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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.
| 201. |
The electron in a hydrogen atom make a transtion `n_(1) rarr n_(2)` where `n_(1) and n_(2)` are the priocipal quantum number of the two states . Assume the Bohr model to be valid . The time period of the electron in the initial state is eight time that in the final state . The possible values of `n_(1) and n_(2)` areA. `n_(1)=4, n_(2)=2`B. `n_(1)=8, n_(2)=2`C. `n_(1)=8, n_(2)=1`D. `n_(1)=6, n_(2)=3` |
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Answer» Correct Answer - A::D As `r_(n)propn^(2), v_(n) prop 1/n`, time `t=(2pir_(n))/(v_(n))` i.e, `t_(n) prop (r_(n))/(v_(n))` or `T_(n) prop n^(3)`, i.e., `(T_(1))/(T_(2))=((n_(1))/(n_(2)))^(3)` or `8=((n_(1))/(n_(2)))^(3)` or `(2)^(3)=((n_(1))/(n_(2)))^(3)` hence `n_(1)=2n_(2)`, `:.` when (a) `n_(1)=4, n_(2)=2` and (d) `n_(1)=6, n_(2)=3` |
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| 202. |
A nucleus `._n^ m X` emits one `alpha-`particle and two `beta-`particles. The resulting nucleus isA. `._(n-4)^(m-6)Z`B. `._(n)^(m-6)Z`C. `._(n)^(m-4)X`D. `._(n-2)^(m-4)Z` |
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Answer» Correct Answer - C When an alpha particle `(._(2)He^(4))` is emitted, the mass number and atomic number of dougther nucleus decreases by four and two respectively. When a `beta` particle `(._(-1)e^(0))` is emitted, the mass number of product remains unchanged, but the atomic number increases by one. Therefore, `._(n)X^(m)` overset (alpha)to `._(n-2)X^(m-4) overset (2beta)to._(n)X^(m-4)` |
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| 203. |
Assume that a neutron breaks into a proton and an electron . The energy reased during this process is (mass of neutron `= 1.6725 xx 10^(-27) kg` mass of proton `= 1.6725 xx 10^(-27) kg` mass of electron `= 9 xx 10^(-31) kg )`A. `0.506MeV`B. `7.10MeV`C. `6.39MeV`D. `5.4MeV` |
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Answer» Correct Answer - A The nuclear reaction is `._(0)n^(1)to._(1)p^(1)+._(-1)e^(0)` Mass defect, `Deltam = (m_(p)-m_(e))-m_(n)` `= (1.6725xx10^(-27)+9xx10^(-31))-1.6725xx10^(-27)` `=9xx10^(-31)kg` Energy released `=(Deltam)c^(2)` `=9xx10^(-31)(3xx10^(8))^(2)J` `DeltaE=(81xx10^(-15))/(1.6xx10^(-13)) MeV` `=0.506 MeV` |
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| 204. |
A radioactive element decays to `1//32th` of its initial activity in 25 decay. Calculate its half life. |
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Answer» Here, `A/(A_0)=1/32, t= 25 days` `T=?` As `A/(A_0)=N/(N_0)=(1/2)^n=1/32 :. n=5` As `T=t/n :. T=25/5=5 days` |
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| 205. |
A radioactive elements has a half life of 2500 yrs. In how many years will its mass decay by 90% of its initial mass? |
| Answer» Correct Answer - `8305.6 yrs` | |
| 206. |
For a radioactive meterial , its activity `A` and rate of charge of its activity `R` are defined as `A = - (dN)/(dt) and R = -(dA)/(dt) ` where `N(t)` is the number of nuclei at time `t` .Two radioactive source P (mean life `tau` ) and Q(mean life `2 tau` ) have the same activity at `t = 2 tau R_(p) and R_(Q)` respectively , if `(R_(p))/(R_(Q)) = (n)/(e)`A. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - B Here, `A_(P)=A_(0)e^(-t/tau)=A_(Q)=A_(0)e^(-t/(2tau))` `R_(P)=A_(0)/tau e^(-t/tau)=R_(Q)=A_(0)/(2tau) e^(-t/(2tau))` at `t=2tau` `(R_(P))/(R_(Q))= (A_(0)/tau e^(-2))/(A_(0)/(2tau) e^(-1)) =2/e` On comparing with `(R_(P))/(R_(Q))=n/e`, we get n=2 |
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| 207. |
Which of the following assertions are correct?A. A neutron can decay to a proton only inside a nucleusB. A proton can change to a neutron only inside a nucleusC. An isolated proton can change into a neutronD. An isolated neutron can change into a proton |
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Answer» Correct Answer - B::D Inside a nucleaus, a proton can change to a neutron. Out side the nucleus, an isolated neutron can change into a proton. |
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| 208. |
Let `m_(p)` be the mass of a poton , `M_(1)` the mass of a `_(10)^(20) Ne` nucleus and `M_(2)` the mass of a `_(20)^(40) Ca` nucleus . ThenA. `M_(2)=2M_(1)`B. `M_(2)gt2M_(1)`C. `M_(2)lt2M_(1)`D. `M_(1) lt10(m_(n)+m_(p))` |
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Answer» Correct Answer - C::D On account of mass defect, `M_(1)lt10(m_(n)+m_(p))` and `M_(2)lt20(m_(p)+m_(n))` As B.E. in case of `._(20)Ca^(40)` is more than B.E. in case of `._(10)Ca^(20)`, therefore, `M_(2)lt2M_(1)` |
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| 209. |
Isotopes of an element are the atoms ........... which have............. but........... . |
| Answer» Correct Answer - of an element; same atomic number; different atomic weights. | |
| 210. |
The isotope `._8O^(16)` has 8 protons, 8 neutrons and 8 electrons, while `._4Be^8` has 4 protons, 4 neutrons and 4 electrons.Yet the ratio of their atomic masses is not exactly 2. Why? |
| Answer» The ratio of mass of `._8O^(16)` and mass of `._4Be^8` is not exactly two energy of the atomic nuclei of the two elements. | |
| 211. |
Write down the names and formulae of the three isotopes of hydrogen. |
| Answer» Hydrogen `(._1H^1)`,deuterium `(._1H^2)` and tritium `(._1H^3)`. | |
| 212. |
How many electrons protons, and neutrons are there in `12 g` of `._(6) C^2` and in `14g` of `._(6) C^(14)` ?. |
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Answer» Correct Answer - `36xx10^(23), 36xx10^(23), 48xx10^(23)` Each atom of `._(6)C^(14)` contains 6 protons, 8 neutrons and 6 electrons and there are `6xx10^(23)` such atoms in 14 gm. of Carbon 14. |
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| 213. |
Why heavy stable nucleus must contain more neutrons than protons? |
| Answer» Coulomb forces between protons are repulsive and nuclear forces are ordinarily attractive. for nuclei to be stable nuclear forces must dominate the repulsive force. Therefore, number of neutrons must be greater than the number of protons. | |
| 214. |
Protons and neutrons exit together in an extermely small space within the nucleus. How is this possible when protons replel each other? |
| Answer» Due to strong attractive nuclear forces. | |
| 215. |
Two elementary particles which have almost infinite life time areA. electron and neutronB. neutron and protonC. electron and protonD. None of the above |
| Answer» Correct Answer - C | |
| 216. |
What is the masss of pion plus `(pi^(+))`? |
| Answer» Mass of a pion plus is 273 times the mass of an electron. | |
| 217. |
If `m_(e)` is mass of an electron, then mass of pion plus `(pi^(+))` particle isA. `207m_(e)`B. `273m_(e)`C. `m_(e)/207`D. `m_(e)/273` |
| Answer» Correct Answer - B | |
| 218. |
A radioactive element `._90 X^238` decay into `._83 Y^222`. The number of `beta-`particles emitted are. |
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Answer» Correct Answer - 1 `._(90)X^(238) to ._(83)Y^(222)` As mass number decreases by `238-222=16`, therefore, number of `alpha` particle emitted `=16/4=4` The charge number reduces to `90-4xx2=82`. To increases charge no. (83-82)=1. One beta particle must be emitted. |
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| 219. |
What is the mass of muon plus `(mu^(+))`? |
| Answer» Mass of a muon plus particle is 207 times the mass of an electron. | |
| 220. |
A radioactive isotope decays in the following sequence : `Aoverset(._0n^1)toA_1overset(alpha)toA_2` If the mass number and atomic number of `A_2` are 176 and 71 respectively, find the mass number and atomic number of `A_1` and A. Which of the three elements are isotopes? |
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Answer» As `alpha=._2He^4`, therefore, mass number of `A_1` `=176+4=180` and charge number of `A_1=71+2=73`. Similarly, mass number of `A=180+181` and charge no. of `A =73+0=73`. Clearly, elements `A_1` and A are isotopes? |
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| 221. |
In a particular fission reaction, a `._92U^(235)` nucleus captures a slow neutron. The fission products are three neutrons, a `._57La^(142)` nucleus and a fission product `._ZX^A`. What is the value of Z? |
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Answer» Applying the principal of conservation of charge number. `Z+57=92 :. Z=92-57=35.` |
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| 222. |
Calculate the energy of the following nuclear reaction: `._(1)H^(2)+._(1)H^(3) to ._(2)He^(4) + ._(0)n^(1)+Q` Given: `m(._(1)H^(2))=2.014102u, m(._(1)H^(3))=3.016049u, m(._(2)He^(4))=4.002603u, m(._(0)n^(1))=1.008665u` |
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Answer» Correct Answer - `17.58MeV` In the given nuclear reaction, Mass of reactants = `m(._(1)H^(2))+m(._(1)H^(3))` `=2.0141202+3.016049` `=5.030151u` Mass of products = `m(._(2)H^(4))+m(._(0)n^(1))` `=4.002603+1.008665` `=5.011268u` Mass defect, `Deltam=5.030151u-5.011268u` `=0.018883u` Energy released `=0.018883xx931MeV` `=17.58MeV` |
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| 223. |
Statement-1: No law is violated in the nuclear reaction `._(0)n^(1)to._(1)H^(1)+._(-1)e^(0)` Statement-2: Mass number and charge number, both are conservedA. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true. |
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Answer» Correct Answer - D Statement-1 is false, but statement-2 is true. The law violated is the law of conservation of spin or angular momentum. |
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| 224. |
What do you mean by Q value of a nuclear reaction? |
| Answer» It is the energy released/absorbed in the nuclear reaction. | |
| 225. |
Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor. |
| Answer» Cadmium of Boron are absorbed of neutrons. They serve as controllers of reaction rate of neutrons in the nuclear reactor. | |
| 226. |
The Coulomb barrier height for `alpha` particle emission is 30.0MeV. What is the barrier height for `._(6)C^(14)`? The required data is `m(._(88)Ra^(223))=223.01850u , m(._(82)Pa^(209))=208.98107u , m(._(86)Rn^(219))=219.00948u, m(._(6)C^(14))=14.00324u , m(._(2)He^(4))=4.00260u` |
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Answer» The coulomb barrier height for a `alpha` decay is equal to the coulomb repulsion between the `alpha` particle and the dougther nucleus of (Z=88-2=86) when they are just touching each other. Hence `u(alpha)=((2e)(86e))/(r_(0)(219^(1//3)+4^(1//3)))` Similarly, the coulomb barrier for `C^(14)` is `u(C^(14))= ((6e)(82e))/(r_(0)(209^(1//3)+14^(1//3)))` Dividing, we get `(u(C^(14)))/(u(alpha))=(6xx82)/(209^(1//3)+14^(1//3))xx(219^(1//3)+4^(1//3))/(2xx86)=(6xx82(6.03+1.59))/((5.93+2.41)xx2xx86)=2.61` `u(c^(14))=u(alpha)xx2.61=30xx2.61=78.3MeV` |
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| 227. |
A nuclear reactor is a powerful device, wherein nuclear energy is utilised for peaceful purpose. It is based upon controlled nuclear chain rection. The nuclear reaction is controlled by the use of control rods (of boron or cadmium) and moderators like heavy water, graphite, etc. The whole reactor is protect with concrete walls 2 to 2.5 meter thick, so that radiation emitted during nuclear reactions may not produce harmful effects. Read the above passage and answer the following questions: (i) Give any two merits of nuclear reactors. (ii) What is radiactive waste? (iii) why do people often oppose the location site of a nuclear reactor? What do you suggest? |
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Answer» (i) Nuclear reactors are used in electric power generation. They are also used to produced radioactive isotopes- which have applications in medicine, industry and agriculture. ii) Radioactive waste consists of fission products and transuranic elements such as plutonium and americium. This waste is extremely hazardous to all forms of life on earth. (iii) People often oppose the location site of a nuclear reactor because any leakage of nuclear radiations can affect adversely miles of area surrounding it. Elaborate safety maesures are needed not only for reactor operations, but also for handling and disposal of radioactive waste. I would suggest that Government must take stringent safety measures and assure people of safeguards in the event of nuclear accidents. At the same time, people must also be educated accordingly. |
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| 228. |
Nuclear fission is the phenomenon of splitting of a heavy nucleaus into two or more ligther nuclei. Nuclear fusion is the phenomenon of fusing two or more ligther nuclei to form a single heavy nucleus. In both these processes, certain mass `(Deltam)` disappears, which appears in the form of nuclear energy, `E=(Deltam)c^(2)`. The release of energy is so sudden that it cannot be controlled. This causes havoc. Nuclear fission is the basis of the an atom bomb and nuclear fusion is the basis of a hydrogen bomb. A powerful device, called Nuclear Reactor has been developed, where in nuclear energy produced is utilised for constructive purposes. (i) How much energy is released in the fission of one nucleus of `U(235)` and in how much time? (ii) Give an estimate of devastation potential of an atomic explosion. (iii) Should we ban nuclear research in this field ? Give your views briefly. |
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Answer» (i) form the fission of a single nucleus of `._(92)U^(235), 200 MeV` energy is released in `10^(-9)` second i.e., almost instantly. (ii) First atomic explosion occurred in 1945 when an atom bomb was dropped on the city of Hiroshima, japan. This resulted in the killing of about 66000 people and grievous injury to an equal number. `2//3` rd of city structures were smashed and so on. (iii) No, the reserch must go on. But we must investigate wasys and means for harnessing the enormous amount of nuclear energy for peacful purpose. Nuclear reactor is an outcome of that reseach. |
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| 229. |
It is found expertimentally that 13.6eV energy is required to separated a hydrogen atom into a proton and an electron. Compute the orbital radius and velocity of electron in a hydrogen atom. |
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Answer» Here, total energy of electron `E=-13.6eV=-13.6xx1.6xx10^(-19)J` `=-2.2xx10^(-18)J` form `E=(-e^2)/(8pi in_0r)` `r=(-e^2)/(2(4pi in_0r)E)=(-9xx10^(-19)xx(1.6xx10^(-19))^2)/(2(-2.2xx10^(-18)))` `=5.3xx10^(-11)m` velocity, `v=e/(sqrt(4pi in_0 mr))` Where `m=9.1xx10^(-31)kg` `:. v=(1.6xx10^(-19))/(sqrt((9.1xx10^(-31)xx5.3xx10^(-11))/(9xx10^9)))` `=2.2xx10^6m//s` |
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| 230. |
A neutron is absorbed by a `._3Li^6` nucleus with subsequent emission of an alpha particle. Write the corresponding nuclear reaction. Calculate the energy released in this reaction. `m(._3Li^6)=6.015126u, m(._2He^4)=4.0026044u` `m(._0n^1)=1.0086654u, m(._1He^3)=3.016049u` Take `1u=931MeV`. |
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Answer» The nuclear reaction is given by `._3Li^6+ ._0n^1 to ._2He^4+ ._1He^3` Mass defect, `Deltam=m(Li^6)+m(._0n^1)-m(He^4)-m(H^3)` `=6.015126+1.0086654-4.00026044-3.016049` `=7.0237914-7.0186534` `Delta m=0.005138u`. Energy realeased =`Delta m xx931MeV` `=0.005138xx931` `=4.783MeV` |
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| 231. |
Explain with the help of a nuclear reaction in each of the following cases, how the neutron to proton ratio change during (i) `alpha`- decay (ii) `beta`- decay. |
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Answer» (i) `alpha`decay is represented as `._92U^(238) to ._90Th^(234)+._2He^4` Initial ratio of `("neutron number") /("proton number")=(238-92)/92=1.586` Final ratio of `("neutron number") /("proton number")=(234-90)/90=1.6` Clearly, the ratio increases slightly in `alpha`-decay. (ii) `beta`- decay may be represented as `._15P^(32) to ._16S^(32)+._(-1)e^0+barv` Initial ratio of `("neutron number") /("proton number")=(32-15)/15=1.13` Final ratio of `("neutron number") /("proton number")=(32-16)/16=1.00` Clearly, the ratio decraeses slightly in `beta`-decay. |
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| 232. |
There is a stream of neutrons with kinetic energy of 0.0327 eV. If half life of neutrons is 700s, what fraction of neutrons will decay before they travel a distance of 10m? Take mass of neutron =`1.675xx10^(-27)kg.` |
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Answer» Here, `KE=0.0327eV=0.0327xx1.6xx10^(-19)J` `T=700s, s=10m, m=1.675xx10^(-27)kg` As `KE=1/2mv^2` `:. v=sqrt((2KE)/m)=sqrt((2xx0.0327xx1.6xx10^(-19))/(1.675xx10^(-27)))` `v=2.5xx10^3m//s` Time taken by these neutrons to travel s=10m is `t=s/v=10/(2.5xx10^3)=4xx10^(-3)s` form `N=N_0(1/2)^(t//T)` `N/(N_0)=(1/2)^(4xx10^-3//700)=0.999952` `:.` Fraction of neutrons decayed `=1-N/(N_0)=1-0.999952=0.000048` |
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| 233. |
Calculate the height of potential barrier for a head on collision of two deuterons. The effective radius of deuteron can be taken to be 2fm. Note that height of potential barrier is given by the Coulomb repulsion between two deuterons when they just touch eachother. |
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Answer» for head on collision, distance between centres of two deuterons=`r=2xxradius` `r=4fm=4xx10^(-15)m` charge of each deuteron `e=1.6xx10^(-19)C` Potential energy `=(e^2)/(4pi in_0r)=(9xx10^9(1.6xx10^(-19))^(2))/(4xx10^(-15))"joule"=(9xx1.6xx1.6xx10^(-14))/(4xx1.6xx10^(-16))KeV=360keV` As `P.E. =2xxK.E.` of each of deutron `=360keV` Coulomb barrier. `:.` K.E. of each deutron=`360/2=180keV` Thus is a measure of height of Coulomb barrier. |
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| 234. |
How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as `._1H^2+._1H^2to ._1H^3+n+3.17MeV` |
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Answer» Number of deuterium atoms in `2.0kg=(6.023xx10^(23)xx2000)/2=6.023xx10^(26)` Energy released when 2 atoms fuse `=3.27MeV` `:.` Total energy released `=3.27/2xx6.023xx10^(26)MeV=1.635xx6.023xx10^(26)xx1.6xx10^(-13)J` `=15.75xx10^(13)J` Energy consumed by the bulb/sec`=100J` `:.` Time for which bulb will glow `=(15.75xx10^(13))/100s=(15.75xx10^(11))/(60xx60xx24xx365)years=4.99xx10^4 "years"` |
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| 235. |
A helium atom consists of two electrons orbiting round a nucleus of charge Z=2. But the electrons do not see the full charge Z=2 of the nucleus. Each electron sees the nucleus slightly screened by the other electron so that the effective charge `Z_(eff)` seen by each electron is less than 2. The ionisation potential for a He atom in its ground state is measured experimentally to be 24.46 eV. Estimate the effective charge of the nucleus as seen by each electron in the helium ground state. |
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Answer» It is known that ground state energy of hydrogen atom =-13.6eV Energy of each electron in He ground state `=-Z_(eff)^2xx13.6eV` `:.` ground state energy of He atom `=2(-Z_(eff)^2)xx13.6eV=-Z_(eff)^2xx27.2eV` In `He^(+)` ground state, Z=2 and there is no screening because there is only one electron. `:.` Energy of `He^(+)` ground state `=-4xx13.6eV` `-5.44eV` `:.` Ionisation potential `=[-54.4+Z_(eff)^2xx27.2],` the exptal value of which is 24.46V `:. -54.4+Z_(eff)^2xx27.2=24.46V,` which gives `Z_(eff)=1.70` |
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