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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.
| 251. |
When a particle is restricted to move along x- axis between `x = 0 and x = 4 ` whwre a is opf nanometer demension , its energy can take only certain spscfic values . The allowed energies of the particles only in such a restiricted regain , correspond to the formation of standing wave with nodes at its end ` x = 0 and x = a `.The wavelength of this standing wave is related to the linear momentum p of the paarticle according to the de Broglie relation .The energy of the particle of mass `m` is reated to its linear momentum as `E = (p^(2))/(2m)` . thus , the energy of the particle can be denoted by a quantum number `n` taking value `1,2,3,....(n= 1, called the ground state)` corresponding to the number of loops in the standing wave use the model described above to answer the following there question for a particle moving in the line ` x = 0 to x = a Take h = 6.6 xx 10^(-34) J s and e = 1.6 xx 10^(-19)C` The speed of the particle , that can take discrete values, is propotional toA. `n^(-3//2)`B. `n^(-1)`C. `n^(1//2)`D. ``n` |
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Answer» Correct Answer - D lambda = (h)/(p) rArr lambda = (h)/(m nu) rArr m nu = (h)/(lambda)` `But (n lambda)/(2) = a rArr lambda = (2a)/(n)` `:. M nu = (nh)/(2a) rArr nu = (nh)/(2am) rArr nu prop n ` |
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| 252. |
Scienists are working hard to develop inclear fusion reactor Nocies of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+ energy` in the core of fasion reactor a gas of heavy hydrogen of `_(1)^(2) H` nucles and electrons is know as plasma . The nuclei move randonity in the reactor to take place Unally , the temperature in the reactor core are too ligh and to natrual will can be used to confine the to pleama for a time l_(0) before the particles by away from the case if `n` is the denasity (number volume ) of determines , the product` nt_(0) `is called Lavson number in one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following botczmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` in the cure of nucleus fusion reactor , the gas become plasma because ofA. strong nucleus force acting between the deuteronsB. coulomb force acting between the deuteronsC. coulmb force acting between deuteron - ecectron pairsD. the hight temperature maintained inside the reactor |
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Answer» Correct Answer - D The collection of `_(1)^(2) H` nuclei and is known as plasma which is formed due to high temperature inside the reactor core |
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| 253. |
Scienists are working hard to develop inclear fusion reactor Nocies of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+ energy` in the core of fasion reactor a gas of heavy hydrogen of `_(1)^(2) H` nucles and electrons is know as plasma . The nuclei move randonity in the reactor to take place Unally , the temperature in the reactor core are too ligh and to natrual will can be used to confine the to pleama for a time l_(0) before the particles by away from the case if `n` is the denasity (number volume ) of determines , the product` nt_(0) `is called Lavson number in one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following botczmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` Result of calulations for fopur different desine of a fasion reactor using `D-D` reaction are given below which of these is most promising based on Lawson crierion ?A. deuteron density `= 2.0 xx 10^(12) cm^(-3)`, confinement time `= 5.0 xx 10^(-3)s`B. deuteron density `= 8.0 xx 10^(14) cm^(-3)`, confinement time `= 9.0 xx 10^(-1)s`C. deuteron density `= 4.0 xx 10^(23) cm^(-3)`, confinement time `= 1.0 xx 10^(-11)s`D. deuteron density `= 1.0 xx 10^(24) cm^(-3)` confinement time `= 4.0 xx 10^(-12)s` |
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Answer» Correct Answer - B for the reading `B` get `nt_(0) gt 5 xx 10^(14) ` which is the Lawson criterion for a reactor to work successfully |
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| 254. |
Scienists are working hard to develop inclear fusion reactor Nocies of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+ energy` in the core of fasion reactor a gas of heavy hydrogen of `_(1)^(2) H` nucles and electrons is know as plasma . The nuclei move randonity in the reactor to take place Unally , the temperature in the reactor core are too ligh and to natrual will can be used to confine the to pleama for a time l_(0) before the particles by away from the case if `n` is the denasity (number volume ) of determines , the product` nt_(0) `is called Lavson number in one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following botczmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` Assume that two deuteron nuclei in the core of fasion reactor at temperacture energy `T` are moving toward each other, each with kinectic energy `1.5 kT` , whenn the seperation between them is large enogh to leglect coulomb potential energy . Also neglate any interaction from other particle in the core . The minimum temperature `T` required for them to reach a separation of `4 xx 10^(-15) m ` is in the rangeA. `1.0 xx 10^(9) K lt T lt 2. 0 xx 10^(9) K`B. `2.0 xx 10^(9) K lt T lt 3.0 xx 10^(9) K`C. `3.0 xx 10^(9) K lt T lt 4.0 xx 10^(9) K`D. `4.0 xx 10^(9) K lt T lt 5.0 xx 10^(9) K` |
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Answer» Correct Answer - A Appling conservation of machanical energy we get Loss of kinetic energy of two deuteron nuclei `= ` Gain in their potential energy `2 xx 1.5kT = (1)/(4 pi s_(0)) (e xx e)/®` `rArr 2 xx 1.5 xx(8.6 xx 10^(-5)(eV)/(k)) xx T = ((1.44 xx 10^(-9) eVm ))/(4 xx 10^(-15) m)` `rArr T = (1.44 xx 10^(-9))/(2 xx 1.5 xx 8.6 xx 10^(-5) xx 4 xx 10^(-15)) = 1.4 xx 10^(9)k` |
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| 255. |
A necear transformation is denoted by `X (n,a) _(3)^(7)Li` Which of the following is the neclues of electron X?A. `_(5)^(10) Be`B. `^(12)C_(6)`C. `_(4)^(11)Be`D. `_(5)^(9)B` |
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Answer» Correct Answer - A `_(z)X^(A) + _(0)n^(1) rarr _(3)Li^(7) + _(2)He^(4)` On comparison, `A = 7 + 4 - 1 = 10 , z = 3 + 2 - 0 = 5 ` it is boren `_(2)B^(10)` |
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| 256. |
Which one of the following atatement i9s `WRONG` in the context of X- rays generated from X- rays tube ?A. Wevelength of characteristic X- rays decrease when the atomic number of the target increaseB. cot - off wavelength of the contimous X - rays depends on the atomic number of the targetC. Intensity of the characteristic X -rays depend on the electrical power given to the X- rays tubeD. cut - off wavelength of the continous X- rays depends on the energy of the electrons in the X-rays tube |
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Answer» Correct Answer - B The continous spectrum depends in the accelerating voltage it has a definite minimum wavelength . Greater the accelerationg vollage for electron , bigher will be the kinetic energy if attains before striking the lorgest bigher will be the frequency of X- rays of continous will be the wavelength . The wavelength of continuous X- rays is indepndent of the atimic number of target matereial. |
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| 257. |
what should be the velocity of an electron so that its momentum becoumes equal to that of a photn of wavelength `5200 Å`A. 700 m/sB. 1000 m/sC. 1400 m/sD. 2800 m/s |
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Answer» Correct Answer - C `m_ev_e = (h)/(lambda)` `=(6.63xx 10^(-34))/((5200xx10^(-10)xx9.1xx10^(-31)` `~~1400 m//s.` |
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| 258. |
From the following equation pick out the possible nuclear fusion reactionsA. `6^(C^(13))+1^(H^1)rarr6^(C^(14))+4.3 MeV`B. `6^(C^(12))+1^(H^1)rarr7^(N^(13))+2 MeV`C. `7^(C^(14))+1^(H^1)rarr8^(O^(15))+7.3 MeV`D. `92^(U^(235))+0^(n^(1))rarr54^(Xe^(140))+36Sr^(94)+0^(n^(1))+0^(n^(1))+y+200 MeV` |
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Answer» Correct Answer - B::C |
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| 259. |
Assuming that about `20 M eV` of energy is released per fusion reaction `._(1)H^(2)+._(1)H^(3)rarr._(0)n^(1)+._(2)He^(4)`, the mass of `._(1)H^(2)` consumed per day in a future fusion reactor of powder `1 MW` would be approximatelyA. `0.001 g`B. `0.2 g`C. `10.0 g`D. `2 g` |
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Answer» Correct Answer - B |
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| 260. |
Consider a nuclear reaction : `Aoverset(lambda_(1))rarrB+C` and `Boverset(lambda_(2))rarrC` A converts into B and to C with decay with decay constant `lambda_(1)` B is alos stable nucleus which futher decays into stable nucleus C with decay constant `lambda_(2).` Mark the correct statement(s)A. `(dN_(C))/(dt)=-(lambda_(1)N_(A)+lambda_(2)N_(B))`B. `(dN_(A))/(dt)=-lambda_(1)N_(A)`C. `(dN_(A))/(dt)=(lambda_(1)N_(A)+lambda_(2)N_(B))`D. `(dN_(B))/(dt)=lambda_(1)N_(A)+lambda_(2)N_(B)` |
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Answer» Correct Answer - B |
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| 261. |
When a metal is illuminated with light of frequency f, the maximum kinetic energy of the photoelectrons is 1.2 eV. When the frequency is increased by 50% the maximum kinetic energy increases to 4.2 eV. What is the threshold frequency for this metal? |
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Answer» Correct Answer - A `K_(max) = E-W 1.2 = E-W …(i) 4.2 = 1.5 E - W …..(ii) Solving these equations, we get `w = 4.8 eV = hf_0` `:. f_0 = (4.8xx1.6xx10^(-19)/6.63xx10^(-34)` `=1.16xx10^15 Hz.` |
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| 262. |
Is it correct to say that `K_(max)` is proportional to f ? If not what would a correct statement of the relationship between `K_(max)` and f? |
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Answer» Correct Answer - A::B `K_(max) = E-W = hf - hf_0` `=h(f - f_0)` `:. K_(max) prop (f - f_0).` |
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| 263. |
A metal is illumimated by light of two different wavelength `248nm` and `310 nm` . The maximum speeds of the photoelecrtron corresponding in these wavwlength are ``u_(1) and u_(2)` respectively . If the ratio u_(1) : u_(2) = 2 : 1 ` and `hc = 1240 eVnm `, the work function of the inetial is rearlyA. `3.7 eV`B. `3.2 eV`C. `2.8eV`D. `2.5 eV` |
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Answer» Correct Answer - A `(hC)/(lambda_(1)) W = (1)/(2) mu_(1)^(2)` `and ((hC)/(lambda_(2)) W = (1)/(2) mu_(2)^(2)`` Dividing the above two equation , we get ` ((hC)/(lambda_(1)) - W)/((hC)/(lambda_(2)) - W) ((hC)/(lambda_(1)) - W)/((hC)/(lambda_(2)) - W) (u_(1)^(2)/(u_(2)^(2)` ` :. `((1240)/(248) -W)/((1240)/(310) -W = (4)/(1)` `:. (1240)/(248) -W = (4 xx 1240)/(310) - 4W` `:. W = 3.7 eV` |
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| 264. |
Consider the following energies (1) The minimum energy needed to excite a hydrogen atom from its ground state `-E_(1)` (2) Energy needed to ionze a hydrogen atom from ground state `=E_(2)` (3) Energy released in `overset(235).U-"fission "=E_(3)` (4) Energy needed to remove a neutron from a `overset(12).C_("cucleus")=E_(4)`A. `E_(1)ltE_(2)ltE_(3)ltE_(4)`B. `E_(1)ltE_(3)ltE_(2)ltE_(4)`C. `E_(1)ltE_(2)ltE_(4)ltE_(3)`D. `E_(2)ltE_(1)ltE_(4)ltE_(3)` |
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Answer» Correct Answer - C |
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| 265. |
Electrons with energy `80 keV` are incdent on the tungsten target of an X - rays tube , k- shell electrons of tangsten have `72.5 keV` energy X- rays emitted by the tube constain onlyA. a contimuous X - rays spectrum (Bremasstrablung) with a miximum wavelength of `0.155Å`B. a contimuous X - rays spectrum (Bremasstrablung) with all wavelengthC. the characteristic X - rays spectrum of tungsten`D. a contimuous X - rays spectrum (Bremasstrablung) with a miximum wavelength of `0.155Å` and the characteristic X - rays spectrum of tungsten. |
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Answer» Correct Answer - D KEY CONCEPT : `lambda_(min) = (he)/(E )` `:.lambda_(min) = (12400)/(80 xx 10^(3)) Å = 0.155 Å` Energy of incident electrons is greater than the ionization energy of electron in `K- shell`, the `K- shell` electrons will be knocked off . Hence, characteristic X - ray spectrum will be obtained |
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| 266. |
Electrons with energy `80 keV` are incident on the tungsten target of an X - rays tube , k- shell electrons of tungsten have `72.5 keV` energy X- rays emitted by the tube contain onlyA. a continuous X-ray spectrum with a minimum wavelength of `0.155 Å`B. a continuous X-ray spectrum with all wavelengthsC. a characteristic X-ray spectrum of tungstenD. a continuous X-ray spectrum with a minimum wavelength of `0.155 Å` and a characteristic X-ray spectrum of tungsten |
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Answer» Correct Answer - D |
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| 267. |
In a sample of radioactive material, what fraction of initial number of active nuclei will remain undistintegrated after half of a half0life of the sample?A. `(1)/(4)`B. `(1)/(2sqrt(2))`C. `(1)/(sqrt(2))`D. `sqrt(2)-1` |
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Answer» Correct Answer - c |
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| 268. |
Let `T` be the mean life of a radioactive sample. `75%` of the active nuclei present in th sample initially will deacy in timeA. 2TB. `(1)/(2)(In 2)T`C. 4TD. `2(In2)T` |
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Answer» Correct Answer - d |
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| 269. |
Find energy released in the alpha decay, Given `_92^238Urarr_90^234Th+_2^4He` `M(`_92^238U)=238.050784u` `M(`_90^234Th)=234.043593u` `M(`_2^4He)=4.002602u` |
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Answer» Correct Answer - B::D Mass defect `summ_1-summ_f=Deltam` `=(238.050784)-(234.043593+4.002602)` `=4.589xx10^-3u` Energy released `=Deltamxx931.48MeV` `=4.27 MeV` |
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| 270. |
Imagine an atom made up of a proton and a bypothetical particle of double the mass of the electron but having the same charge as the electron . Apply the Bohr atom model and consider all possible transitions of this bypothrtical particle that will be emited level . The longest wavwlength photon that will be emidet has longest wavwlength `lambda` (given in terms of the bytherg constant `R` for the hydrogen aton) equal toA. `9(5R)`B. `36//(5R)`C. `18//(5R)`D. `4//(5R)` |
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Answer» Correct Answer - C KEY CONCEPT : `lambda prop (1)/(m)` For ordinary hydrogen atom, wavwlength `(1)/(lambda) = R [(1)/(2^(2)) - (1)/(2^(2))] = (5 R)/(36) ` or `lambda(36)/(5R)` with hypothencal particle , required wavwlength `lambda^(1) = (1)/(2) xx (36)/(5R) = (18)/(5R)` |
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| 271. |
An imaginary particle has a charge equal to that of an electron and mass 100 times tha mass of the electron. It moves in a circular orbit around a nucleus of charge + 4 e. Take the mass of the nucleus to be infinite. Assuming that the Bhor model is applicable to this system. (a)Derive an experssion for the radius of nth Bhor orbit. (b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second orbit. |
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Answer» (a) we have `(m_q v^2)/(r_n) = (1)/(4pi epsilon_0) (Ze^2`)/(r_n^2`) …(i) The quantization of angular momentum gives `m_pur_n = (nh)/(2pi) …(ii)` Solving Eqs. (i) and (ii), we get `r = (n^2h^2epsilon_0)/(Zpim_pe^2)` `Substituting `m_p = 100m` where, m = mass of electorn and Z= 4 we get `r_n = (n^2 h^2 epsilon_0)/(400pi me^2)` (b) As we know, `E_1^H = - 13.60 eV` and `E_n prop ((Z^2)/(n^2)) m` For the given particle,`E_4 = (-13.60)(4)^2 )/((4))^2 xx100` and E_2 = (-13.60)(4)^2/((2))^2` xx100 `=- 5440eV` `DeltaE=E_4-E_2` `=4080 eV` `:. lambda(in Å) = (12375)/(DeltaE(in eV))` `=(12375)/(4080)` `=3.0 Å` |
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| 272. |
Let the potential energy of the hydrogen atom in the ground state be zero . Then its energy in the excited state will beA. 10.2eVB. `13.6 eV`C. `23.8 eV`D. `27.2 eV` |
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Answer» Correct Answer - a |
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| 273. |
What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?A. 10.2 eVB. 13.6 eVC. 23.8 eVD. 27.2 eV |
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Answer» Correct Answer - C Otherwise `U_1 =- 27.2 eV.` Therefore, we have increased it by 27.2 eV. It implies that we have increased it by 27.2 eV in all states. `U_2 = -6.8 eV` `:.U_2 = (-6.8+27.2) eV = +20.4 eV` `E_2 = U_2+K_2 =(20.4 +3.4) eV` = 23.8 ev. |
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| 274. |
Find the binding anergy of an electron in the ground state of a hydrogen like atom in whose spectrum the thrid Balmer line is equal to 108.5 mm.A. `54.4 eV`B. `13.6 eV`C. `112.4 eV`D. None of these |
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Answer» Correct Answer - A `Delta E = (12375)/(1085) = 11.4 eV` Third Balmer line is corresponding to the transition, n =5 to n =2 `E_5 - E_2 =11.4` `:. (E_1)/((5))^2 - (E_1)/((2))^2 11.4` `E_1=-54.28 eV` `:. |E_1| = 54.28 eV.` |
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| 275. |
The longest wavelength of the Lyman series for hydrogen atom is the same as the wavelength of a certain line in the spectrum of `He^+` when the electron makes a trensiton from `n rarr 2`. The value of n isA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - B Longest wavelength of Lyman series means, minimum energy corresponding n =2 to n=1. `:. (E_2 - E_1)_H = (E_n - E_2)_(He+)` `:. (-13.6)/((2))^2 +(13.6)/((1))^2 = (-13.6(Z)^2/(n^2 + (13.6(Z)^2/((2))^2` Putting Z =2, we get n =4. |
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| 276. |
Two identical photo-cathodes receive light of frequencies `v_1` and `v_2`. If the velocities of the photoelectrons (of mass m) coming out are `v_1` and `v_2` respectively, thenA. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`B. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)`C. `v_1- v_2 = [((2h)/(m))(v_1-v_2))]^(1//2)`D. `v_1^2 - v_2^2=(2h)/(m) (v_1 - v_2)` |
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Answer» Correct Answer - B `(1)/(2) mv_1^2=hv_1 - W` `(1)/(2) mv_2^2 = hv_2 -W` From these two equation, we can see that `V_1^2 -v_2^2= (2h)/(m) (v_1 - v_2)` |
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| 277. |
The frequency of the first line in Lyman series in the hydrogen spectrum is v. What is the frequency of the corresponding line in the spectrum of doubly ionized Lithium?A. vB. 3 vC. 9 vD. 2 v |
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Answer» Correct Answer - C Frequency `prop`energy and energy `prop Z^2`. |
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| 278. |
Which enrgy state of doubly ionized lithium `(Li^(++)` has the same energy as that of the gorund state of hydrogen?A. n=1B. n=2C. n =3D. n=4 |
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Answer» Correct Answer - C `E prop (Z^2)/(n^2)` `:. (Z)^2/(n^2) =1` or n = Z =3. |
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| 279. |
As energy of `24.6 eV` is required to remove one of the required to remove both the electrons from a nuture brfore alon isA. `38.2`B. `94.2`C. `51.8`D. `79.0` |
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Answer» Correct Answer - D When one `e^(bar)` is removed from neutral helium atom , it becames a one `e^(bar)` speciel . For one `e^(bar)`speciel .we know `E_(n) = (- 13.62^(2))/(n^(2)) ev//atom` For helium ion ,`Z = 2` and for first arbit `n = 1` `:. E_(1) = (-13.6)/((1)^(2)) xx 2^(2) = - 54.4 eV` `:. ` Energy required to remove this `e^(bar) = 54.4 + 24.6 = 79 eV` |
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| 280. |
The de-Broglie wavelength of electron in gound state of an hydrogen atom isA. `1.06 Å`B. `1.52 Å`C. `0.53 Å`D. `3.33 Å` |
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Answer» Correct Answer - D |
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| 281. |
The binding energy of an electron in the gorund state of He atom is equal to `E_0 = 24.6 eV.` Find the energy required to remove both electrons form the atom. |
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Answer» Energy required to remove first electron is 24.6 eV. After removing first electron form this atom, it will become `He^+` `E_1 =- (13.6)(2)^2 (as E prop Z^2 and Z=2)` = - 54.4 eV `:. Energy required to remove this second electron will be 54.4 eV.` `:. Total energy required to remove both electrons` =24.6 +54.4 =79 eV. |
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| 282. |
what is the ratio of de-Broglie wavelength of electron in the second and third Bohr orbits in the hydrogen atoms?A. `2//3`B. `3//2`C. `4//3`D. `3//4` |
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Answer» Correct Answer - A `(2pir_2) = 2lambda_2` (2pir_3) = 3lambda_3` `:. (lambda_2)/(lambda_3) = (3r_2)/(2r_3)` Now, `r prop n^2` `:. (r^2)/(r^3) = ((2)/(3))^2` Substituting in Eq. (i), we get `lambda_2)/(lambda_3) = (2)/(3)` |
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| 283. |
Ratio of the de Broglie wavelenght of molecules of helium and hydrogen at temperatur `27^(@)` C and `327^(@)` C respectively isA. 2B. 1C. 3D. 4 |
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Answer» Correct Answer - B |
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| 284. |
In the Bohr model of the hydrogen atom, what is the de-Broglie wavelength for the electron when it is in (a) the n=1 level? (b) Then n=4 level? In each case, compare the de-Broglie wavelength to the circumference `2pir_n` of the orbit. |
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Answer» Correct Answer - A::B::C (a) `E_1 = -13.6 eV `:. KE = |E_1| = 13.6 eV` `lambda_1 (in Å`) = sqrt((150)/(KE(in eV)) for an electron` `=sqrt(150)/(13.6) =3.32 Å` `2pir_1 = 2pi(0.529 Å) ~~ 3.32 Å` (b) `E_4 =(E_1)/((4))^2 = (-13.6)/((4))^2 = - 0.85 eV` `lambda_4 = sqrt((150)/(0.85)) as KE = |E_4| = 0.85 eV` `~~13.3 Å `2pir_4 = 2pi(n)^2 r_1 (as r prop n^2)` `=53.15 Å = 4lambda_4.` |
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| 285. |
Positron is the antiparticle of an electron .It has the same mass as an electron but the opposite charge An electron and a positron moving towards each other with equal and opposite velocities.A. can annihilate into one photon. Conserving both energy and momentumB. cannot annihilate into one photon because energy cannot be conservedC. cannot annihilate into one photon because momentum cannot be conservedD. cannot annihilate into one photon because charge cannot be conserved |
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Answer» Correct Answer - C |
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| 286. |
As per Bohr model , the minimum energy (in eV) required to remove electron from the ground state of doubly ioinized `Li` alom `(Z = 3)` is |
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Answer» Correct Answer - D For hydrogen and hydrogen like atoms, `E_n = -13.6((Z)^2)/((n^2)))` Therefore, ground state energy of doubly ionized lithium atom (Z =3, n =1) will be `E_1 = (-13.6) ((3)^2/((1))^2` =- 122.4 eV `:.` lonization energy of an electron in ground state of doubly ionized lithium atom will be 122.4 eV. |
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| 287. |
The de Broglie wavelength of an electron moving with a velocity of `1.5xx10^(8)ms^(-1)` is equal to that of a photon find the ratio of the kinetic energy of the photon to that of the electron.A. 2B. 4C. `(1)/(2)`D. `(1)/(4)` |
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Answer» Correct Answer - D |
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| 288. |
The activity of a sample reduces from `A_(0) "to" (A_(0))/(sqrt(3))` in one hour. The activity after f3 hours more will beA. `(A_(0))/(3sqrt(3))`B. `(A_(0))/(9)`C. `(A_(0))/(9sqrt(3))`D. `(A_(0))/(27)` |
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Answer» Correct Answer - B |
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| 289. |
Let `A_(n)` be the area enclosed by the `n^(th)` orbit in a hydrogen atom. The graph of `l n (A_(n)//A_(t))` against In `(n)`A. will pass through originB. will be a straight line with slpe 4C. will be a parabolaD. will be a circle |
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Answer» Correct Answer - A::B |
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| 290. |
Let `A_(0)` be the area enclined by the orbit in a hydrogen atom .The graph of in `(A_(0) //A_(1))` against `ln(n)`A. Will not pass through originB. Will be a straight line with slope 4C. will be rectangular hyperbolaD. Will be parabola |
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Answer» Correct Answer - B `A = pir^2` `A prop r^2 or A prop n^4 (as r prop n^2)` `:. (A_n)/(A_1) = ((n))^4` `in ((A_n)/(A_1)) = 4 in (n)` Therefore, in ((A_n)/(A_1)) versus in n graph is a straight line of slop 4. |
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| 291. |
In the hydrogen atom, an electron makes a transition from n=2 to n=1. The magnetic field produced by the circulating electron at the nucleusA. Decreases 16 timesB. Increases 4 timesC. Decreases 4 timesD. Increases 32 times |
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Answer» Correct Answer - D `B = (mu_0i)/(2pi)` or `B prop (i)/(r )` See the hint of Q.No 4 of same section. `iprop (1)/(n^3) and r prop n^2` `:. B =(1)/(n^5)` `:. (B_1)/(B_2) = ((n_2)/(n_1))^5 = ((2))^5 = 32` |
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| 292. |
An electron in a hydrogen atom makes a transiton from first excited state ot ground state. The magnetic moment due to circulating electronA. 21B. 10C. 15D. None of these |
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Answer» Correct Answer - B `(M)/(L) = (q)/(2m) = constant` `M prop L` and `L = (nh)/(2pi) or L prop n` `:. M prop n.` |
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| 293. |
The excitation energy of a hydrogen -like ion in its first excited state is `40.8 eV` Find the energy needed to remain the electron from the ionA. will not pass through originB. will be a straight line with slope 4C. will be a rectangular hyperbolaD. will be a parabola |
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Answer» Correct Answer - A `E_2 - E_1 = 40.8 eV` `:. (E_1)/((2))^2 E_1= 40.8 eV` or ` -(3)/(4) E_1 = 40.8 eV` or `E_1= -54.4 eV` |E_1| = 54.4 eV` |
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| 294. |
Intensity and frequency of incident light both are doubled. Then, what is the effect on stopping potential and saturation current. |
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Answer» By increasing the frequency of incident light energy of incident light will increase. So, maximum kinetic energy of photoelectrons will also increase. Hence, stopping potential will increase. Further, by doubling the frequency of incident light energy of each photon will be doubled. So, intensity itself becomes two times without increasing number of photons incident per unit are per unit time. Therefore, saturation current will remain unchanged. |
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| 295. |
When a metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential is `5 V_0`. When the same surface is illuminated with the light of wavelength `3lambda`, the stopping potential is `V_0`. Then, the work function of the metallic surface isA. `hc//6lambda`B. `hc//5lambda`C. `hc//4lambda`D. `2hc//4lambda` |
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Answer» Correct Answer - A `e(5V_0) = (hc)/(lambda) -W….(i)` `eV_0 = (hc)/(3lambda) -W …(ii)` Solving these equations, we get `W =(hc)/(6lambda).` |
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| 296. |
When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 V and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then (a) the stopping potential will be 0.2 V (b) the stopping potential will be 0.6 V (c ) the saturation current will be 6.0 mA (d) the saturation current will be 2.0 mA |
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Answer» Correct Answer - A::B::C::D Stopping potential depends on two factors - one the energy of incident light and the other the work - function of the metal. By increasing the distance of source from the cell, neither of the two change. Therefore, stopping potential remains the same. (d) Saturation current is directly proportional to the intensity of light incident on cell and for a point source, intensity ` `lprop 1//r^2` When distance is increased form 0.2 m to 0.6 m (three times). the instensity and hence the saturation current will decrease 9 times, i.e. the saturation current will be reduced to 2.0 mA. `:.` The correct option are (b) and (d). |
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| 297. |
A non-monochromatic light is used in an experiment on photoelectric effect. The stopping potentialA. is related to mean wavelengthB. is not related to shortest wavelengthC. is related to the maximum K.E. of emitted photoelectronD. intensity of incident light |
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Answer» Correct Answer - C |
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| 298. |
When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic surface. This phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons depends on the wavelength of incident light and is independent of the intensity of light. Number of emitted photoelectrons depends on intensity. `(hv-phi)` is the maximum kinetic energy of emitted photoelectron (where `phi` is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference across the tuve. Wavelength of charasteristic X-ray depends on the atomic number. Q. If frequency `(upsilongtupsilon_0)` of incident light becomes n times the initial frequency (v), then KE of the emitted photoelectrons becomes (`v_0` threshold frequency).A. n times to the initial kinetic energyB. more than n times to the initial K.E.C. less than n times to initial K.E.D. K.E. of emitted photoelectrons remain unchanged |
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Answer» Correct Answer - A |
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| 299. |
The binding energies per nucleon for deuteron (`_1H^2`) and helium (`_2He^4) are 1.1 MeV and 7.0 MeV respectively. The energy released when two deuterons fuse to form a helium nucleus (`_2He^4`) is…….. |
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Answer» Correct Answer - B::C `2_1H^2rarr_2He^4` Binding energy of two deuterons, `E_1=2[2xx1.1]=4.4 MeV` Binding energy of helium nucleus, `E_2=4(7.0)=28.0 MeV` `:.` Energy released `Delta E=E_2-E_1` `=(28-4.4)MeV=23.6 MeV` |
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| 300. |
`_(87)^(223) `Ra is a radioactive substance having half life of `4` days Find the problability that a nucleas undergoes after two half livesA. `1`B. `(1)/(2)`C. `(3)/(4)`D. `(1)/(4)` |
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Answer» Correct Answer - C For a nucleus to disintegrate in two half life , the probability is `(3)/(4) ` as `75% `of the nucleus will disintegrate in this time |
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