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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.
| 1. |
Which of the following "is"//"are" correct?a)`alpha`-rays are more penetrating then `beta`-rays.b)`alpha`-rays have greater ionizing power than `beta`-rays.c)`beta`-particles are not present in the elements, yet they are emitted from the nucleus.d)`alpha`-rays are not emitted simultaneously with `alpha`- and `beta`-rays.A. `alpha` rays are more penetrating then `beta`-raysB. `alpha`-rays have greater ionizing power then `beta`-raysC. `beta`particle are not present in the nucleus, yet they are emitted from the nucleus.D. `gamma`-rays are not emitted simultaneously with `alpha` and `beta`-rays |
| Answer» Correct Answer - b,c,d | |
| 2. |
A radioactive substance is being consumed at a constant of `1 s^(-1)`. After what time will the number of radioactive nuclei becoem `100`. Initially, there were 200 nuceli present.A. 1 secB. 2 secC. In (2) secD. `(1)/(In (2)) sec` |
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Answer» Correct Answer - b `N = N_(0) e^(-lambda t)` `100 = 200 e^(-1 xx t)` `(1)/(2) = e^(-t)` t = In (2) sec |
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| 3. |
A radiaocatice isotope is being produced at a constant rate `X`. Half-life of the radioactive substance is `Y`. After some time, the number of radioactive nuceli become constant. The value of this constant is . |
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Answer» At the stage of radioactive equilibrium, Rate of formation of nuclide = Rate of decay of nuclide `x = lambda N` `N = (x)/(lambda) = (x)/(In 2)//y) = (xy)/(In 2)` |
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| 4. |
Moderator is a material which is used to slow down the neutrons produced during nuclear fission. The neutrons from the source are of high speed and energy. Heavy water or graphite moderators slow down the speed of the neutrons. The energy of fast moving neutrons decreases from 2MeV to 0.02535 eV, it corresponds to the velocity of 220 m `"sec"^(-1)`. At this velocity, the neutrons are in thermal equilibrium with the moderator. such neutrons are called thermal neutrons. Thermal neutrons cause further fission reaction. The essential characterstices of moderators are: (i) Its molar mass must be low, (ii) It should not absorb neutrons. (iii) It should undergo elastic collisions with neutrons. A good moderator should:A. not be a gas onlyB. not have appertite for neutrons onlyC. be light in mass number onlyD. be all the above three |
| Answer» Correct Answer - d | |
| 5. |
Cyclotron is used to accelerateA. protonsB. deutronsC. neutronsD. electrons |
| Answer» Correct Answer - c | |
| 6. |
On large scale, tritium is produced by which of the following nuclear reaction?A. `._(3)^(6)Li + ._(0)^(1)n to ._(2)^(4)He + ._(1)^(3)T`B. `._(1)^(2)D + ._(1)^(2)D to ._(1)^(3)T + ._(1)^(1)H`C. `._(7)^(14)N + ._(0)^(1)n to ._(6)^(12)C + ._(1)^(1)T`D. `._(7)^(14)N + ._(1)^(1)H to ._(1)^(3)T` + other fragments |
| Answer» Correct Answer - b | |
| 7. |
The number of `alpha`-and `beta`-particles emitted in the nuclear reaction, `._(90)Th^(228) to ._(83)Bi^(212)`, respectively areA. `4 alpha` and `1 beta`B. `3 alpha` and `7 beta`C. `8 alpha` and `1 beta`D. `4 alpha` and `7 beta` |
| Answer» Correct Answer - a | |
| 8. |
Identify the nuclear reaction that differs from the rest:A. Positron emissionB. K-captureC. `beta`-decayD. `alpha`-decay |
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Answer» Correct Answer - d Only `gamma`-emission does not change the n/p (Neutron/Proton, ratio) of the parent element. |
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| 9. |
The number of `alpha`-particles emitted per second by 1g fo `.^(226)Ra` is `.7 xx 10^(10)`. The decay constant is:A. `1.39 xx 10^(-11) sec^(-1)`B. `13.9 xx 10^(-11) sec^(-1)`C. `139 xx 10^(-10) sec^(-1)`D. `13.9 xx 10^(-10) sec^(-1)` |
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Answer» Correct Answer - a `("No. of atoms disintegration per second")/("Total number of atoms present")` or `(3.7 xx 10^(10))/((6.02 xx 10^(23))/(226)) = (226 xx 3.7 xx 10^(10))/(6.02 xx 10^(23)) = lambda` |
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| 10. |
Two elements `P` and `Q` have half-line of `10` and `15` minutes repectively. Freshly preapared sample of mixuture containing equal number of atoms is allowed to decay for `30` minutes. The ratio of number of atoms of `P` and `Q` in left in mixture is:A. 0.5B. 2C. 1D. 3 |
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Answer» Correct Answer - a In 30 minutes, there will be 3 half lives of P and 2 half lives of Q `:.` Number of P atoms will be 1/8th and number of Q atoms will be 1/4th of original atoms. Then `("Number of atoms of P")/("Number of atoms of Q") = (1//8)/(1//4) = (1)/(2)`, i.e., 0.5 |
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| 11. |
Consider the following nuclear reactions: `""_(92)""^(238)Mto.""_(y)""^(x)N+2.""_(2)""^(4)He, ""_(x)""^(y)Nto ""_(B)""^(A)L+2beta""^(+)` The number of neutrons in the element L isA. 142B. 144C. 140D. 146 |
| Answer» Correct Answer - b | |
| 12. |
The radiation from naturally occuring radioactive substance as seen after deflection by a magnetic field in one direction are :A. definitely `alpha`-ryasB. definitely `beta`-raysC. both `alpha` and `beta`raysD. either `alpha` or `beta`-rays |
| Answer» Correct Answer - d | |
| 13. |
Nucleus of an atom resembles with a drop of liquid. Density of nucleus is very high, i.e., `10^(8)` tonne/cc or 130 trillion tonnes `m^(-3)`. This density is about a trillion times greater than that of water. Density of nuclei of all elements are same, it is independent of atomic number or atomic mass. However, the radius of nucleus depends on the mass number . Surface However, the radius of nucleus depends on the mass number. Surface tension of nucleus is also very high. i.e., about `1.24 xx 10^(18)` times, the suface tension of water. The radius of `._(6)^(12)C` nucleus is:A. `5 xx 10^(-15) m`B. `1.4 xx 10^(-15) m`C. `3.5 xx 10^(-15) m`D. `6 xx 10^(-15) m` |
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Answer» Correct Answer - c `r = r_(0) A^(1//2)` where A = mass number `r_(0) = 1.4 xx 10^(-15) m` |
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| 14. |
The time of decay for the nuclear reaction is given by `t = 5t_(1//2)`. The relation between mean life `tau` and time of decay t is given by:A. `2 tau` In 2B. `5 tau` In 2C. `2 tau^(4)` In 2D. `(1)/(tau^(4))` In 2 |
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Answer» Correct Answer - b `t = 5t_(1//2)` `t = 5 xx (In 2)/(lambda)` `t = 5 tau` In 2 |
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| 15. |
Nucleus of an atom resembles with a drop of liquid. Density of nucleus is very high, i.e., `10^(8)` tonne/cc or 130 trillion tonnes `m^(-3)`. This density is about a trillion times greater than that of water. Density of nuclei of all elements are same, it is independent of atomic number or atomic mass. However, the radius of nucleus depends on the mass number . Surface However, the radius of nucleus depends on the mass number. Surface tension of nucleus is also very high. i.e., about `1.24 xx 10^(18)` times, the suface tension of water. Radius of nucleus is directly proportional to:A. `A^(2)`B. `A^(1//3)`C. `[A]^(3)`D. A |
| Answer» Correct Answer - c | |
| 16. |
Which among the following relation is correctA. `t_(1//2) = 2t_(3//3)`B. `t_(1//2) = 3t_(3//4)`C. `t_(3//4) = 2t_(1//2)`D. `t_(3//4) = 3t_(1//2)` |
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Answer» Correct Answer - c `(0.639)/(t_(1//2)) = (2.303)/(t) log ((N_(0))/(N))` `(0.693)/(t_(1//2)) = (2.303)/(t_(3//4)) log (100)/(75)` `t_(3//4) = 2t_(1//2)` |
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| 17. |
Half life of a radioactive element is 10 days. What percentage of the element will remain undecayed after 100 days?A. 0.1B. 0.001C. 0D. 0.99 |
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Answer» Correct Answer - b In ten times of half life 99.9%, the element undergoes decay, then percentage of undecayed radioactive element will be 0.1% |
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| 18. |
80% of the radioactive nuclei present in a sample is found to remain undecayed after one day. The percentage of undecayed nuclei left after two days will beA. 64B. 20C. 46D. 80 |
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Answer» Correct Answer - a `lambda = (2.303)/(t) log ((N_(0))/(N))` `= (2.303)/(1) log ((100)/(80))` `lambda = (2.303)/(2) log ((100)/(N))` `(2.303)/(1) log ((100)/(80)) = (2.303)/(2) log ((100)/(N))` `((5)/(4))^(2) = (100)/(N)` `N = 64` |
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| 19. |
Half life of a radioactive sample is `2x` years. What fraction of this sample will remain undecayed after `x` years?A. `(1)/(2)`B. `(1)/(sqrt(2))`C. `(1)/(sqrt(3))`D. 2 |
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Answer» Correct Answer - b `lambda = (2.303)/(t) log ((N_(0))/(N))` `(0.693)/(t_(1//2)) = (2.303)/(t) log_(10) ((N_(0))/(N))` `(0.639)/(2x) = (2.303)/(x) log_(10) ((N_(0))/(N))` `(1)/(2) log_(10) 2 = log ((N_(0))/(N))` `(N)/(N_(0)) = (1)/(sqrt(2))` Fraction undecayed `= (1)/(sqrt(2))` |
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| 20. |
Initial amount of the radioactive element with half life 10 days is 16 g. What amount in gm of this element will remain after 40 days? |
| Answer» Correct Answer - 1 | |
| 21. |
`.^(131)I` has half life period 13.3 hour. After 79.8 hour, what fraction of `.^(131)I` will remain ? |
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Answer» `N = N_(0) ((1)/(2))^(n)` `(N)/(N_(0)) = ((1) /(2))^(6) = (1)/(4)` |
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| 22. |
When nucleus of an electrically neutral atom undergoes a radioactive decay process, it will remain neutral after the decay if the process is (a) An `alpha`- decay (b) `A beta^(o+)`-decay (c ) `A gamma`-decay (d) `A K`- capture processA. an `alpha`-decayB. a `beta`-decayC. a`gamma`-decayD. a K-capture process |
| Answer» Correct Answer - c,d | |
| 23. |
One mole of a present in a closed vessel undergoes decay as `""_(z)""^(m)Ato""_(Z-4)""^(m-8)B+2""_(2)""^(4)He`. The volume of He collected at NTP a fter 20 days is (`t""_(1//2)=10`days)a)11.2 litreb)22.4 litrec)33.6 litred)67.2 litre |
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Answer» We know that, `N = N_(0) ((1)/(2))^(n)` N = remaining mole of A `N = 1 ((1)/(2))^(2) = (1)/(4)` Number of decayed moles `= 1 - (1)/(4) = (3)/(4)` Number of moles of helium formed `= 2 xx` Number of decayed moles of `A = 2 xx (3)/(4) = (3)/(2)` Volume of helium at STP `= (3)/(2) xx 22.4 = 33.6` litre |
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| 24. |
If the amount of radioactive substance is increased three times, the number of atoms disintegrated per ubit time would :A. be doubleB. not be changeC. be tripleD. be `(1)/(3)` rd of the original number of atoms |
| Answer» Correct Answer - c | |
| 25. |
The half life of polonium is 140 days. In what time will 15 g of polonium be disintegrated out of its initial mass of 16 g? |
| Answer» Polonium left is `(1)/(16)th` of the initial i.e., 4 half lives. | |
| 26. |
Two radioactive elements X and Y have half lives 6 min and 15 min respectively. An experiment starts with 8 times as many atoms of X as Y. how long it takes for the number of atoms of X left to equal the number of atoms of Y left?A. 6 minB. 12 minC. 48 minD. 30 min |
| Answer» Correct Answer - d | |
| 27. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. 3 hrsB. 9 hrsC. 24 hrsD. 12 hrs |
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Answer» Correct Answer - d `N = N_(0) ((1)/(2))^(n) , (N)/(N_(0)) = ((1)/(2))^(n)`, `(1)/(64) = ((1)/(2))^(n)`, n = 6 half lives ` :.` time `= 2 xx 6 = 12` hrs |
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| 28. |
Half - lives of two radioactive . Initially . The samples have equal number of nuclie After `80` minutes ,the ratio of decyed number of `A and B` nuclei will beA. `1 : 16`B. `4 : 1`C. `1 : 4`D. `5 : 4` |
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Answer» Correct Answer - d Amount remaining after n half lives `(N) = N_(0) ((1)/(2))^(n)` `:.` Ratio of decayed nuclei of A to that of B will be `(N_(0) - N_(A))/(N_(A) - N_(B)) = (N_(0)[1-((1)/(2))^(4)])/(N_(0)[1-((1)/(2))^(2)]) = (1-(1)/(16))/(1- (1)/(4))` `= ((15)/(16))/((3)/(4)) = (5)/(4) ` |
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| 29. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. 6 hrsB. 12 hrsC. 24 hrsD. 128 hrs |
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Answer» Correct Answer - b `N = N_(0) ((1)/(2))^(n)` `(N)/(N_(0)) = ((1)/(2))^(n)` `(1)/(64) = ((1)/(2))^(n)` `n = 6` `:. T = 2 xx 6 = 12` hours |
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| 30. |
Half-life of a radioactive substance `A` is two times the half-life of another radioactive substance `B`. Initially, the number of `A` and `B` are `N_(A)` and `N_(B)`, respectively . After three half-lives of `A`, number of nuclei of both are equal. Then, the ratio `N_(A)//N_(B)` is .A. `(1)/(2)`B. `(1)/(8)`C. `(1)/(3)`D. `(1)/(6)` |
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Answer» We know that, the amount remaining after n half lives can be calculated as: `N = N_(0) ((1)/(2))^(n)` Remaining amount of `A = N_(A) ((1)/(2))^(3)` Remaining amount of `B = N_(B) ((1)/(2))^(6)` `N_(A) ((1)/(2))^(3 = N_(B) ((1)/(2))^(6)` `(N_(A))/(N_(B)) = (8)/(64) = (1)/(8)` |
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| 31. |
Calculate the average life of a radioactive substance whose half life period is 1650 years. |
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Answer» Average life `= 1.44 xx t_(1//2)` `1.44 xx 1650 = 2376` years |
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| 32. |
Assertion (A) : The average life of a radioactive element is infinity Reason (R ) : As a radioactive element disinegrates, more of it is formed in nature by itselfa)If both (A) and (R ) are correct, and (R ) is the correct explaination of (A)b)If both (A) and (R ) are correct, but (R ) is not the correct explaination of (A)c)If (A) is correct,but (R ) is incorrectd)If both (A) and (R ) are incorrect.A. If both (A) and (R ) are correct and (R ) is the correct explanation for (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A )C. If both (A) and (R ) are incorrect.D. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - C | |
| 33. |
The energy released in an atom bomb explosion is mainly due toA. Conversion of heavier to lighter atomsB. Products having lesser mass than initial substanceC. release of neutronsD. release of electrons |
| Answer» Correct Answer - b | |
| 34. |
Assertion `(A) :` Hydrogen bomb is more powerful than atomic bomb. Reason `(R) :` In hydrogen bomb, fusion reaction is initiated.a)If both `(A)` and `(R)` are correct , and `(R)` is the correct explanation of `(A)`b)If both `(A)` and `(R)` are correct, but (R) is not the correct explanation of `(A)`c)If `(A)` is correct, but `(R)` is incorrect.d)If `(A)` is incorrect, but `(R)` is correct.A. If both (A) and (R ) are correct and (R ) is the correct explanation for (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A )C. If both (A) and (R ) are incorrect.D. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - B | |
| 35. |
A chemist prepares `1.00 g` of pure `._(6)C^(11)`. This isotopes has half life of 21 min, decaying by the equation: `._(6)C^(11)`.`=>` `._(5)B^(11)`. + `._(-1)e^(0)`. a. What is the rate of disintegration per second (dps) at starts ? b. What is the activity and specific activity of `._(6)C^(11)` at start? c. How much of this isotope `(._(6)C^(11))` is left after 24 hr its preparation? |
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Answer» Applying, `-(dN)/(dt) = lambda N_(0)` `= (0.693)/(21 xx 60) xx (1 xx 6.02 xx 10^(23))/(11)` `= 3 xx 10^(19)` dps (b) Activity `= (3 xx 10^(19))/(3.7 xx 10^(10))` (1 curie `= 3.7 xx 10^(10)` dps) `= 8.108 xx 10^(8)` curie Sp. activity ` = 3 xx 10^(19) xx 10^(3) = 3 xx 10^(22)` dis/kg s = `8.108 xx 10^(11)` curie (c ) Applying, `N = N_(0) ((1)/(2))^(n) [n = (t)/(t_(1//2)) = (24 xx 60)/(21) = 68.57]` `N = 1 xx ((1)/(2))^(68.57) = 2.29 xx 10^(-21) g` |
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| 36. |
The half life period of a first order reaction is 60 min. What percentage will be left after 240 min.A. 0.175B. 0.15C. 0.125D. 0.1 |
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Answer» Correct Answer - c `n = (45)/(15) = 3` = No. of half lives `N = N_(0) ((1)/(2))^(n) = 100 xx ((1)/(2))^(3) = 12.5%` |
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| 37. |
Gamma rays are:A. high energy electronsB. low energy electronsC. high energy electromagnetic wavesD. high energy positrons |
| Answer» Correct Answer - c | |
| 38. |
The correct order of ionising capcity of `alpha, beta` and `gamma`-rays isA. `alpha gt beta gt gamma`B. `beta gt alpha gt gamma`C. `gamma gt alpha gt beta`D. `beta gt gamma gt alpha` |
| Answer» Correct Answer - a | |
| 39. |
A radioactive isotope having a half life of 3 days was received after 12 days. It was found that there were 3 gm of the isotopes in the container. The initial weight of the isotope when packed wasa)12 gmb)24gmc)36 gmd)48 gmA. 48 gB. 36 gC. 24 gD. 12 g |
| Answer» Correct Answer - a | |
| 40. |
A certain radioactive isotope decay has `alpha`-emission, `._(Z_(1))^(A_(1))X to ._(Z_(1)-2)^(A_(1)-4)Y` half life of X is 10 days. If 1 mol of X is taken initially in a sealed container, then what volume of helium will be collected at STP after 20 days?A. 22.4 LB. 11.2 LC. 16.8 LD. 33.6 L |
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Answer» Correct Answer - c After 20 days 0.75 mol helium will be formed. `:.` Volume of helium at STP `= 0.75 xx 22.4 = 16.8 L` |
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| 41. |
Uranium `._(92)U^(238)` decayed to `._(82)Pb^(206)`. They decay process is `._(92)U^(238) underset((x alpha, y beta))(rarr ._(82)Pb^(206))` `t_(1//2)` of `U^(238) = 4.5 xx 10^(9)` years Atomic mass of `U^(238)` is 238.125 amu. Its packing fraction will beA. 5.25B. 0.125C. 12.5D. 1.25 |
| Answer» Correct Answer - a | |
| 42. |
`._(84)^(218)Po` `(t_(1//2) = 183 sec)` decay to `._(82)^(214)Pb` (`t_(1//2) = 161`) sec by `alpha` emission, while `._(82)^(214)Pb` decay by `beta`-emission. In how much time the number of nuclei of `._(82)^(214)Pb` will reach to the maximum?A. 182 secB. 247.5 secC. 308 secD. 194.8 sec |
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Answer» Correct Answer - b `._(84)^(218)Po overset(lambda_(1) = (0.693)/(183) = 3.786 xx 10^(-3) sec^(-1))(to)` `._(82)^(214)Pb overset(lambda_(2) = (0.693)/(161) = 4304 xx 10^(-3) sec^(-1))(to)` `t_("max") = (2.303)/(lambda_(1) - lambda_(2)) log (lambda_(1))/(lambda_(2))` `(2.303)/(3.786 xx 10^(-3) - 4.304xx 10^(-3)) log (3.786 xx 10^(-3))/(4.304 xx 10^(-3))` `= - (2.303) xx 5.183 xx 10^(-4)) (-0.05569)` `= 247.5 sec` |
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| 43. |
In the radioacitve decay `._(Z)X^(A) rarr ._(z + 1)Y^(A) rarr ._(z - 1)P^(A - 4) rarr (z - 1)Z^(*A - 4)` The sequence of emission is a. `alpha, beta, gamma` b. `gamma, alpha, beta` c. `beta, alpha, gamma` c. `beta, gamma, alpha`A. `alpha, beta gamma`B. `beta, alpha, gamma`C. `gamma , alpha, beta`D. `beta, gamma, alpha` |
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Answer» Correct Answer - b `._(Z)^(A)X - ._(-1)^(0)e to ._(Z + 1)^(A)Y, ._(Z + 1)^(A)Y - ._(2)^(4)He to ._(Z - 1)^(A - 4)Z`, `._(Z - 1)^(A - 4)Z - gamma to ._(Z - 1)^(A - 4) Z**` |
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| 44. |
(A) The position of an element an element in periodic in table after emission of one `alpha` and two `beta`-partilce remians unchanged. (R ) Emission of one `alpha` and two `beta` particles gives isotope of the parent element which acquires same position in the periodic table.A. If both (A) and (R ) are correct and (R ) is the correct explanation for (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A )C. If both (A) and (R ) are incorrect.D. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - A | |
| 45. |
A radioactive nuclide is produced at a constant rate of `alpha` per second . It’s decay constant is `lambda`. If `N_(0)` be the no. of nuclei at time t=0 , then max. no. nuclei possible are :A. `(alpha)/(lambda)`B. `N_(0) + (alpha)/(lambda)`C. `N_(0)`D. `(lambda)/(alpha) + N_(0)` |
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Answer» Maximum number of nuclei will be present when Rate of decay = rate of formation `lambda N = alpha` `N = (alpha)/(lambda)` |
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| 46. |
A nuclide has mass number (A) and atomic number (Z). During a radioactive process if: (A) both A and Z decrease, the process is called `alpha`-decay (B)A remains unchanged and Z decreases by one, the process is called `beta^(+)` or positron decay of K-electron capture (C ) both A and Z remain unchanged the process is called `gamma`-decay (D)both A and Z increase, the process is called nuclear isomerism. The correct answer is:A. 1,2 and 3B. 2,3, and 4C. 1,3, and 4D. 1,2, and 4 |
| Answer» Correct Answer - a | |
| 47. |
Statement-1 : Phosphours-32 decays to sulphur-32 with emission of a `beta`-particle. Because Statement-2: The neutron to proton ratio is less than 1.0 for all light stable nuclides.A. Statement-1 is true, Statement-2 is ture, statement-2 is a correct explanation for statement -1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is ture, statement-2 is falseD. Statement-2 is false, statement-2 is true |
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Answer» Correct Answer - c `._(15)^(32)P to ._(15)^(32)S + ._(1)^(0)e` `beta`-emission involves following nuclear process: `._(0)^(1)n to ._(1)^(1)H + ._(-1)^(0)e` It will not take place when (n/p) ratio is less than 1 |
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| 48. |
The radioactive nuclide `._(90)^(234) Th` shows two successive `beta-` decay followed by one `alpha-` decay. The atomic number and mass number respectively of the resulting atom is:A. 92234B. 94230C. 90230D. 92230 |
| Answer» Correct Answer - c | |
| 49. |
`._(4)Be^(7)` captures a K electron into its nucleus .What is the mass number and atomic number of the nuclide formed ? |
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Answer» Correct Answer - 7 In K-electron capture, a proton of nucleus changes into neutron. `._(1)^(1)H ._(-1)^(0)e to ._(0)^(1)n` `._(4)^(7)Be + ._(-1)^(0)e to ._(3)^(7)Li` |
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| 50. |
A positron is emitted from `._(11)Na^(23)` . The ratio of the atomic mass and atomic number of the resulting nuclide isA. 22/10B. 22/11C. 23/10D. 23/12 |
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Answer» Correct Answer - c `._(1)^(1) H to ._(0)^(1)n + underset("Positron")(._(+1)^(0)e)` |
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