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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.
| 51. |
The de-Broglie wavelength of an electron in the first Bohr orbit isA. equal to one-fourth the circumference of the first orbitB. equal to half the circumference of the first orbitC. equal to twice the circumference of the first orbitD. equal to the circumference of the first orbit |
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Answer» Correct Answer - D |
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| 52. |
Hydrogen atoms are excited from ground state of the principle quantum number `4`. Then the number of spectral lines observed will beA. `3`B. `6`C. `5`D. `2` |
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Answer» Correct Answer - B Number of spectial lines `N_(E) = (n(n - 1))/(2) = (4(4 - 1))/(2) = 6` |
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| 53. |
To explain his theory, Bohr usedA. conservation of linear momentumB. conservation of angular momentumC. conservation of quantum frequencyD. conservation of energy |
| Answer» Correct Answer - B | |
| 54. |
Which of the following statements are true?A. The shortest wavelength of X-rays emitted from an X-ray tube depends on the corrent in the tube.B. Characteristic X-ray spectra is simple as compared to optical spectra.C. X-ray cannot be different by means of an ordianary grating.D. Three exist a sharp limit on the short wavelength side for each continous X-ray spectrum |
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Answer» Correct Answer - B::C::D Statement (a) is false. The shortest wavelength of the X-ray emitted depend on the energy of the electrons incident on the target . This , in turn , depends on the potential through which they have fallen . In fact `lambda_(min) = (hc)/(eV)` Statement (b) is true. X-ray spectra of all heavy elements are similar in character. Statement ( c) is also true. The shortest wavelength of the X-rays `("compared to the greating constant of optical grating")` makes it difficult to observe X-ray differaction with ordianery gratings. Statement (d) is also true. The sharp limit on the short wavelength side is dependent on the voltage applied to the incident electron and is given by : `lambda_(min) = (hc)/(eV)` |
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| 55. |
Minimum energy required to takeout the only one electron from ground state of `He^(+)` isA. `13.6 eV`B. `54.4 eV`C. `27.2 eV`D. `6.8 eV` |
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Answer» Correct Answer - B `E_(n) = - (13.6z^(2))/(n^(2)) eV rArr E_(1) = -(13.6xx(2)^(2))/((1)^(1)) =- 54.4 eV` |
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| 56. |
A narrow electron beam passes undeviated through an electric field `E = 3 xx 10^(4) "volt"//m` and an overlapping magnetic field `B = 2 xx 10^(-3) Weber//m^(2)`. If electric field and magnetic field are mutually perpendicular. The speed of the electron isA. `60 m//s`B. `10.3 xx 10^(7) m//s`C. `1.5 xx 10^(7)m//s`D. `0.67 xx 10^(7) m//s` |
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Answer» Correct Answer - B If elctron beam passes undeviated then magnetic force should nalance electirc force `v = (E)/(B) = (3 xx 10^(4))/(2 xx 10^(-3)) = 1.5 xx 10^(7) m//sec`. |
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| 57. |
A cathode ray tube contains a pair of parallel metal plates `1.0 cm` apart and `3.0 cm` long. A narrow horizontal beam of electron with a velocity of `3 xx 10^(7) ms^(-1)` is passed down the tube midway betwwen the two plates. When a potential difference of `550 V` is maintained across the plates, it is found that the electron beam is so deflected that it just strikes the end of one of the plates. then the specific change of an electron (that is, the ratio of its change to mass) in `C//kg` is :A. `3.6 xx 10^(-14)C//kg`B. `1.8 xx 10^(-11) C//kg`C. `3.6 xx 10^(-12)C//kg`D. `1.8 xx 10^(-9)C//kg` |
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Answer» Correct Answer - B Here `y = (1)/(2) cm = 0.5 xx 10^(-2)m` `l = 3 cm = 3 xx 10^(-2)m` and `v = 3 xx 10^(7) ms^(-1)` Potential difference, `V = 550` volts `l = vt` or `t = (1)/(v)` `:. Y = (1)/(2) at^(2) = (1)/(2) ((eE)/(m)) (t^(2))/(v^(2)) = (1)/(2)e ((V)/(dm)) xx (l^(2))/(v^(2))` `0.5 xx 10^(-2) = (1)/(2) ((e)/(m)) ((550)/(1 xx 10^(-2))) xx ((3 xx 10^(2))/(3 xx 10^(7)))` `(e)/(m) = (2 xx 0.5 xx 10^(-2) xx 1 xx 10^(-2) xx 9 xx 10^(14))/(550 xx (3 xx 10^(-2))^(2))` `= 1.8 xx 10^(11) C//kg` |
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| 58. |
The angular momentum of the `alpha`- particles which are scattered through large angle by the heavier nuclei, is conserved because of theA. nature of repulsive forceB. conservation of kinetic energyC. conservation of potential energyD. there is no external torque |
| Answer» Correct Answer - D | |
| 59. |
An atom emits a spectral line of wavelength `lambda` when an electron makes a transition between levels of energy `E_(1) and E_(2)`. Which expression correctly relates `lambda E_(1) and E_(2)` ?A. `lambda = (h c)/(E_(1) + E_(2))`B. `lambda = (2 h c)/(E_(1) + E_(2))`C. `lambda = (2 h c)/(E_(1) - E_(2))`D. `lambda = (h c)/(E_(1) - E_(2))` |
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Answer» Correct Answer - D By quantum thoery of radiation, the energy change `delta E` between energy level is proportional to the frequency of electromagnetic radiation `f` and is given by. `Delta E = h f = (h c)/)lambda)` `Hence, `lambda = (h c)/(Delta E) = (h c)/(E_(1) - E_(2))` |
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| 60. |
Transition between three energy energy levels in a particular atom give rise to three Spectral line of wevelength , in increasing magnitudes. `lambda_(1), lambda_(2)` and `lambda_(3)`. Which one of the following equations correctly ralates `lambda_(1), lambda_(2)` and `lambda_(3)`? |
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Answer» `(1)/(lambda_(1)) = R_oo Z^(2) [(1)/(2^(2)) - (1)/(3^(2))]` (i) `(1)/(lambda_(2)) = R_oo Z^(2) [(1)/(1^(2)) - (1)/(2^(2))]` (ii) `(1)/(lambda_(3)) = R_oo Z^(2) [(1)/(1^(2)) - (1)/(3^(2))]` (iii) On adding (i) and (ii), we get `(1)/(lambda_(1)) + (1)/(lambda_(2)) = R_oo Z^(2) [(1)/(1^(2)) - (1)/(3^(2))]` Thus,`(1)/(lambda_(1)) + (1)/(lambda_(2)) = (1)/(lambda_(3))` `lambda_(3) = (lambda_(1) lambda_(2))/(lambda_(1) + lambda_(2))` |
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| 61. |
A hydrogen atom in a having a binding of `0.85eV`makes transition to a state with excited energy `10.2eV`(a) identify the quantum number n of theupper and the lower energy state involved in the transition (b) Find the wavelength of the emitted radiation |
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Answer» a. Let `n_(1)` be the initial state of electron, Then. `E_(1) = - (13.6)/(n_(1)^(2)) e V` Here, `E_(1) = - 0.85 e V` `:. - 0.85 = - (13.6)/(n_(1)^(2))` or `n_(1) = 4` b. Let `n_(2)` be the final excitation state of electron. Since excitation energy is always measured with respect to the ground state, therefore `Delta E = 13.6 [1 - (1)/(n_(2)^(2))]` here `Delta E = 10.2 e V` `:. 10.2 = 13.6 [1 - (1)/(n_(2)^(2))]` or `n_(2) = 2` Thus the electron jumps from `n_(2) = 4` to `n_(2) = 2` c. The wavelength of the photon emitted for a transition between `n_(2) = 4` to `n_(2) = 2` is given by `(1)/(lambda) = R_(oo) [(1)/(n_(2)^(2)) - (1)/(n_(1)^(2))]` `(1)/(lambda) = 1.097 xx 10^(7) [(1)/(2^(2)) - (1)/(4^(2))]` `lambda = 4860 Å` |
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| 62. |
If an electrons jumps from `m^(th)` orbit to the nth orbit `(mgtn)` the energy of the atom changes from `E_(m)` . This extra energy `E_(m)-E_(n)` is emitted as a photon whose wavelength is given by `1/lambda=RZ^(2)(1/n^(2)-1/m^(2))` Where `R=1.09xx10^(7)m^(-1)` (Rydberg constant) A photon ejected from the transition of electron from m^(th) excited state of `He^(+)` ion to `n^(th)` state is allowed to fall on a photoelectric material with work function,`phi=7eV`. [Given `h=4.14xx10^(-15)eVs` and `c=3xx10^(8)ms_(1)]` The photo -electric effect will take place for the transition of electron fromA. `3^(rd)` orbit to `2^(nd)`B. `4^(th)` orbit to `3^(rd)` orbitC. `5^(th)` orbit to `3^(rd)` orbitD. Photoelectron can not ejected |
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Answer» Correct Answer - A `therefore1/lambda=1.09xx10^(7)xx4(1/4-1/3) Rightarrow lambda=165nm` ` Rightarrow phi =7eV=(hc)/lambda_(0)=(hc)/phi=177nm` ` therefore lambda lt lambda_(0)` Photoelectric effect will occure |
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| 63. |
In a Rutherford scattering experiment when a projectile of change `Z_(1)` and mass `M_(1)` approaches s target nucleus of change `Z_(2)` and mass `M_(2)`, te distance of closed approach is `r_(0)`. The energy of the projectile isA. directly proportional to `M_(1)xxM_(2)`B. directly proportional to `Z_(1)Z_(2)`C. inversely proportional to `Z_(1)`D. Directly proportional to mass `M_(1)` |
| Answer» Correct Answer - B | |
| 64. |
An ionised H-molecules consists of an electron and wo protons. The protons are seperated by a small distance of the order of angstrom. In the ground state,A. the electron would not move in circular orbits.B. the energy would be `(2)^(2)` time that’s of a H atomC. the electrons, orbit would go around the pro-tons .D. the molecules will soon decay in a protons and a H-atom |
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Answer» Correct Answer - A::C In ionized hydrogen molecules, the net force acting on the electron due to two and proton and is not cen-tral so , the electron would not move in circular or -bit and electron orbits would go around two pro-tons |
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| 65. |
Two hydrogen-like atoms `A` and `B` are of different masses and each atom contains equal numbers of protons and neutrons. The difference in the energies between the first Balmer lines emitted by `A and B` , is `5.667 e V`. When atom atoms `A and B` moving with the same velocity , strike a heavy target , they rebound with the same velocity in the process, atom `B` imparts twice the momentum to the target than that `A` imparts. Identify the atom `A and B`. |
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Answer» Let `Z_(A) and Z_(B)` be the atomic numbers and `m_(A) and m_(B)` be the mass numbers of hydrogen-like atoms `A and B`, respectively . Energy of nth state of hydrogen-like atoms is `E_(n) = - (Z^(2) Rhc)/(n^(2)) = - (Z^(2) xx 13.6)/(n^(2)) e V` Energy emitted for 1 line of Balmer series for atom`A`. ` Delta E_(1) = - Z_(A)^(2) xx 13.6 ((1)/(2^(2)) - (1)/(3^(2))) e V` Energy emitted for 1 line of Balmer series for atom`B`. ` Delta E_(2) = Z_(B)^(2) xx 13.6 ((1)/(2^(2)) - (1)/(3^(2))) e V` Given ` Delta E_(1) - Delta E_(2) = 5.667 e V, therefore `5.667 e V = (Z_(B)^(2) - Z_(A)^(2)) xx 13.6 ((1)/(2^(2)) - (1)/(3^(2))) e V` `(Z_(B)^(2) - Z_(A)^(2)) = (5.667 xx 36)/(13.6 xx 5) = 3` (i) Let `u` be the intial velocities of each from `A` and `B` , respectively, then according to principal convervation of momentum for `A` `m_(A) u = M v_(1) - m_(A) u` or `M v_(1) - 2 m_(A) u` (ii) Similarly, for `B M v_(2) = 2 m_(B) u` (iii) Given , `M v_(2) = 2 mv_(1)`, therefore. `2 m_(B) u = 2 (2 m_(A) u)` This gives , ` `m_(B) = 2 m_(A)` (iv) As number of protons and neutrons in each of `A and B` same separately . `:. m_(B) = 2 Z_(B)` and `m_(A) = 2 Z_(A)` (v) Substituting this in (iv) , we get `2 Z_(B) = 2 (2 Z_(A)), i.e., Z_(B) = 2 Z_(A)`, Solving Eq. (i) and Eq. (v), we get `Z_(A) = 1` and `Z_(B) = 2` i.e., atom `B` is singly ionized helium. |
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| 66. |
Consider aimimg a beam of free electrons to wards free atoms . When they scatter, an electron and a protons cannot combine be produced a H-atom,A. beacause of energy conservationB. with simulataneously releasing energy in the form or radiation .C. because of momentum conservationD. because of angular momentum conservation . |
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Answer» Correct Answer - A::B When a beam of gree electron is aimed to towards free protons, the electron get `5t` scattered on ac-count of energy conservation . An electron and a proton can combine to produce ga H-atom only if they simulataneously relase energy in the form of radiations. |
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| 67. |
Two hydrogen-like atoms `A` and `B` are of different masses and each atom contains equal numbers of protons and neutrons. The difference in the energies between the first Balmer lines emitted by `A and B` , is `5.667 e V`. When atom atoms `A and B` moving with the same velocity , strike a heavy target , they rebound with the same velocity in the process, atom `B` imparts twice the momentum to the target than that `A` imparts. Identify the atom `A and B`.A. `12.1 eV`B. `13.6 eV`C. `14.3 eV`D. `15.1 eV` |
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Answer» Correct Answer - D `Delta E = { 13.6 Z_(B)^(2) ((1)/(1^(2)) - (1)/(2^(2))) - 13.6 Z_(A)^(2) ((1)/(1^(2)) - (1)/(2^(2)))} eV` `81.6 eV = (13.6 xx3)/(4) (Z_(B)^(2) - Z_(A)^(2)) = 8` (i) Using conservation of momentum. For `A, m_(A) u = MV_(1) - m_(A) (u)/(2)` `(3)/(2)m_(A) u = MV_(1)` For `B, (3)/(2)m_(B) u = MV_(2)` But `MV_(2) = 3 MV_(1)` `m_(B) - 3m_(A)` Since both `A and B` carry same number of proton and neutron , we have `Z_(B) = 3Z_(A)` But `Z_(B)^(2) - Z_(A)^(2) = 8` `9 Z_(A)^(2), Z_(BN)^(2) = 8` `Z_(A) = 1, Z_(B) = 3` Hence ,`A` is `_(1)^(2) H` and `B` is `_(4)^(6)Li` Now the difference in energy between the first Balmer lines emitted by `A and b` `Delta E = 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z_(B)^(2) - 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z_(A)^(2)` `= 13.6 xx (5)/(36) xx (Z_(B)^(2) - Z_(A)^(2))` `= 13.6 xx 5 xx (8)/(36) = 15.1 eV` |
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| 68. |
Two hydrogen-like atoms `A` and `B` are of different masses and each atom contains equal numbers of protons and neutrons. The difference in the energies between the first Balmer lines emitted by `A and B` , is `5.667 e V`. When atom atoms `A and B` moving with the same velocity , strike a heavy target , they rebound with the same velocity in the process, atom `B` imparts twice the momentum to the target than that `A` imparts. Identify the atom `A and B`.A. `_(1)^(1) H`B. `_(1)^(2) H`C. `_(3)^(6) Li`D. `_(2)^(4) Li` |
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Answer» Correct Answer - B `Delta E = { 13.6 Z_(B)^(2) ((1)/(1^(2)) - (1)/(2^(2))) - 13.6 Z_(A)^(2) ((1)/(1^(2)) - (1)/(2^(2)))} eV` `81.6 eV = (13.6 xx3)/(4) (Z_(B)^(2) - Z_(A)^(2)) = 8` (i) Using conservation of momentum. For `A, m_(A) u = MV_(1) - m_(A) (u)/(2)` `(3)/(2)m_(A) u = MV_(1)` For `B, (3)/(2)m_(B) u = MV_(2)` But `MV_(2) = 3 MV_(1)` `m_(B) - 3m_(A)` Since both `A and B` carry same number of proton and neutron , we have `Z_(B) = 3Z_(A)` But `Z_(B)^(2) - Z_(A)^(2) = 8` `9 Z_(A)^(2), Z_(BN)^(2) = 8` `Z_(A) = 1, Z_(B) = 3` Hence ,`A` is `_(1)^(2) H` and `B` is `_(4)^(6)Li` Now the difference in energy between the first Balmer lines emitted by `A and b` `Delta E = 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z_(B)^(2) - 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z_(A)^(2)` `= 13.6 xx (5)/(36) xx (Z_(B)^(2) - Z_(A)^(2))` `= 13.6 xx 5 xx (8)/(36) = 15.1 eV` |
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| 69. |
Which of the following products, in a hydrogen atom , are independent of the principal quantum number `n` ? The symbels have their usual meanings.A. `omega^(2) r`B. `(E)/(v^(2))`C. `v^(2) r`D. `(E)/( c)` |
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Answer» Correct Answer - B::C `v prop (1)/(n), E prop (1)/(n^(2)), and r prop n^(2)` |
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| 70. |
The force acting on the electron in a hydrogen atom depends on the principal quantum number asA. `F prop n^(2)`B. `F prop (1)/(n^(2))`C. `F prop n^(4)`D. `F prop (1)/(n^(4))` |
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Answer» Correct Answer - D `F = (m v^(2))/( r)` But `v prop (1)/(n) and r prop n^(2)` `implies F prop (1)/(n^(4))` |
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| 71. |
The force acting on the electron in a hydrogen atom depends on the principal quantum number asA. `Fprop1//n^(2)`B. `Fprop1//n^(4)`C. `Fprop1//n^(5)`D. Does not depend on n+ |
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Answer» Correct Answer - B |
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| 72. |
An electron is not deflected on passing through a certain region , becauseA. there is magnetic field in that region and the electron enters into it in any direction .B. there may be magnetic field but the velocity of electron may be parallel to the direction of magnetic fieldC. electron is an charge-less particleD. there is electric field and the electron enters into it in any direction |
| Answer» Correct Answer - B | |
| 73. |
In Millikan oil drop experiment, a charged drop of mass `1.8 xx 10^(-14) kg` is stationary between its plates. The distance between its plates is `0.90 cm` and potential difference is `2.0` kilo volts. The number of electrons on the drop isA. `500`B. `50`C. `5`D. `0` |
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Answer» Correct Answer - C `QE = mg rArr Q = (mg)/(E) rArr n = (mgd)/(Ve)` `rArr n = (1.8 xx 10^(-14) xx 10 xx 0.9 xx 10^(-2))/(2 xx 10^(3) xx 1.6 xx 10^(-19)) = 5` |
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| 74. |
An electron starts from rest and travels `0.9`m in an electric field of `200`V//m. After this , it enters a magnetic field at right angle to its direction of motion . If the radius of circular path of the electron is `9`cm, the magnetic field induction is (Given e=`1.6xx10^(-19)`C,` m=9xx10^(-31)`g) `A. `5xx10^9-4wb//m^(2)`B. `5xx10^(-5)wb//m^(2)`C. `5xx10^(-3)wb//m^(2)`D. `5xx10^(-2)wb//m^(2)` |
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Answer» Correct Answer - A `v^(2)-U^(2)=2as,V^(2)=2(Eq)/mSRightarrowV=sqrt(2(Eq)/mS)r=(mV)/(Bq)RightarrowB=(mV)/(rq)` |
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| 75. |
An electron enters perpendicular to a uniform magnetic field with a speed o `10^(8) cm//s`. The particle ecperiences a force due to the magnetic field and the speed of the electronA. will decreaseB. will increaseC. will remain constantD. may increase or decrease |
| Answer» Correct Answer - C | |
| 76. |
When subjected to a transverse electric field, cathode rays moveA. down the potential gradientB. up the potential gradientC. along a hyperbolic pathD. along a circular path |
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Answer» Correct Answer - B In an electirc field, a force opposite to the direction of electric field acts on negatively charged particles (i.e, from lower potential to higher potential). |
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| 77. |
A charged particle enters a region of uniform magnetic field at an angle of `85^@` to the magnetic line of force. The path of the particle is a circle. Is this statement true or false?A. circularB. ellipticalC. spiralD. a straight line |
| Answer» Correct Answer - C | |
| 78. |
Cathode rays are made to pass between the poles of a magnet as shown in figure. The effect of magnetic field is A. to deflect them toward the south poleB. to deflect them perpendicular to the plane of the paper and toward the observerC. to deflect them toward the norht poleD. to increase the velocity of the rays |
| Answer» Correct Answer - B | |
| 79. |
A well collimated parallel pencil of cathode rays falls through a potential difference `3kV` & enters the spacing between two parallel metallic plates, parallel to their length the spacing between the plates being `0.5cm`. The pencil strikes a fluorescent screen, mounted perpendicular to the length of the plates at the farther end of the plates & produces fluorescent spot. if now a potential difference of 3V is applied across the two plates, calculate the linear deflection of the spot on the screen. Given the length of the plates is 10 cm. |
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Answer» Correct Answer - `[0.05cm]` |
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| 80. |
A charged particle of charge `Q` and mass`m` moves with velocity `v` in a circular path due to transverse magnetic field, B, then its frequency isA. `(QB)/(2pim`B. `(QvB)/(2pim)`C. `(QmvB)/(2pi)`D. `(vb)/(2piQm)` |
| Answer» Correct Answer - A | |
| 81. |
Cathode rays enter into a unifrom magnetic field perpendicular to the direction of the field. In the magnetic field their path will beA. a parabolaB. a circleC. a straight lineD. an ellipse |
| Answer» Correct Answer - A | |
| 82. |
A particle carrying a charge moves perpendicular to a uniform magnetic field of induction `B` with a momentum `p` then the radius of the circular path isA. `Be//p`B. `pe//B`C. `p//Be`D. `Bep` |
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Answer» Correct Answer - C `P=mv`,`r=(mv)/(Be)` |
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| 83. |
If the shorts series limit of the balmer series for hydrogen is `3644 Å`, find the atomic number of the element which gives X-ray wavelength down to `1 Å`. Identify the element. |
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Answer» The short series limit of the Balmar series is corresponding to transition `n = oo to n = 1` which is given by `(1)/(lambda) = R((1)/(2^(2)) - (1)/(oo^(2))) = ( R)/4` or `R = (4)/(lambda) = (4)/(3644) (Å) ^(-1)` The short wavelength corresponds to transition from `n = oo to n = 1` which is given as `(1)/(lambda) = R (Z- 1)^(2) [(1)/(1^(2)) - (1)/(oo^(2))]` or `(Z- 1)^(2) = (1)/(lambda_( c) R) = (1)/(1 Å xx (4)/(3644) (Å)^(-1)) = (3644)/(4) = 911` or `Z - 1 = 30.2` or `Z = 31.2 = 31`. Thus, the atomic number of the element is `31` which is gallium. |
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| 84. |
A material whose `K` -absorption edge is `0.2 Å` is irradiated by X-ray of wavelength `3644 Å` , find the maximum energy of the photoelectrons that are emitted from the `K` shell. |
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Answer» The binding energy for `K` shell in e V is `E_(k) = (hc)/(lambda_(k)) = (12431)/(0.2) e V = 62.155 keV` The energy of the inclined photon in `eV` is `E = (hc)/(lambda) = (12431)/(0.15) = 82.873 keV` Therefore , the maximum energy of the photoelectrons emitted from the `K` shell is `E_(max) = E - E_(K) = (82.873 - 62.155) keV` `= 20.718` |
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| 85. |
Figure shows the enegry levels `P, Q, R, S` and `G` of an atom where `G` is the ground state. A red line in the emission spectrum of the atom can be obtaned by an energy level change from `Q` so `S`. A blue line can be obtained by following energy level change A. `P` to `Q`B. `Q` to `R`C. `R` to `S`D. `R` to `G` |
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Answer» Correct Answer - D If `E` is the energy radiated in transition then `E_(R rarr G) gt E_(Q rarr S) gt E_(R rarr S) gt E_(Q rarr R) gt E_(P rarr Q)` For getting blue line energy radiated should be maximum `(E prop (1)/(lambda))`. Hence (d) is the correct option. |
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| 86. |
For a cartain metal, the `K` obsorption edge is at `0.72 Å`. The wavelength of `K_(alpha), K_(beta). And K_(gamma)` lines of of `K` series are `0.210 Å , 0.192 Å, and 0.180 Å, respectively. The eneggies of `K, L and M` orbit are `E_(K), E_(L) and E_(M)`, respectively. ThenA. `E_(K) = - 13.04 keV`B. `E_(L) = - 7.52 keV`C. `E_(M) = - 3.21 keV`D. `E_(K) = - 13.04 keV` |
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Answer» Correct Answer - A::B::C Energy of `K` absorption edge `E_(K) = (1242 eV nm)/(0.0172 nm) = 72.21 xx 10^(3) eV = 72.21 KeV` Energy of `K_(alpha)` line is `E_(K_alpha) = (hc)/(e lambda_(alpha)) = (1242 eV nm)/(0.021 nm) = 59.14 KeV` Similarly, `E_(K_beta) = (1242)/(0.0192) = 64.69 KeV` `E_(K_gamma) = (1242)/(0.0180) = 69 KeV` Energy of `K` shell ` = (E_(K_alpha) - E_(K))` `= (59.14 - 72.21) KeV = - 13.04 KeV` Energy of `L` shell ` = E_(K_beta) - 72.21 KeV` `= 64.69 KeV - 72.21 KeV = - 7.52 KeV` Energy of `M` shell ` = (E_(K_gamma) - E_(K)) = (1242 eV nm)/(0.018 nm) - 72.21 KeV` `= 69 KeV - 72.21KeV = - 3.21 KeV` |
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| 87. |
Figure shows the enegry levels `P, Q, R, S` and `G` of an atom where `G` is the ground state. A red line in the emission spectrum of the atom can be obtaned by an energy level change from `Q` so `S`. A blue line can be obtained by following energy level change A. `P "to"Q`B. `Q"to"R`C. `R"to"S`D. `R"to"G` |
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Answer» Correct Answer - D |
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| 88. |
Simplified model of electron energy levels for a cartain atom is shown in figure. The atom is bombarded will fast moving electrons. The impact of one of these electrons can cause the removal of electron from K-level , thus, creating a vacancy in the K-level. This vacancy in K-level is filled by an electron from l-level and the energy reased in this transition can either appear as electromagnetic waves or may all be used knock out an electron from M-level of the atom. Based on above information, answer the following question: The minimum potential difference through which bombarding electron beam must be acclerated from rest to cause the ejection of electron from K-level is.A. `18750 V`B. `400 kV`C. `16 kV`D. `21.6 kV` |
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Answer» Correct Answer - A The minimum enegry required to remove the electron from `K`- level is `3 xx 10^(-15) J`. Let `V` be the potential difference required , then `eV = 3 xx 10^(-15) J`. `V = (3 xx 10^(-15))/(1.6 xx 10^(-16)) = 18750 V` |
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| 89. |
Figure shown five energy levels of an atom , one being much lower than the other four. Five transitios between the levels are idicated, each of which produces a photon of definite energy and frequency. Which one of the spectrum below best corresponding to the set of transition indicated?B. C. D. |
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Answer» Correct Answer - D The first three transition from the left fall in the Lyman series of the hydrogen spectrum which corresponds to ultraviolet radiation. The fourth transion fall in the Balmer series of the spectrum which corresponds to the visible light emission . Thus, frequencies of the last two transition are closed to each other on the extreme left of the frequencies spectrum whereas the frequencies of the first three transition are closer to one another and fall on the right corner of the frequencies spectrum. The spectrum of the transition is thus best represented in diagram |
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| 90. |
Figure represent some of the lower energy level of the hydrogen atom in simplified from. If the transition of an electron from `E_(4) to E_(2)` were accociated with the emission of blue light , which one of the following transition could be accociated with the emission of red light? A. `E_(4) to E_(1)`B. `E_(3) to E_(1)`C. `E_(3) to E_(2)`D. `E_(1) to E_(3)` |
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Answer» Correct Answer - C For emission of a photon with greater wavelength energy gap should be less. Blue light falls in the Balmer series and it is obtained when the atom makes a transition from `E_(4) to E_(3)`. Red light also falls in the Balmer series and it has a lower frequency compared to blue light. By quantum theory of radiation , the energy change `E` is proportional to the frequency of electromagnnetic radiation `f by E = h f`. Thus red light is associated with a smaller energy change from a lower energy level (compared to `E_(4)` to the first excited state `E_(2)`. Hence, the only possible transition that result in the emission of red light is the `E_(3) and E_(2)` transition. |
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| 91. |
An electron in a hydrogen atom makes a transition from first excited state to ground state . The equivelent currect due to circulating electron :A. increases 2 timeB. increases 4 timesC. increases 8 timesD. remains the same |
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Answer» Correct Answer - C `i=q/T and Tpropn^(3)Rightarrowipropn^(3) iprop1/n^(3)`` |
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| 92. |
For a certain hypothetical one electron atom, the wavelength `(in Å)` for the spectral lines for transitions originating at n=p and terminating at n=1 are given by `lambda = (1500 p^2)/(p^2 - 1), where p = 2,3,4` (a)Find the wavelength of the least energetic and the most energetic photons in this series.(b) Construct an energy level diagram for this element showing the energies of the lowest three levels. (c ) What is the ionization potential of this element?A. `3.96 V`B. `9.23 V`C. `6.34 V`D. `8.28 V` |
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Answer» Correct Answer - D The ionization potential corresponds to energy required to librate an electron from its ground state. i.e., ionization energy `= 8.28 eV` Hence, ionization potential `= 8.28 V` |
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| 93. |
For a certain hypothetical one electron atom, the wavelength `(in Å)` for the spectral lines for transitions originating at n=p and terminating at n=1 are given by `lambda = (1500 p^2)/(p^2 - 1), where p = 2,3,4` (a)Find the wavelength of the least energetic and the most energetic photons in this series.(b) Construct an energy level diagram for this element showing the energies of the lowest three levels. (c ) What is the ionization potential of this element?A. The wavelength of the least energetic and the most energetic photons in this series is `2000 Å, 1500Å`.B. Difference between energies of fourth and third orbit is `0.40 eV`.C. Energy of second orbit is `6.2 eV`D. The ionisation potential of this element is `8.27 eV`. |
| Answer» Correct Answer - A::B::C::D | |
| 94. |
Assume that the de Broglie wave associated with an electron can from a standing wave between the atom arrange in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave if the distance d between the atoms of the array is `2 A^0` A similar standing wave is again formed if d is increased to `2.5Å` . Find the energy of the electrons in electron volts and the least value of d for which the standing wave type described above can from .A. `302 eV`B. `151 eV`C. `75.5 eV`D. `75.5xx10^(6) eV` |
| Answer» Correct Answer - B | |
| 95. |
An electron in hydrogen atom after absorbing an energy photo jumps from energy state `n_(1)` to `n_(2)`. Then it returns to ground state after emitting six different wavelength in emission spectrum. The energy of emitted photons is either equal, to less than or greater than the absorbed photons . then `n_(1)` and `n_(2)`A. `n_(2)=4,n_(1)=3`B. `n_(2)=5, n_(2)=3`C. `n_(2)=4, n_(1)=2`D. `n_(2)=4, n_(1)=1` |
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Answer» Correct Answer - C From `n_(2)=4` , six lines are obtained in emission spectrum Now:`E_(4)_(1)=E_("obserbed")` ` E_(4)to3 lt E_("absorbed")` and `E_(4)_(-1)E_(3)_(1),E_(2-1) gt E_("absorbed")` Hence , `n-(1) =2` and `n_(2)=4` |
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| 96. |
A charge particle is moving in a uniform magnetic field ib a circular path. The energy of the particle is doubled . If the initial radius of he circular path was `R`, the radius of the new circular path after the energy is doubled will beA. `R//2`B. `sqrt2R`C. `2R`D. `R//sqrt2` |
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Answer» Correct Answer - B `rpropsqrtKERightarrowr_(2)/r_(1) =sqrt((KE_(2))/(KE_(1)))` |
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| 97. |
Two separate monochromatic light beams `A` and `B` of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength are `lambda_(A)` and `lambda_(B)` respectively. Assuming that all the the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam `A` to that from `B` isA. `(A)/(B)`B. `(D)/(A)`C. `(A)/(B)`D. `(B)/(A)` |
| Answer» Correct Answer - A | |
| 98. |
A monochromatic beam of electromagnetic radiation has an intensity of `1W//m^2`. Then the average number of photons per `m^3` for a 10MeV `gamma` rays isA. 4166B. 3000C. 5000D. 2083 |
| Answer» Correct Answer - D | |
| 99. |
Dtermine the accelerating potential necessary to given an electron a de Broglie wavelength of `1 7Å`, which is the size of the interatomic spacing of atoms in a crystal |
| Answer» `V=(h^(2))/(2m_(0)e lambda^(2))=151V`. | |
| 100. |
Pertain to the following statement and figure The figure above shown level diagram of the hydrogen atom. Serveral transition are market as I, II, III, … The diagram is only indicative and not to scale. Which transition will occur when a hydrogen atom is irradiated with radiation of wavelength `103 nm`?A. IB. IIIC. IVD. V |
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Answer» Correct Answer - D Wavelngth of radiation`= 103 nm = 1030 nm = 1030 Å` `Delta E = (12400)/(1030 Å) = 12.0 eV` So, difference of energy should be `12.0 eV ("apporox")` Hence `n_(1) = 1 `and `n_(2) = 3` `(-13.6) eV (-1.51) eV` Therefore transition is `V`. |
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