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.
| 351. |
Figure shows a hypothetical binding energy per nucleon (B.E/A); versus (mass number) A curve. Which of the following statements is correct?(1) Fusion of nuclei, having mass number in the range 20 < A < 50, will release energy (2) Fission of nuclei, having mass number in the range 50 < A < 100, will release energy (3) Fusion of nuclei, having mass number in the range 60 < A < 100, will release energy (4) Fission of nuclei, having mass number in the range 125 < A < 175, will release energy, if the two fragments are identical |
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Answer» Correct option : (3) Explanation : Energy is released in a nuclear (fission or fusion) reaction, if the binding energy per nucleon of the products, is more than that for reactants. This is so only far statement (3). In all the other statements, B.E/nucleon, of products, is less than that of the reactants. In such a process energy, has to be absorbed. |
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| 352. |
What is unified atomic mass unit? Express it in J/c2 and MeV/c2 . |
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Answer» The unified atomic mass unit is an accepted but non-SI unit of mass. It is defined to be equal to 1/12 of the mass of a free atom of the isotope of carbon with mass number 12 which is at rest and in its ground state. It is denoted by u. Its value in SI unit is obtained experimentally. 1 u = 1.660538782 (83) × 10-27 kg, with the standard uncertainty in the last two digits given in v parenthesis. Taking, 1 u = 1.660538782 × 10-27 kg c = 2.99792458 × 108 m/s, e = 1.602176462 × 10-19 C and using the relation E = me2 { ≡ m ≡ E / c2 ), we get, 1 u = 1.49241783 × 10-10 J/c2 ≅ 1.492 × 10-10 J/c2 and 1 u = 931.494042 MeV / c2 ≅ 931.5 MeV/c2 |
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| 353. |
The half-life of a radioactive material is 4 days. Find the time required for 1/4 of the initial number of radioactive nuclei of the element to remain undisintegrated. |
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Answer» For t = nT, N = N0/ 2n. In this case, n = 2. ∴ f = 2T = 2 × 4 = 8 days is the required time. |
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| 354. |
State the nature and properties of β-particles. |
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Answer» Nature of β-particles : A β-particle is an electron or a positron. Properties of β-particles : 1. β-particles have a moderate ionizing power. It is about 100 times less than that of α-particles, but 100 times more than that of γ-rays. 2. They have a moderate penetrating power. It is about 100 times more than that of α-particles, but 100 times less than that of γ-rays. 3. They are deflected by electric and magnetic fields. Their deflection is more than the deflection of α-particles in the same field but in the opposite direction. 4. They affect photographic plates. 5. They cause fluorescence in a fluorescent material such as zinc sulphide. 6. Their energies and speeds are very high. Their speed is of the order of 10 m/s. Some β-particles have speeds of the order of 0.99 c, where c is the speed of light in free space. 7. They cause more biological damage than α-particles, because they have more penetrating power. [Note : When a nucleus emits an electron, one of its neutrons changes to a proton; the electron is accompanied by a neutral and almost massless particle called antineutrino \(\vec v_e\) with which the electron shares its energy and momentum. Hence, β-particles are emitted with speeds ranging from about 0 to 0.99 c The properties of a positron are identical to those of an electron except that it carries a positive charge of the same magnitude as an electron. A positron emission in a β+ decay is accompanied by a neutrino ve. A positron is an anti-particle of an electron, and an antineutrino is an antiparticle of neutrino. The existence of a small neutral particle, emitted simultaneously with the electron in β -decay, was proposed in 1931 by Wolfgang Pauli (1900 -1958), Austrian-US theoretical physicist. It was confirmed experimentally in 1956 by Frederic Reines and Clyde Lorrain Cowan, Jr., US physicists. This neutral particle, that appears in β+ -decay, was called the neutrino. It travels with a speed very close to that of light in free space. The existence of positron (in 1928) and other antiparticles was predicted by Paul Adrien Maurice Dirac (1902 – 84), British theoretical physicist. All these predictions were eventually confirmed experimentally; the positron was discovered in 1932 by Carl David Anderson, US physicist.] |
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| 355. |
In one mean lifetime of a radioactive element, the fraction of the nuclei that has disintegrated is [e is the base of natral logarithm.](A) \(\cfrac1e\)(B) 1 – \(\cfrac1e\)(C) e (D) e – 1. |
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Answer» Correct option is (B) 1 – \(\cfrac1e\) |
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| 356. |
State the nature and properties of γ-rays. |
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Answer» Nature of γ-rays : 1. γ-rays are electromagnetic waves of very short wavelength (about 10-12 m to 10-14 m). 2. They are uncharged. Properties of γ-rays : 1. γ-rays produce feeble ionization. Their ionizing power is 104 times less than that of a-particles and 100 times less than that of β-particles. 2. They have the maximum penetrating power. It is about 100 times that of βparticles and 104 times that of αparticles. 3. They are not deflected by electric and magnetic fields as they are uncharged. 4. They affect photographic plates. 5. They cause fluorescence in a fluorescent material such as zinc sulphide. 6. Their speed in free space is 3 × 108 m/s( the same as that of light waves and X-rays in free space). 7. γ-rays can be diffracted by crystals. In recent times, γ-ray diffraction has emerged as a powerful tool in structural and defect studies of crystals. 8. They destroy living cells and tissues and are used for destroying cancer cells. [Note : γ-rays (not established clearly in Berquerel’s work) were discovered in 1900 by Paul Villard (1860-1934), French physicist.] |
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| 357. |
Why are α- and β-particle emissions often accompanied by γ-rays? |
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Answer» A given nucleus does not emit α- and βparticles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α- and βparticle emissions are often accompanied by γ-rays. |
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| 358. |
(a) What is α-decay? What is the consequence of an α-decay on a radioactive element? What is the Q value or Q factor in this case ?Q = [mU – mTh – mα]c2(b) What is β-decay ? What is the consequence of a β-decay on a radioactive element? What is the Q value or Q factor in this case ? |
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Answer» (a) A radioactive transformation in which an aparticle is emitted is called α-decay. In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4. Example : \(^{38}_{92}U\) → \(^{234}_{90}Th\) + \(^4_2α\) Q = [mU – mTh – mα]c2 (b) A radioactive transformation in which a βparticle is emitted is called β-decay. In a β -decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged. Example : \(^{23}_{90}Th\) → \(^{234}_{91}Pa\) + \(^0_{-1}e\) + \(\bar v_e\) where \(\bar v_e\) is the antineutrino emitted to conserve the momentum, energy and spin. Q = [mTh – mpa – me]c2 In a β+ -decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged. Example : \(^{30}_{15}P\) → \(^{30}_{14}Si\) + \(^0_{+1}e\) + ve where ve is the neutrino emitted to conserve the momentum, energy and spin. Q = [mp – mSi – me]c2 [Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.] |
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| 359. |
Define decay constant or distintegration constant of a radioactive element. If λ is the decay constant of a radioactive element, show that about 37% of the original nuclei remains undecayed after a time interval of λ-1. |
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Answer» The decay constant or disintegration constant of a radioactive element is defined as the ratio of the disintegration rate at an instant to the number of undecayed nuclei of the element present at that instant. Let N0 be the number of nuclei of a radioactive element present at time t = 0 and N, the number of undecayed nuclei at time t. From the radioactive law, N = N0e-λt where λ is the decay constant. At t = λ-1, the fraction of undecayed nuclei is \(\cfrac{N}{N_0}\) = e-λ × λ-1 = e-1 = \(\cfrac1e\) Since e ≅ 2.718, \(\cfrac{N}{N_0}\) = \(\cfrac1{2.718}\) = 0.3679 Therefore, about 36.79% ≈ 37% of the original nuclei remains undecayed after a time λ-1 Since λ is the probability that a nucleus of the element will decay in one second, λ-1 gives the mean-life or the mean life time τ of the radioactive element measured in second; τ = λ-1. |
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| 360. |
In successive radioactive decay if the decrease in mass number is 32 and the decrease in atomic number is 8, how many (i) α particles (ii) β particles are emitted in the process ? |
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Answer» 1. Number of α-particles emitted = 32/4 = 8 2. Number of β-particles emitted = 16 – 8 = 8. |
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| 361. |
Calculate the nuclear radii of `._(56)Ba^(140)` and `._(8)O^(17)`. Given `R_(0) = 1.5 fm`. |
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Answer» Correct Answer - 7.888 fm, 3.857 fm |
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| 362. |
Find the nuclear radii of 206Pb and 208Pb. |
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Answer» Data : R0 = 1.2 × 10-15 m Nuclear radius, R=R0A1/3 (i) For 206Pb, A = 206 ∴ R = (1.2 × 10-15) (206)1/3 = 1.2 × 10-15 × 5.906 = 7.087 × 10-15 m = 7.087 fm (ii) For 208Pb, A = 208 ∴ R = (1.2 × 10-15)(208)1/3 = 1.2 × 10-15 × 5.925 = 7.11 × 10-15 m = 7.11 fm |
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| 363. |
The wavelength of `K_(a)` , X-rays for lead isotopes `Pb^(208),Pb^(206),Pb^(204)" are"lambda_(1),lambda_(2)" and"lambda_(3),` resperctively. ThenA. `lambda_(1)=lambda_(2)gtlambda_(3)`B. `lambda_(1)gtlambda_(2)gtlambda_(3)`C. `lambda_(1)ltlambda_(2)ltlambda_(3)`D. `lambda_(2)=sqrt(lambda_(1)lambda_(3))` |
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Answer» Correct Answer - D Since, atomic numer are same `lambda_(1)=lambda_(2)=lambda_(3)` ` or " " lambda_(2)=sqrt(lambda_(1)lambda_(3))` |
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| 364. |
Number of spectral lines in hydrogen atom isA. 8B. 6C. 15D. `oo` |
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Answer» Correct Answer - D Number of spectrl lines in hydrogen atom is `oo` |
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| 365. |
The energy of an element with a vacancy in K-shell is 35.2 V, in L-shell is 5.25 keV and in M-shell is 0.55 keV higher than the energy of the atom of with no vacancy. Find the frequency of `K_(alpha),K_(beta) and L_(alpha)` X-rays of that element. |
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Answer» It is given that, `E_(K)=35.2keV,E_(L)=5.25keV,E_(M)=0.55 keV` `rArr" " E_(K)-E_(L)=(hc)/(lambda_(K_(A_(alpha))))=hv_(K_(alpha))` `rArr(35.2-5.25)xx10^(3)xx1.6xx10^(-19)=6.6xx10^(-34)v_(K_(alpha))` `rArr" " v_(K_(alpha))=(29.95xx10^(3)xx1.6xx10^(-19))/(6.6xx10^(-34))=7.2xx10^(-34)Hz` Similarly, `E_(K)-E_(M)=hv_(K_(beta))` `rArr" " v_(K_(beta))=((35.2-0.55)xx10^(3)xx1.63xx10^(-19))/(6.6xx10^(-34))=8.4xx10^(18)Hz` Also, `E_(L)-E_(M)=hv_(k_(alpha))` `rArr" " ((5.25-0.55)xx10^(3)xx1.6xx10^(-19))/(6.6xx10^(-34))=1.1xx10^(18)Hz` |
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| 366. |
The wavelength of `K_(alpha)` X-ray line for an element is `0.42Å` . Find the wavelength of `K_(beta)` line emitted by the same element. |
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Answer» The characterstic wavelength for particular element is given by `(1)/(lambda)=R(Z-b)^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` For `K_(alpha`)line `(1)/(lambda_(alpha))=(Z-b)^(2)R((1)/(1^(2))-(1)/(n_(2)^(2)))=(Z-b)^(2)R((3)/(4))` For `K_(beta)` line, `(1)/(lambda)=(Z-b)^(2)((1)/(1^(2))-(1)/(3^(2)))=(Z-b)^(2)R((8)/(9))` `rArr(lambda_(beta))/(lambda_(alpha))=((3//4))/((8//9))=(27)/(32)` It is given that `lambda_(alpha)` is `0.42Å` `:." " lambda_(beta)=0.42xx(27)/(32)=0.35Å` |
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| 367. |
The wavelength of `K_(alpha)` line in copper is `1.54 Å`. The ionisation energy of `K` electron in copper in Joule is |
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Answer» For `K_(alpha)`-line we can write, `(1)/(lambda)=R(Z-1)^(2)((1)/(1^(2))-(1)/(2^(2)))=R(Z-1)^(2)xx(3)/(4)rArrR(Z-1)^(2)=(4)/(3lambda_(alpha))` Ionisation energy of given by `E=Rhc(Z-1)^(2)((1)/(1^(2))-(1)/(oo))=hc(4)/(3lambda_(alpha))` `=(6.6xx10^(-34)xx3xx10^(8)xx4)/(3xx1.54xx10^(-10))=1.7xx10^(-15)J` |
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| 368. |
The wavelength of `K_(a)` line copper is `1.5Å`. The ionisation energy of K electron in copper isA. `11.2xx10^(-17)J`B. `12.9xx10^(-16)J`C. `1.7xx10^(-15)J`D. `10xx10^(-16)J` |
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Answer» Correct Answer - B Required inisation energy , `E=(hc)/(lambda)` `(=6.6xx10^(-34)xx3xx10^(8))/(1.5xx10^(-10))=12.9xx10^(-16)J` |
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| 369. |
Determine the wavelength of the first Lymanline,the transition from `n=2"to"n=1`.In what region of the electromagnetic spectrum does this line lie ? |
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Answer» Strategy : We use , `hf=E_u-E-l`, with the energies obtained from energy level diagram to find the energy and the wavelength of the transition. The region of the electromagnetic spectrum is found using the electromagnetic spectrum. In this case , `hf=E_2-E_1` ={-3.4 eV -(-13.6 eV)} =10.2 eV `=(10.2 eV) (1.60xx10^(-19) J//eV)=1.63xx10^(-18)` J Since `lambda`=c/f, we have `lambda=c/f="hc"/(E_2-E_1)` `=((6.63xx10^(-34) J.s)(3.00xx10^8 m//s))/(1.63xx10^(-18)J)` `=1.22xx10^(-7)` m or 122 nm , which is in the UV region of the electromagnetic spectrum `lambda=1/((2.44xx10^6 m^(-1))=4.10xx10^(-7)`m or 410 nm. This is the fourth line in the Balmer series and is violet in color. |
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| 370. |
A sphere of work function `phi=4.6 eV` is suspended in a vacuum number by an insulating thread. Radiation of wavelength `lambda =0.2 mum` stricke on the sphere. The maximum electric pontential of the sphere will beA. 4.6 VB. 6.2 VC. 1.6 VD. 3.2 V |
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Answer» Correct Answer - A `E_("Photon")=(12400)/(lambda(Å))=(12400)/(2000)=6.2 eV` `KE_("max")=E_("Photon")-phi=1.6 eV` Maximum potential of sphere =1.6 eV |
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| 371. |
The wavelength of first line of Balmer series is `6563Å`. The wavelength of first line of Lyman series will beA. `1215.4Å`B. `2500Å`C. `7500Å`D. `600Å` |
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Answer» Correct Answer - A `(lambda_("Lyman"))/(lambda_("Blamer"))=(((1)/2^(2)-(1)/(3^(2))))/(((1)/(1^(2))-(1)/(2^(2))))=(5)/(27)" " [:. (1)/(lambda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))]` `rArrlambda_("Lyman")=(5)/(27)xxlambda_("Balmer")=(5)/(27)xx6563=1215.4Å` |
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| 372. |
If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie wavelength of the incident electrons to the shortest wavelength X-ray produced? `((e)/(m)"for electron"=1.8xx10^(11)Ckg^(-1))`A. `0.1`B. `0.2`C. `0.3`D. `1.0` |
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Answer» Correct Answer - A For incident electron ,`(1)/(2)mv^(2)=eV` or `" " p^(2)=2meV` `:.` de-Broglie wavelength , `lambda_(1)=(h)/(p)=(h)/(sqrt(2mev))` Shortest X-rays wavelength, `lambda_(2)=(hc)/(eV)` `:. " " (lamda_(1))/(lambda_(2))=(1)/(c)sqrt(((V)/(2))((e)/(m)))` `" " =sqrt((10^(4))/(2)xx1.8xx10^(11))/(3xx10^(8))=0.1` |
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| 373. |
The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum isA. `1//2`B. `2/137`C. `1//137`D. `1//237` |
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Answer» Correct Answer - C `v_(n)=(Ze^(2))/(ehepsi_(0))(1)/(n)` The velocity of electron in the first orbit (n=1) of H-atom (Z-1) is `" " v_(1)=(e^(2))/(2hepsi_(0))=(e)/(137)` `:." " (v_(1))/(c)=(1)/(137)` |
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| 374. |
The electron of a hydrogen atom revolves the proton in a circuit nth of radius `r_(0) = (in_(0) n^(2)h^(2))/(pi m e^(2)) ` with a speed `upsilon_(0) =(e^(2))/(2 in_(0) nh)` The current the to circulating charge is proportional toA. `e^(2)`B. `e^(3)`C. `e^(5)`D. `e^(6)` |
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Answer» Correct Answer - C Time period of electron , `T=(2mpir)/(v)=((2pixxepsi_(0)n^(2)h^(2))/(pime^(2)))/((e^(2))/(2epsi_(0)nh))` `T=(4epsi_(0^(2))n^(3)h^(3))/(me^(4))` Current, `i=(e)/(T)=(e)/((4epsi_(0)^(2)n^(3)h^(3))/(me^(4)))rArri=(me^(5))/(4epsi_(0)^(2)n^(3)h^(3))i.e., iprope^(5)` |
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| 375. |
The second line in Paschen series is obtained when the electron makes transition fromA. fourth orbit to third orbitB. seventh orbit to third orbitC. six orbit to third orbitD. fifth orbit to third orbit |
| Answer» Correct Answer - D | |
| 376. |
The balmer series for the H-atom can be ob-servedA. if we measure the frequencies of light emitted when an excited atom falls to the ground stateB. if we measure the frequencies of light emitted due to transitions between excited states and the first excited stateC. in any transition in a H-atomD. as a sequence of frequencies with the higher frequencies getting closely packed |
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Answer» Correct Answer - B::D Balmer series for the H-atom can be observed if we measure the frequencies of light emitted due to transitions between higher excited states and the first excited state and as a sequence of frequencies with the higher frequencies getting closely packed. |
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| 377. |
Radius `(r_(n))` of electron in nth orbit versus atomic number (Z) graph isA. B. C. D. |
| Answer» Correct Answer - B | |
| 378. |
Let `E=(-1me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))` be the energy of the `n^(th)` level of H-atom state and radiation of frequency `(E_(2)-E_(1))//h` falls on it ,A. it will not be absored at allB. some of atoms will move to the first excited stateC. all atoms wil be excited to the `n = 2` stateD. no atoms will make a transition to the `n = 3` state |
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Answer» Correct Answer - B::D When all the H-atoms are in the ground state and radiation of frequency `((E_(2)- E_(1)))/(h)` falls on it, some of atoms will move to the first excited state and no atoms will make a transition to the `n = 3` state. |
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| 379. |
In which of the following systems will the radius of the first orbit `(n = 1)` be minimum ?A. Hydrogen atomB. Deutrium atomC. Singly ionised heliumD. Doubly ionised lithium |
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Answer» Correct Answer - D `r_(n)prop(n^(2))/(Z)or r_(n)prop(1)/(Z)` |
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| 380. |
An electron moves in a circular orbit at a distance from a proton with kinetic energy E to escape to infinity, the minimum energy which must be supplied to the electron isA. EB. 2EC. 0.5ED. `Esqrt(2)` |
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Answer» Correct Answer - A If kinetic energy is E, then totla enery will be -E i.e., energy E must be supplied to the electron. |
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| 381. |
In which of the following systems will the radius of the first orbit `(n = 1)` be minimum ?A. doubly ionized lithiumB. singly ionized heliumC. deuterium atomD. hydrogen atom |
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Answer» Correct Answer - A Radius of first orbit , `r prop(1)/(Z)`, For doubly ionized lithium, `Z(=3)` will be maximum, hence for doubly ioized lithium, r will be minimum. |
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| 382. |
Let `E=(-1me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))` be the energy of the `n^(th)` level of H-atom state and radiation of frequency `(E_(2)-E_(1))//h` falls on it ,A. it will not be absorbed at allB. some of atoms will move to the first excited stateC. all atoms will be excited to the `n=2` stateD. all atoms will make a transition to the `n=3` state |
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Answer» Correct Answer - B The given energy of `n^(th)` level of hydrogen atom is `E_n=-(me^4)/(8epsilon_0^2n^2h^2)` Since all the H-atom are in ground state `(n=1)` then the radiation of given freqency `(E_2-E_1)/(h)` falling on it may be absorbed by some of the atoms and move them to the first excited state `(n=2)`. |
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| 383. |
What is a nuclear reaction? Give one example. |
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Answer» A reaction between the nucleus of an atom and a bombarding particle leading to the production of a new nucleus and, in general, the ejection of one or more particles is known as a nuclear reaction. Example : \(^4_2\alpha\) + \(^{27}_{13}AI\) \(\longrightarrow\) \(^{30}_{15}P\) + \(^4_0n\) + Q Here, \(^4_2\alpha\) is the bombarding particle, \(^{27}_{13}AI\) is the target nucleus, \(^{30}_{15}P\) is the product nucleus \(^4_0n\) is the energy released in the reaction. It is carried (usually) by the ejected particle. |
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| 384. |
What are the quantities conserved in a nuclear reaction? |
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Answer» The total momentum, energy, spin, charge and number of nucleons are conserved in a nuclear reaction. |
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| 385. |
Find the maximum frequency, the X-ray emitted by an X-ray tube operating at `30 kV` |
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Answer» For maximum frequency , the total kinetic energy (in eV) should be converted into an X-ray photon. Thus, hf=eV `f="eV"/h=(1.6xx10^(-19)xx30xx10^3)/(6.63xx10^(-34))` `=7.2xx10^8` Hz |
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| 386. |
The current flowing through the X-ray tube which is operating at 25 kV is 1 mA. Calculate the number of electrons hitting target per second . Also, find the energy falling on the target per second. |
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Answer» Current flowing in X-ray tube =1m A Potential difference =25 kV Current (I) is given by I="ne" where, n =number of electrons hitting the target per second. `:. " "n=(I)/(e)=(1xx10^(-3))/(1.6xx10^(-6))=6.25xx10^(15)` Now, the energy falling on the target per second is equal to the power consumed by the X-ray tube i.e.,P=VI Here, `V=24 kV=25xx10^(3)V` `I=1mA=1xx10^(-3)A` `:. " " P=(25xx10^(3))(1xx10^(-3))=25W` |
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| 387. |
Name the two quantised conditions proposed by Bohr in the atom model. |
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Answer» Bohr proposed : (i) quantised of energy states related to the transition of electrons from one orbit to another. (ii) quantisation of orbit or angular momentum. |
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| 388. |
The formula of washing soda is .............. , while that of sodium sulphate is ............ |
| Answer» `Na_(2)CO_(3)*10H_(2)O, Na_(2)SO_(4)` | |
| 389. |
Which of the following is a tetraatomic molecule ?A. `O_(2)`B. `O_(3)`C. `NO_(2)`D. `SO_(3)` |
| Answer» Correct Answer - D | |
| 390. |
The number of atoms present in a molecule of a substance is calledA. molecularityB. atomicityC. valencyD. reactivity |
| Answer» Correct Answer - B | |
| 391. |
Which of the following is the correct formula for sodium phosphate ? `Na_(3)PO_(4), Na(PO_(4))_(3),NaHPO_(4),Na_(2)(PO_(4))_(3)` |
| Answer» `Na_(3)PO_(4)` | |
| 392. |
Name the instrument which can be used to view the atoms of an element . |
| Answer» Electron tunneling microscope (ETM) | |
| 393. |
Mass of one molecule of ammonia isA. `(1)/(6.022xx 10^(23))`B. (34)/(6.022xx10^(23))`C. `17)/(6.022xx10^(23))`D. none of these |
| Answer» Correct Answer - C | |
| 394. |
The valency of nitrogen in ammonia `(NH_(3))` isA. 2B. 0C. 3D. 4 |
| Answer» Correct Answer - C | |
| 395. |
What is the ratio by number of atoms in a molecules of ammonia ? |
| Answer» N atom=1,H atom=3,ratio 1:3 | |
| 396. |
Which of the following chemical formulae is (are) correct ?A. `NaPO_(4)`B. `BiCl_(2)`C. `MgCl`D. `Ca_(3)(PO_(4))_(2)` |
| Answer» Correct Answer - D | |
| 397. |
Which of the following does contain maximum number of molecules ?A. 1g `SO_(2)`B. 1 g ` CO_(2)`C. 1 g`CH_(4)`D. 1 g `N_(2)` |
| Answer» Correct Answer - C | |
| 398. |
The indivisibility of atom was proposed byA. RutherfordB. DaltonC. BohrD. Einstein |
| Answer» Correct Answer - B | |
| 399. |
If `""^12C` is used as standard , the relative mass of an atom of sodium is 23 and the relative mass of an atom of phosphorus is 31 . Find the relative masses of a samples of sodium and phosphorus with respect to each other when each contains `3.0 xx 10^(25)` atoms . |
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Answer» The relative masses of samples that contain equal numbers of atoms will be the same as the relative masses of individual atoms , thus , `("mass of the sample of sodium")/("mass of the sample of phosphorus") = (23)/(31)` . |
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| 400. |
Which of the following statement is (are) incorrect about an atom?A. An atom can exist independently .B. Atoms of elements combine to form molecules.C. Matter is made up of molecules, not atoms .D. An atom is either positively or negatively charged. |
| Answer» Correct Answer - A::C::D | |