InterviewSolution
Saved Bookmarks
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.
| 151. |
The half-life period of a radioactive element is 140 day. After 560 days, 1 g of the element will reduce toA. `1//8 g`B. `1//16 g`C. `1//4 g`D. `1//2 g` |
|
Answer» Correct Answer - b Number of half-lives(n) = `("Total time")/(t_(1//2)) =(560)/(14) =4` Amount left `= ((1)/(2))^(n)xx"initial amount" = ((1)/(2))^(4)xx1g = (1)/(16) g` |
|
| 152. |
The half life period of a radioctive element is 140 days. Afte 560 days, one gram of the element will reduced to :A. `1/2g`B. `1/4`gC. `1/8`gD. `1/16`g |
|
Answer» Correct Answer - d `1overset(140)to1/2overset(140)to1/4overset(140)to1/8overset(140)to1/16` |
|
| 153. |
The half`-` life periof of a radioactive element is 140 days. After 560 days, one gram of the element will reduce toA. `(1)/(2)g`B. `(1)/(4)g`C. `(1)/(8)g`D. `(1)/(16)g` |
|
Answer» Correct Answer - D `T=nxxt_(1//2)` `:. n=(T)/(t_(1//2))=(560)/(140)=4` Now, ` N_(t)=N_(0)((1)/(2))^(n)=1xx((1)/(2))^(4)=(1)/(16)g` |
|
| 154. |
`.^(23)Na` is the more stable isotope of Na. Find out the process by which `._(11)^(24)Na` can undergo radioactive decayA. `beta^(-)` emssionB. `alpha`- emissionC. `beta^(+)` emssionD. K electron capture |
|
Answer» Correct Answer - A n/p ratio of `.^(24)Na` nuclide is 13/11 i.e., greater than unity and hence radioactive. To achieve stability, it would tend to adjust its n/p ratio to the proper value of unity. This can be done by breaking a neutron into proton and electron. `._(0)n^(1) rarr ._(+1)P^(1) + ._(-1)e^(0) or beta^(-)` The proton will stay inside the nucleus whereas electron which cannot exist in the nucleus, will be emitted out as `beta-` ray |
|
| 155. |
`.^(23)` Na is the more stable isotope of Na. Find out the process by which `._(11)^(24)` Na can undergo radioactive decay.A. `beta""^(-)`emissionB. `alpha` emissionC. `beta""^(+)`emissionD. K electron capture |
|
Answer» Correct Answer - a N/P is above stability zone |
|
| 156. |
`Na^(23)` is more stable isotope of `Na`. Find out the process by which `._(11)Na^(24)` can undergo radioactive decay.A. `beta^(c-)-` emissionB. `alph-`emissionC. `beta^(o+)-`emissionD. `K` electron capture. |
|
Answer» Correct Answer - A Isotopic `_(11)Na^(24)` is less stable than `._(11)Na^(23)` because it shows radioactive decay Less stability of `Na^(24) w.r.t Na^(23)` also based upon `13//11(n/p)` ratio. Higher the value higher will be unstability. So it is disintegrated to attain stability). `underset(Less stabl e)(._(11)Na^(24)) rarr underset(Stabl e)(._(11)Na^(23))+ underset("Neutron")(._(0)n^(1)` This neutron on decomposition gives proton and `beta-` particle`(._(-1)e^(0))` `._(0)n^(1) rarr underset(Prot on)(._(1)H^(1)) ` or `._(1)P^(1)+underset(beta-partic l e)(._(-1)e^(0)` Hence, isotopic sodium is changed into sodium by means of emission of `beta-` particle and a proton `i.e., ` by `beta-` emission. |
|
| 157. |
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. |
| Answer» Correct Answer - B::C::D | |
| 158. |
The half life period of a radioactive elements does not depend upon:A. TemperatureB. PressureC. Initial amount of radioactive element takenD. Nature of radioactive element |
| Answer» Correct Answer - A::B::C | |
| 159. |
If `X^(a)` species emit firstly a positron then two `alpha` and `beta` last one `alpha` is also emitted and finally convert in `Y^(c )` species so correct the relation isA. a=c+12,d=b-5B. a=c-8,d=b-1C. a=c-6,d=b-0D. a=c-4,d=b-2 |
|
Answer» Correct Answer - A `_(92)U^(235)` nucleus absorbs a neutron and then disintegrate in `_(54)Xe^(139),38Sr^(94) and X.` Thus `_(92)U^(235)+_(0)n^(1)rarr_(54)Xe^(139)+_(38)Sr^(94)+3_(0)n^(1)` |
|
| 160. |
`""""_(90)Thh""^(234)` disintegrates to give `""""_(82)Pb""^(206)` as final product. How many alpha and beta particles are emmited during the process? |
| Answer» `""_(90)Th""^(234)` distinegrates to give `""_(82)Pb""^(206)` as the final product. How may alpha and beta particles are emmitted during this process? | |
| 161. |
A radioactive substance has `t_(1//2)` 60 minutes. After 3 hrs, what precentage of radioactive substance will remainA. 0.5B. 0.75C. 0.25D. 0.125 |
|
Answer» Correct Answer - D Amount left `= (N_(0))/(2^(3)) = (100)/(8) = 12.5%` |
|
| 162. |
Choose the incorrect one.A. 1 curie `= 3.7xx10^(10)ds^(-1)`B. 1 rutherford `= 10^(6) ds^(-1)`C. 1 becquerel `= 1 ds^(-1)`D. 1 fermi `= 10^(3) ds^(-1)` |
|
Answer» Correct Answer - d The nuclear raduis is expressed in fermi (Fm) units. 1 fermi `= 10^(-13) cm = 10^(-15) m` |
|
| 163. |
If 5 g of a radioactive substance has `t_(1//2) = 14 h, 10 g` of the same substance will have `t_(1//2)` equal toA. 70 hB. 14 hC. 28 hD. 50 h |
|
Answer» Correct Answer - b `t_(1//2) = (0.693)/(lambda)` and it is independent of initial amount or mass. |
|
| 164. |
The decay constant of a radioactive substance is 0.173 `(year)""^(-1)`. ThereforeA. nearly 63% of the radioactive substance will decay in `(1/0.173years)`B. half life of the radio active substance is `(1/0.173years)`C. One sixth of the radioactive substance will be left after 8 yearD. all the above statements are true |
|
Answer» Correct Answer - a `lamda=2.303/tlog.N""_(0)/N)` |
|
| 165. |
The number of `beta`-particles emitted during the change `""_(a)""^(c)Xto""_(d)""^(b)Y` isA. `(a-b)/4`B. `d+[(a-b)/2]+c`C. `d+[(c-b)/2]-a`D. `d+[(a-b)/2]-c` |
|
Answer» Correct Answer - c For `alpha-Zdarrby2` For `alpha-Adarrby4` For `alpha-A` and Z doesn t change `beta-Zuarr` by one unit, A-doesn’t change `alpha=(c-b)/4,beta=2Xalpha-Zp-Zd, 2X(c-b)/(4)-[a-d]` =`d+[(c-b)/(2)]-alpha` |
|
| 166. |
Two substances A and B are present such that `[A_(0)]=4[B_(0]` and half-life of A is 5 minutes and that of B is 15 minutes. If they start decaying at the same time following first order kinetics after how much time the concentration of both of them would be same ?A. 15 minB. 60 minC. 12 minD. 180 min |
| Answer» Correct Answer - d | |
| 167. |
A radioactive element `X` has an atomic number of 100. It decays directly into an element `Y` which decays directly into an element `Z`. In both the processes either one `alpha` or one `beta-` particle is emitted. Which of the following statement could be true?A. `Y` has an atomic number of `102`B. `Z` has an atmic number of `101`C. `Z` has an atomic number of `97`D. `Z` has an atomic number of `99` |
|
Answer» Correct Answer - D `X` and `Y` can decay one `alpha` each or one `beta` each or X-decays, `1alpha`, Y-decays `1 beta` or X-decays `1 beta` or Y-decays `1alpha`. In either case (a), (b) and (c) cannot be true. Ans. (D) |
|
| 168. |
When radioactive minerals like clevelte, monozite and pitch blende are heated to 1273 K in vacuo the noble gas obtained isA. RnB. KrC. HeD. Ne |
|
Answer» Correct Answer - c |
|
| 169. |
If the subsidiary quantum number of a subenergy level is `4`, the maximum and minimum values of the spin multiplicities are:A. `9, 1`B. `10, 1`C. `10, 2`D. `4, -4` |
|
Answer» Correct Answer - C `l=4`, number of degenerate orbitals `=2l+1=9` Maximum total spins `=9xx1/2` Maximum multiplicity `=2S+1=2xx9/2+1=10` minimum multiplicity `=1/2` minimum multiplicity `=2xx1/2+1=2` |
|
| 170. |
In any subshell, the maimum number of electrons having same value of spin quantum number is :A. `sqrt(l(l+1))`B. `l+2`C. `2l+1`D. `4l+2` |
|
Answer» Correct Answer - 3 Maximum number of electron with same spin is equal to maximum number of orbitals, i.e., `(2l+1)` |
|
| 171. |
If the velocity of the electron in first in first of `H` atom is `2.18xx10^(6) m//s`, what is its value in third orbit? |
|
Answer» Correct Answer - `7.27xx10^(5) m//s` `v_(3)=v_(1)xx(Z/n) rArr v_(3)=2.18xx10^(6)xx(1/3)=7.27xx10^(5) m//s` |
|
| 172. |
Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is.A. `4.41xx10^(-16) J atom^(-1)`B. `-4.41xx10^(-17) J atom^(-1)`C. `-2.2xx10^(-15) J atom^(-1)`D. `8.82xx10^(-17) J atom^(-1)` |
|
Answer» Correct Answer - 2 I.E. of `He^(+)=19.6xx10^(-18)J atom^(-1)` I.E. `=-E_(1)` `E_(1)` for `He^(+)` is `=-19.6xx10^(-18) J atom^(-1)` `((E_(1))_(He^(+)))/((E_(1))_(Li^(3+)))=((Z_(He^(+)))^(2))/((Z_(Li^(2-)))^(2))rArr (-19.6xx10^(-18))/((E_(1))_(Li^(2-)))=4/9` `E_(1)(Li^(2+))=(-19.6xx9xx10^(-18))/(4)=-44.1xx10^(-18)=-4.41xx`0^(-17) J atom^(-1)` |
|
| 173. |
For the given series reaction in `n^(th)` step, find out the number of protons & energy. `_(92).^(238)U rarr Ba+Kr+3_(0)n^(1)+"Energy" (E)` |
|
Answer» Correct Answer - `3^(n), 3^(n-1) E` Nuclear fission is a series reaction `3, 9, 27`……. Neutron and `E, 3E, 9E…….` energy are emitted. Hence answer is `3^(n), 3^(n-1) E`. |
|
| 174. |
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom . Calculate the ionisation energy of sodium in kJ `mol^(-1)`.A. `494.65`B. `400`C. `247`D. `600` |
|
Answer» Correct Answer - A I.E. of one sodium atom `=(hc)/(lambda)` & I.E. of one mole `Na` atom `=(hc)/lambda N_(A)=(6.62xx10^(34)xx3xx10^(8)xx6.02xx10^(23))/(242xx10^(-9))=494.65 kJ.mol.` |
|
| 175. |
Complete the following table: `|{:("Particle","Mass No.","Atomic No.","Protons","Neitrons","Electrons"),("Nitrogen atom",-,-,-,7,7),("Calcium ion",-,20,-,20,-),("Oxygen atom",16,8,-,-,-),("Bromide ion",-,-,-,45,36):}|` |
|
Answer» For nitrogen atom. No. of electron `=7` (given) No. of neutrons `=7` (given) `:.` No. of protons `=Z=7` ( :. Atom is electrically neutral) Atomic number `=Z=7` Mass No. (A)=No. of protons + No. of neutrons `=7+7=14` For calcium ion. No. of netrons `=20` (Given) Atomic No. (Z)`=20` (Given) `:.` No. of protons `=Z=20`, But in the formation of calcium ion, two electrons are lost from the extranuclear part according to the equation `Cararr Ca^(2+)+2e^(-)` but the composition of the nucleus remains unchanged. `:.` No. of electrons in calcium ion `=20-2=18` Mass number (A)= No. of protons + No. of neutrons `=20+20=40` For oxygen atom. Mass number (A) = No. of protons + No. of neutrons `=16` (Given) Atomic No. `(Z)=8` (Given) No. of protons `=Z=8`, No. of electrons `=Z=8` No. of neutrons `=A-Z=16-8=8` For bromide ion. No. of neutrons `=45` (given) No. of electrons `=36` (given) But in the formation of bromide ion, one electron is gained by nuclear part according to equation `Br+e^(-) rarr Br^(-),` But the composition of nucleus remains ubchanged. `:.` No. of protons in bromide ion = No. of electrons in bromine atom `=36-1=35` Atomic number (Z) = No. of protons `=35` Mass number (A)= No. of neutrons + No. of protons `=45+35=80` |
|
| 176. |
Certain sum glasses having small of `AgCl` incroporated the lenses, on expousure to light of appropriate wavelength turns to gray colour to reduce the glare following the reactions: `AgCloverset(hv)(rarr) Ag("Gray")+Cl` If the heat of reaction for the decomposition of `AgCl` is `248 kJ mol^(-1)`, what maximum wavelength is needed to induce the desired process? |
|
Answer» Energy needed to change `=248xx10^(3) J//mol` If photons is used for this purpose, then according to Einstein law one molecule absorbs one photon. Therefore, `:. N_(A).(hc)/(lambda)=248xx10^(3)` `lambda=(6.626xx10^(-34)xx3.0xx10^(8)xx6.023xx10^(23))/(248xx10^(3))=4.83xx10^(-7) m` |
|
| 177. |
The approximate wavelength associated with a gold-ball weighting 200 g and moving at a speed of 5m/h is of the order of :A. `10^(-1) m`B. `10^(-20) m`C. `10^(-30)m`D. `10^(-40)m` |
|
Answer» Correct Answer - C `lambda=(h)/(mv)=(6.625xx10^(-34))/(0.2xx5)xx3600~~10^(-30) m` |
|
| 178. |
In H-atom if `r1` is the radius fo first Bohr orbit de-Broglie wavelength of an elecrton in ` 3^(rd)` orbit is :A. `3pi x`B. `6pi x`C. `(9x)/(2)`D. `x/2` |
|
Answer» Correct Answer - B `r_(1)=0.529 Å` `r_(3)=0.529xx(3)^(2)Å=9x` So, `lambda=(2pir)/(n)=(2pi(9x))/(3)=6 pi x` |
|
| 179. |
The wavelength of a spectral life for an electronic transition inversely proportional to:A. number of electrons undergoing transitionB. the nuclear charge of the atomC. the velocty of an electron undergoing transitionD. the difference in the energy involved in the transition |
| Answer» Correct Answer - D | |
| 180. |
The radius `""_(Z)M""^(4)` nucleus is (outer most configuration `3s""^(2)3p""^(1)` and A+Z=40)A. 4.2 FMB. `1.4xxroot(3)40`C. `1.4xxroot(2)40`D. `1.4xx40FM` |
|
Answer» Correct Answer - a `R=R""_(0)(A)""^(1//3)(R""_(0)=1.4F.M)` |
|
| 181. |
`d_(xy)` orbital has lobes between `x`- and `y`- axes. The wave function of two lobes are positive and those of other two are negative. The positive wave function signifies that :A. both `x` and `y` are positiveB. both `x` and `y` are negativeC. either `x` ir `y` us negativeD. None of these |
| Answer» Correct Answer - A, B | |
| 182. |
Which quantum number defines the orientation of orbital in the space around the nucleus?A. Principal quantum number `(n)`B. Angular momentum quantum numberC. Magnetic quantum number `(m_(l))`D. Spin quantum number `(m_(s))` |
| Answer» Correct Answer - 3 | |
| 183. |
An element is isobaric with the inert gas atom `._(18)A^(40)`. The electronic arrangement of the element is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(2) 4s^(2)`. How many neutrons does each atom of the element carry in its nucleus?A. 22B. 20C. 18D. 16 |
|
Answer» Correct Answer - B Atomic number of inert gas atom = 20 Atomic mass of inert gas atom = 40 (isobaric to `Ar^(40)`) `:.` Number of neutrons = 40 - 20 = 20 |
|
| 184. |
A 3p-orbital has :A. Two non-spherical nodesB. Two spheric nodesC. One spherical and non spherical nodesD. One spherical and two non spherical nodes |
|
Answer» Correct Answer - C Spherical node `=n-l-1` non sherical `=l` |
|
| 185. |
Consider an electron in the `n^(th)` orbit of a hydrogen atom in the Bohr model . The circumference of the orbit can be expressed in terms of the de Broglie wavelength `lambda` of the electron as :A. `(0.529) n lambda`B. `sqrt(n) lambda`C. `(13.6) lambda`D. `n lambda` |
| Answer» Correct Answer - D | |
| 186. |
Point out the anugular momentum of an electron in, (a) `4s` orbital (b) `3p` orbital (c ) `4^(th)` orbit )according to Bohr model) |
|
Answer» Correct Answer - (a)`0`, (b) `h/sqrt(2pi)`, (c) `(2n)/(pi)` Orbital angular momentum `=sqrt(l(l+1))h/(2pi)` For `4 s` orbital, `l=0 :.` Angular momentum`=sqrt(0(0+1))h/(2pi)=0` For `3p` orbital, `l=1 :.` Angular momentum `=sqrt(1(1+1))h/(2pi)=sqrt(2) h/(2pi)` For `4^(th)` orbit, Angular momentum `=(nh)/(2pi)=(4h)/(2pi)=(2h)/(pi)` |
|
| 187. |
A freshly prepared radioactive source of half -life 2 hours emits radiations 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 hoursB. 12 hoursC. 24 hoursD. 128 hours |
|
Answer» Correct Answer - B `N = (N_(0))/(64) = (N_(0))/(2^(6)) :. N = 6` Thus total time `= 2xx 6 = 12hr` |
|
| 188. |
If it is assumed that `._(92)^(235)U` decays only by emitting `alpha and beta-`particles, the possible product of the decay isA. `._(89)^(235)Ac`B. `._(89)^(227)Ac`C. `._(89)^(230)Ac`D. `._(89)^(231)Ac` |
|
Answer» Correct Answer - B `n_(alpha) = (DeltaA)/(4)` (`DeltaA` change in mass no) `DeltaA` is divisible by 4 in options (b) and (d) In option (d), `n_(alpha) =1` `:. n_(beta) = Delta Z - 2n_(alpha) = 3 - 2 =1`, which is not possible |
|
| 189. |
After the faliure of Bohr atomic theory but its ability to explain to the atomic spectral a need was felt for the new model that could incorporate, the concept of stationary orbit, de Broglie concept, Heisenberg uncertainity principle. The concept that in corporate above facts is called quantum mechanics of the atomic model wave mchanical model. it includes set of quantum number and `|psi^(2)|` a mathematical expression of the probability of finding an electron at all points in space. This probability function is the best indication available of how the electon behaves, for as a consequence of the Uncertainity Principle, the amount we can know about the electron is limited. While quantum mechanics can tell us the exact probability of finding an electron at any two particular points, it does not tell us how the electron moves from one of these points to the other. Thus the idea of an electron orbit is lost, it is replaced with a description of where the electron is most likely to be found. This total picture of the probability of finding an electron at various points in space is called an orbital. There are various types of orbitals possible, each corresponding to one of the possible combinations of quantum numbers. These orbitals are classified according to the value of n and l associated with them. In order to avoid confusion over the use of two numbers, the numerical velues of l are replaced by letters, electrons in orbitals with `l=0` are called s-electrons those occupying orbitals for which `l=1` are p-electrons and those for which `l=2` are called d-electrons. The numerical and alphabetical correspondences are summarized in table. Using the alphabetical notation for l, we would say that in the ground state of hydrogen atom `(n=1, l=0)` we have a `1s`-electron, or that the electron moves in a `1s`-orbital. The relation of the spherical polar co-ordinates `r, theta` and `phi` to Cartesian coordinates x, y and z. To make the concept of an orbital more meaningful, it is helpful to examine the actual solution of the wave function for the one-electron atom. Because of the spherical sysmmetry of the atom, the wave functions are most simply expressed in terms of a spherical polar-coordinate system, shown in fig., which has its orbit at the nucleus. It is found that the wave function can be expressed as the product of two functions, one of which (the 'angular part' X) depends only the angle `theta` and `phi` the other of which (the 'radial part' R) depends only on the distance from the nucleus. Thus we have `phi(r, theta, phi)=R(r)X(theta, phi)` Angular and radial parts of hydrogen atom wave function `{:("Angular part" X(theta, phi),,"Radial part" R_(n, l)(r)),(X(s)=(1/(4pi))^(1//2),,R(1s)=2(z/a_(0))^(3//2) e^(-sigma//2)),(X(p_(x))=(3/(4pi))^(1//2) sin theta cos phi,,R(2s)=1/(2sqrt(2))(z/a_(0))^(3//2) (2-sigma)e^(-sigma//2)),(X(p_(y))=(3/(4pi))^(1//2) sin theta sin phi,,R(2p)=1/(2sqrt(6))(z/a_(0))^(3//2) sigmae^(-sigma//2)),(X(p_(z))=(3/(4pi))^(1//2) costheta,,),(X(d_(2)^(2))=(5/(16pi))^(1//2)(3 cos^(2) theta-1),,),(X(d_(xz))=(15/(4pi))^(1//2) sin theta cos theta cos phi,,R(3s)=1/(9sqrt(3))(z/a_(0))^(3//2) (6-6sigma+sigma^(2))e^(-sigma//2)),(X(d_(yz))=(15/(4pi))^(1//2) sin theta cos theta sin phi,,R(3p)=1/(9sqrt(6))(z/a_(0))^(3//2) (4-sigma)sigmae^(-sigma//2)),(X(d_(x^(2)-y^(2)))=(15/(4pi))^(1//2) sin^(2)theta cos2phi,,R(3d)=1/(9sqrt(30))(z/a_(0))^(3//2)sigma^(2) e^(-sigma//2)),(X(d_(xy))=(15/(4pi))^(1//2) sin^(2)theta sin 2phi,,),(,,sigma=(2Zr)/(na_(0)),a_(0)=h^(2)/(4pi^(2)me^(2))):}` This factrorization helps us to visualize the wave function, since it allows us to consider the angular and radial dependences separately. It contains the expression for the angular and radial parts of the one electron atom wave function. Note that the angular part of the wave function for an s-orbital it alwats the same, `(1//4 pi)^(1//2)` regardless of principal quantum number. It is also true that the angular dependence of the p-orbitals and of the d-orbitals is independent of principle quantum number. Thus all orbitals of given types (s, p, or d) have the same angular behavour The table shows, however, that the radial part of the wave function depends both on the principal quantum n and on the angular momentum quantum number l. To find the wave function for a particular state, we simply multiply the appropriate angular and radial parts togather called normalized wave function. The probability of finding an electron at a point within an atom is proportional to the square of orbital wave function, i.e., `psi^(2)` at that point. Thus, `psi^(2)` is known as probability density and always a positive quantity. `psi^(2) dV ("or" psi^(2). 4pir^(2)dr)`. represents the probability for finding electron in a small volume dV surrounding the nucleus. The electron probility density for `1s`-orbital best represented by the relation.A. `(1)/(2sqrt(pi))(Z/a_(0))^(3//2)xxe^(r/a_(0)`B. `1/pi(Z/a_(0))^(3)xxe^(-(2zr)/a_(0))`C. `1/pi(Z/a_(0))^(3//2)xxe^(r/a_(0))`D. `2/pi(Z/a_(0))^(3)xxe^(-(2zr)/a_(0))` |
| Answer» Correct Answer - B | |
| 190. |
A wooden artifact sample gave activity `32-beta` particles per second while the freshly cut wood gave activity of `64 beta` particles per second in Geiger Muller counter. Calculate the age of the wooden artifact `(t_(1//2) "of" C^(14) = 5760` years) |
|
Answer» `(0.693)/(t_(1//2)) = (2.303)/(t_("age")) "log" ((N_(0))/(N))` `(0.693)/(5760) = (2.303)/(t_("age")) "log" ((64)/(32))` `t_("age") = 5760` years |
|
| 191. |
A radioactive nucleus decays by emitting one alpha and two beta particles, the daughter nucleus is `…………………..` of the parent |
| Answer» Correct Answer - Isotope | |
| 192. |
The half-life period of `C^(14)` is 5760 years. A piece of woods when buried in the earth had `1% C^(14)`. Now as charcoal it has only `0.25% C^(14)`. How long has the piece of wood been buried? |
|
Answer» `t_(1//2)` of `C^(14) = 5760` `:. Lambda = (0.693)/(5760) "year"^(-1)` `N_(0C)^(14) = 1%` `N_(C^(14) = 0.25%` `:.t = (2.303)/(lambda) "log" (N_(0))/(N)` `= (2.303 xx 5760)/(0.693) "log" (1)/(2) = (2.303 xx 5760)/(0.693) "log" 2^(2)` `t = 11520` year |
|
| 193. |
An old chair of wood show `""_(6)C""^(14)` activity which is 80% of the activity which is 80% of the activity found today. Calculate the age of the old chair of wood. (`t""_(1//2)of .""_(6)C""^(14)=5770 yr`)?A. `t=2.303/5770log"100/80`B. `t=5770/0.301log"100/80`C. `t=5770/0.693log"100/80`D. `t=0.693/5770log"100/80` |
|
Answer» Correct Answer - b `t=2.303/lamdalog.(N""_(0)/N)` |
|
| 194. |
`C^(14)` has a half`-` life of 5760 years. `100mg` of the sample containing `.^(14)C` is reduced to `25mg` ina)11520yearsb)2880yearsc)1440yearsd)17128yearsA. 11520 yearsB. 2880 yearsC. 1440 yearsD. 17280 years |
| Answer» Correct Answer - A | |
| 195. |
The number of neutrons in the parent nucleus which gives `N^(14)` on beta emission is `…………………………`. |
| Answer» `._(6)C^(14) rarr ._(7)N^(14)+._(-1)e^(0)` (Number of neutron`=14-6=8)` | |
| 196. |
The number of neutrons in the parent nucleus which gives `N^(14) " on " beta-`emission and the parent nucleus isA. `8, C^(14)`B. `6, C^(12)`C. `4, C^(13)`D. None of these |
|
Answer» Correct Answer - A `._(6)C^(14) rarr ._(7)N^(14) + ._(+1)e^(0)` No. of neutrons in `C^(14) = 14 - 6 = 8` |
|
| 197. |
87.5% decomposition of a radioactive substance completes in 3 hours. What is the half-life of that substanceA. 2 hoursB. 3 hoursC. 90 minutesD. 1 hours |
|
Answer» Correct Answer - D `(0.693)/(t_(1//2)) = (2.303)/(180) xx "log" (100)/(12.5)` `t_(1//2) = (0.693 xx 180)/(2.303 xx 3 xx 0.3010) = 60` mini = 1 hr. |
|
| 198. |
An old piece of wood has 25.6T as much `C^(14)` as ordinary wood today has. Find the age of the wood. Half-life period of `C^(14)` is 5760 years. |
|
Answer» Suppose the amount of `C^(14)` present in the wood originally (i.e., the same which the wood today has) `= a`. Then the amount of `C^(14)` present now in the old wood `(25.6)/(100) a = 256a` The time `t` in which `C^(14)` changed from `a` to `0.256a` will then be given by `t = (2.303)/(K) "log" (a)/(0.256a)` But `K = (0.693)/(t_(1//1)) = (0.693)/(5760) = 1.203 xx 10^(-4) "year"^(-1)` `:. t = (2.303)/(1.203 xx 10^(-4)) "log" (1)/(0.256) = 11329` years |
|
| 199. |
Assertion : `._(11)^(22)Na` emits a positive giving `._(12)^(22)Mg` Reaosn : In `beta` emission neutron is transformed into proton |
|
Answer» Correct Answer - D `._(11)Na^(22) rarr ._(12)Mg^(22) + ._(-1)beta^(0)` Thus this change involves a `beta`-particle emission and not a positron. Also, proton emission converts proton into neutron as : `._(1)P^(1) rarr ._(0)n^(1) + ._(+1)beta^(0)` |
|
| 200. |
An artificial radioactive isotope gave `._(7)^(14)N` after two successive `beta-`particle emission. The number of neutrons in the parent nucleus must beA. 9B. 14C. 5D. 7 |
|
Answer» Correct Answer - A `._(x)X^(y) overset(2beta)rarr ._(7)N^(14)` `._(x =7-2)X^(y =14) = ._(5)X^(14)` Total no. of neutrons `= 14 - 5 = 9` |
|