectral seriesA. Limiting line of Balmer seriesB. Limiting line of Lyman seriesC. First line of Lyman seriesD. First line of Balmer series |
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Answer» Correct Answer - B `bar(v) = R [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` |
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| 982. |
Calculate the total spin and magnitic moment for atoms having atomic numbers `7`, `24`, `34` and `36`. |
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Answer» The electronic configruation are : ` ._7 N : 1s^2 , 2 s^2 2 p^3` unpaired electron `=3` ` ._(24) Ce : 1 s ^2 , 2s^ 2 2p^6 , 3 p^7 3d^5 . 4s^1` unpaired electron `= 6` ` ._( 34) Se : 1 s^2 , 2a^(2)2p^6 , 2p^2 3p^(6)3d^5 , 4s^1` unpaired electron =6 ` ._(24) : 1a^2 , 2a^(2) 2p^6 , 3s^2 3p^6 3d^(10) , 4s^2 4p^4` unpaired electron `=2` ` ._(36) Kr : 1s^2 , 2s^2 2p^6 , 2s^2 sp^6 3d^(10) , 4s^2 4p^6` unpaired electron `= ` ` :.` Total spin for an atom ` = += 1//2 ` no. of unpairred electron For`._7N`, it is `= +- 3 //2 ` For` ._(24)Cr`, it is `= +- 3` For` ._(34)Se`, it is `= +- 1` For `._(36)` Kr, it is `=0` Also magnetic moment ` = sqrt ([n (n+2)])` Bohr magnetion For `._7N`, it is `= sqrt (( 15))` For` ._(24)` Cr, it is` = sqrt (( 48))` For `._(34 )` Se it is `= sqrt 8` For `._(36)` Kr , it is `= sqrt ((0))`. |
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| 983. |
The radius of `Al_(23)^(27)` nucleus will be :A. `1.2 xx 10^(-15)m`B. `27 xx 10^(-15)m`C. `10.8 xx 10^(-15)m`D. `3.6 xx 10^(-15)m` |
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Answer» Correct Answer - D |
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| 984. |
Velocity of de Broglie wave is given by :A. `(c^(2))/(v)`B. `(hv)/(mc)`C. `(mc^(2))/(h)`D. `vlamda` |
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Answer» Correct Answer - B `lamda=(h)/(mv)=(h)/(p)` `p=(h)/(lamda)` `mv=(hv)/(c)` `v=(hv)/(mc)`. |
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| 985. |
`A` photon of wavelenth `3000` A strikes `a` metal surface, the work function of the metal being `2.20eV`. Calculate `(i)` the energy of the photon in `eV(ii)` the kinetic energy of the emitted photo electron and `(iii)` the velocity of the photo electron. |
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Answer» (i) Energy of the photon `E=hv=(hc)/(lambda)=((6.6xx10^(-34)Js)(3xx10^(8)ms^(-1)))/(3xx10^(-7)m)=6.6xx10^(-19)J` `1eV = 1.6xx10^(-19)J` Therefore `E =(6.6xx10^(-19)J)/(1.6xx10^(-19)J//eV)=4.125eV` (ii) Kinetic energy of the emitted photo electon `"Work function" " "=2.20 eV` Therefore, `KE " "=2.475-2.20` `" "=1.925eV=3.08xx10^(-19)J` `(iii)` Velocity of the photo electron `KE=(1)/(2) mv^(2)=3.08xx10^(-19)J` `"Therefore"`, velocity `(v) = sqrt((2xx3.08xx10^(-19))/(9.1xx10^(-31)))=8.22xx10^(5)ms^(-1)` |
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| 986. |
The de Broglie wavelenth of `1 mg` grain of sand blown by a `20 m s^-1` wind is :A. `3.3xx10^(-29)m`B. `3.3xx10^(-21)m`C. `3.3xx10^(-49)m`D. `3.3xx10^(-42)m` |
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Answer» Correct Answer - A `lamda=(h)/(mv)=(6.626xx10^(-34))/(1xx10^(-6)xx20)=3.313xx10^(-29)m` |
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| 987. |
The mass of photon having wavelength `1 nm` is :A. `2.21xx10^(-35)kg`B. `2.21xx10^(-33)g`C. `2.21xx10^(-33)kg`D. `2.21xx10^(-26)kg` |
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Answer» Correct Answer - C `lamda=(h)/(mc)` `m=(h)/(lamdac)=(6.626xx10^(-34))/(1xx10^(-9)xx3xx10^(8))=2.21xx10^(-33)kg`. |
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| 988. |
The mass of photon having wavelength `1 nm` is :A. `2.21 xx 10^-35 kg`B. `2.21 xx 10^-33 g`C. `2.21 xx 10^-33 kg`D. `2.21 xx 10^-26 kg` |
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Answer» Correct Answer - C ( c) `m = (h)/(lamda c)` `m = (h)/(lamdac) = (6.626 xx 10^-34)/(1 xx 10^-9 xx 3 xx 10^8)` =`2.21 xx 10^-33kg`. |
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| 989. |
The energy absorbed by each molecule `(A_2)` of a substance is `4.4 xx 10^-19 J` and bond energy per molecule is `4.0 xx 10^-19 J`. The kinetic energy of the molecule per atom will be.A. `2.0xx10^(-20)` JB. `2.2xx10^(-19)` JC. `2.0xx10^(-19)` JD. `4.0xx10^(-20)` J |
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Answer» Correct Answer - A K.E. per atom =`((4.4xx10^(-19))-(4.0xx10^(-19)))/2` `=(0.4xx10^(-19))/2=2.0xx10^(-20)` |
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| 990. |
The energy absorbed by each molecule `(A_2)` of a substance is `4.4 xx 10^-19 J` and bond energy per molecule is `4.0 xx 10^-19 J`. The kinetic energy of the molecule per atom will be.A. `2.0 xx 10^-20 J`B. `2.2 xx 10^-19 J`C. `2.0 xx 10^-19 J`D. `4.0 xx 10^-20 J` |
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Answer» Correct Answer - A (a) `KE` of molecule = energy absorbed by molecule `- BE` per molecule =`(4.4 xx 10^-19) - (4.0 xx 10^-19) J = 0.4 xx 10^-19 J` `KE "per atom" = (0.4 xx 10^-19)/(2) J = 2.0 xx 10^-20 J`. |
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| 991. |
The energy absorbed by each molecule `(A_2)` of a substance is `4.4 xx 10^-19 J` and bond energy per molecule is `4.0 xx 10^-19 J`. The kinetic energy of the molecule per atom will be.A. `2.0xx10^(-20) J`B. `2.2xx10^(-19)J`C. `2.0xx10^(-19)J`D. `4.0xx10^(-20)J` |
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Answer» Correct Answer - A (a) Kinetic energy (KE) of molecule = energy absorbed by molecule-bond energy per molecule `(4.4xx10^(-19))-(4.0xx10^(-19))J=0.4xx10^(-19) J` KE per atom `=(0.4xx10^(-19))/(2) J = 2.0 xx10^(-10)J` |
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| 992. |
If `n = 6`, the correct sequence for filling of electrons will be.A. `ns rarr (n-2) f rarr np rarr (n-1)d`B. `ns rarr np rarr (n-1) d rarr (n-2)f`C. `ns rarr (n-2)f rarr (n-1)d rarr np`D. `ns rarr (n-1)d rarr (n-2)f rarr np` |
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Answer» Correct Answer - C `ns rarr (n-2) f rarr (n-1) d rarr np` (for n = 6). |
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| 993. |
Which of the following electron transitions in a hydrogen atom will require the largest amount of energy?A. From n=1 to n=2B. From n=2 to n=3C. From n`=infty` to n=1D. From n=4 to n=5 |
| Answer» Correct Answer - A | |
| 994. |
Correct statement isA. `K=4s^1 , Cr=3d^4 4s^2 , Cu=3d^10 4s^2`B. `K=4s^2 , Cr=3d^4 4s^2 , Cu=3d^10 4s^2`C. `K=4s^2 , Cr=3d^5 4s^1 , Cu=3d^10 4s^2`D. `K=4s^1 , Cr=3d^5 4s^1 , Cu=3d^10 4s^1` |
| Answer» Correct Answer - D | |
| 995. |
The energy liberated when an excited electron returns to its ground state can have:A. any value from zero to infinityB. only negative valuesC. only specified positive valuesD. None of the above |
| Answer» Correct Answer - C | |
| 996. |
Which of the following has maximum energy ?A. B. C. D. |
| Answer» Correct Answer - B | |
| 997. |
If `n = 6`, the correct sequence for filling of electrons will be.A. `ns rarr np rarr (n-1) d rarr (n-2) f`B. `ns rarr (n-2) f rarr (n-1) d rarr np`C. `ns rarr (n-1)d rarr (n-2) f rarr np`D. `ns rarr (n-2) f rarr np rarr (n-1) d` |
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Answer» Correct Answer - A (a) `np` is filled after `ns` in each shell. |
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| 998. |
Which of the following is false?A. The angule momentum of an electron due to its spinni9ng is given as `sqrt(s(s + 1))((h)/(2pi))`, where s can take a value of `1//2`B. The angule momentum of an electron due to its spinni9ng is given as `m_(s)((h)/(2pi))`, where `m_(s)` can take a value of `+1//2`C. The azinuthal quantum number cannot have negative valuesD. The potential energy of an electron in an orbit is twice in magnitude as campaired to its kinetic energy |
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Answer» Correct Answer - B b The expression `m_(1) ((h)/(2pi))` is that of z- component of angular momentum c. The azimuthal quantum number has the value `0 , 1,2, …..(n - 1)` d. The expression are `KE = (1)/(2) mv^(2) =- (1)/(2)(Ze^(2))/((4 pi epsilon_(0))r)` `PE = - (Ze^(2))/((4 pi epsilon_(0))r)` |
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| 999. |
Which of the following is true ?A. Neutron is a positively charged electronB. The magnetic moment of an atom is related to the number of unpaired electron in its electronic configurationC. Bohr theory can be succesifully modified to explain the electronic spectrum of multielectron atomD. The angle momentum of an eklectron in an atom is gives by `n ((h)/(2pi))` |
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Answer» Correct Answer - B a. The neutrino has charge and seems to have rest mass equal to zero it is emitted along with the emission of a positron (possitive charge of` + le` and mass equal to electron) for example `._(10)^(19) Ne rarr _(9)^(19)F + _(1)^(0)e + V` Antineutrico is emitted along with the emmission of `beta ` particle b. The expression of magnetic moment is `mu_(m) = sqrt(n(n + 2)) mu_(B)` d. The correct expressions is `L = sqrt(l(l + 2)) ((h)/(2pi))` |
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| 1000. |
Which of the following is true ?A. Diapositive zine exbibits paramagenetism due to loss of two electron dfrom a `3d` orbitals of nutrqal atomB. In `beta` emmision from a nucless , the atomic number of the diagram element decreases by `1`C. The emission of one `alpha` particle from a radioactive atom result in the decreases of atomic n umber by `2` and mass number by `4`D. The successive atom result in the decrease of atomic number by `11` |
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Answer» Correct Answer - C a. The electronic configuration of `Zn^(2+)` (atomic number `30)` is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)` There are no unoaired eklectrons , hence it is diamagnetic b. In `beta` emmission , the atomic number of gaughter elements is increased by `1` due to teh basic conversion of neutron into proton in the nucleius c. An `alpha` - particle is `_(2)^(4) He^(2+)` .Hence atomic number and number of daughter element dectrased by `2` and `4` respectgively . d. the will occur an increases in atomic number by `2` `overset(m)underset(n)` A overset(-beta)` to`overset(m)underset(n + 1)B overset(-beta)to overset(m) underset(n + 2)`C` |
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