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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.
| 551. |
The energy of `n^(th)` orbit is given by `E_(n) = ( -Rhc)/(n^(2))` When electron jumps from one orbit to another orbit then wavelength associated with the radiation is given by `(1)/(lambda) = RZ^(2)((1)/(n_(1)^(2)) - (1)/ (n_(2)^(2)))` The ratio of wavelength `H_(alpha)` of Lyman Series and `H_(alpha)` of Pfund Series isA. `54.7 :1`B. `0.0183 :1`C. `61.4 :1`D. `0.0163 :1` |
| Answer» Correct Answer - 4 | |
| 552. |
calculate the energy assoclated with the first orbit of `He^(+)` . What is the radius of this orbit? |
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Answer» `E_(n)=-((2.18xx10^(-18)J)Z^(2))/n^(2)"atom"^(-1)` for `He^(+),n=1 , Z=2` ` E_(1) =-((2.18xx10^-18 J)(2^(2)))/1^(2)=-8.72xx10^(-18)J` the radius of the orbit is given by equation (2.15) `r_(n) = ((0.0529nm)n^(2))/Z` since n=1 , and Z = 2 ` r_(n) =((0.0529 nm)1^(2))/2 = 0.02645 nm ` |
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| 553. |
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is `800 p m`, calculate the characteristic velocity associated with the neutron. |
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Answer» From de Broglie’s equation, `lambda = (h)/(mv)` `v = (h)/(mlambda)` Where, v = velocity of particle (neutron) h = Planck’s constant m = mass of particle (neutron) `lambda =` wavelength Substituting the values in the expression of velocity (v), `v = (6.626 xx 10^(-34) Js)/((1.67493 xx 10^(-27) kg)(800 xx 10^(-12) m))` `= 4.94 xx 10^(2) ms^(-1)` `v = 494 ms^(-1)` `:.` Velocity associated with the neutron `= 494 ms^(-1)` |
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| 554. |
What is the maximum number of electrons that can be associated with the following set of quantum numbers ? `n = 3, l = 1 and m = -1`A. 10B. 6C. 4D. 2 |
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Answer» Correct Answer - D The given set of quantum number represents one 3d orbital. An orbital can have maximum 2 electrons |
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| 555. |
The maximum number of unpaired electrons can present in `p_(x)` orbital is |
| Answer» Correct Answer - 2 | |
| 556. |
How many electrons can fit in the orbital for which `n = 3 and l = 1` ?A. 2B. 6C. 10D. 14 |
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Answer» Correct Answer - A `n = 3, l = 1` represents 3p-subshell which has three orbitals. In any of these orbitals, maximum electrons that can be fitted is 2 |
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| 557. |
According to quantum mechanics, each electron in an atom is described by___different quantum numbers.A. fourB. threeC. fiveD. two |
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Answer» Correct Answer - A Three of which `(n,l`, and `m_(1))` specify the wave function that gives the probability of finding the electron at various points in space. Note that three different quantum numbers are needed because there are three dimensions to space. A fourth quantum number `(m_(s))` refers to a magnetic property of electrons called spin. |
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| 558. |
The number of electrons in an atom with atomic number 105 having `(n +l) = 8` isA. 15B. 17C. 19D. 21 |
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Answer» Correct Answer - B The electronic configuration of element with atomic number 105 is `1s^(2), 2s^(2) 2p^(6), 3s^(2) 3p^(6) 3d^(10), 4s^(2) 4p^(6) 4d^(10) 4f^(14), 5s^(2) 5p^(6) 5d^(10) 5f^(14), 6s^(2) 6p^(2) 6d^(3) 7s^(2)` For `5f, (n + l) = 5 + 3 = 8`, Electrons present = 14 For `6d, (n + l) = 6 + 2 = 8`, Electrons present = 3 |
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| 559. |
The mathematical approach of quantum mechanics involves treating the electon in an atoms as a __________wave.A. standingB. matterC. waterD. sound |
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Answer» Correct Answer - A A standing wave is a wave that does not travel and therefore, has a least one point at which it has zero amplitude, called a node. For instance, a standing wave is created when a huitar string vibrates when it is plucked. The electron can be decribed by the same kind of standing wave mathematics that is applied to the vibrating guitar string. |
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| 560. |
According to `(n + l)` rule, after np level is filled with electrons, the next electron will enter intoA. `(n + 1) p`B. `(n +1) s`C. `(n -1) d`D. `nd` |
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Answer» Correct Answer - B For example, after 3p, electron enters into 4s level |
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| 561. |
What will be the uncertainty in velocity of a bullet with a mass of 10 g whose position is known with `pm0.01mm`?A. `5.275xx10^(-33)ms^(-1)`B. `5.275xx10^(-25)ms^(-1)`C. `5.275xx10^(-5)ms^(-1)`D. `5.275xx10^(-28)ms^(-1)` |
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Answer» Correct Answer - D `Deltax=pm0.01mm=1xx10^(-5)m, m=10g=1xx10^(-2)kg Deltav=h/(4pim.Deltax)=(6.626xx10^(-34)Js)/(4xx3.14xx10^(-2)kgxx10^(-5)m) =5.275xx10^(-28)ms^(-1)` |
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| 562. |
What is the minimum uncertainty in theh position of a bullet of mass 5g that is known to have a speed somewhere between 550,00000 and 550,00001 `m s^(-1)`? |
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Answer» Correct Answer - `1.05 xx 10^(-32) m` Here, maximum `Delta v = 550,00001 - 550,00000 = 1 m s^(-1)`. |
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| 563. |
The Schrddot(o)dinger equation has been solved exactly only for (i) `H` (ii) `He^(+)` (iii) `Li^(2+)` (iv) `b^(3+)`A. `(i),(ii),(iii),(iv)`B. `(ii),(iii),(iv)`C. `(i),(ii),(iii)`D. `(i),(ii),(iv)` |
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Answer» Correct Answer - C It is solved exactly only for one-electron species. Simplifying assumptions are necessary to solve the equation for more complex atoms and molecular. However, scientists have used their intuition, ingenuity, and modern computers to apply this equation to more complex sysytem. |
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| 564. |
Correct set of four quantum numbers for the valence (outermost) electron of rubidium `(Z = 37)` isA. `5, 0, 0, + (1)/(2)`B. `5, 1, 0, +(1)/(2)`C. `5, 1, 1, +(1)/(2)`D. `6, 0, 0, +(1)/(2)` |
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Answer» Correct Answer - A The electronic configuration rubidium (Z=37) is `[Kr]^(36)5s^(1)` For 5s electron , `n=5, l=0, m=0, s=+(1)/(2)` |
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| 565. |
The correct set of four quantum numbers for the valence electron of rubidium (Z = 37) isA. 5, 1, 1, `+1//2`B. 6, 0, 0, `+1//2`C. 5, 0, 0, `+1//2`D. 5, 1, 0, `+1//2` |
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Answer» Correct Answer - C Electronic configuration of rubidium `(Z = 37)` is `[Kr]^(36) 5s^(1)` Hence, for the valence electron `(5s^(1)), n = 5, l = 0, m = 0, s = 1//2` |
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| 566. |
Which of the following is not ture for the quantum mechanical model of the atom?A. Even though there is one electron in an `H` atom, there are many atomic orbitals in the atom as many wave functions are possible as the solution of the wave equation.B. All the information about the electron in an atom is stored in its orbital wave function `psi` and quantum mechanics makes it possible to extract this information out of `psi`.C. The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function, i.e., `|psi|^(2)` at that point.D. `|psi|^(2)`is known as the probability density and can never be positive. |
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Answer» Correct Answer - D It is always positive. Form the value of `|psi|^(2)` at different points within an atom, it is possible to predict the region around the nucleus where the electron will most probably be found. |
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| 567. |
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible. (i) `n=0, l=0, m_(l)=0, m_(s)=+1//2 " " ` (ii) `n=1, l=0, m_(l)=0, m_(s)=-1//2` (iii) `n=1, l=1, m_(l)=0, m_(s)= +1//2 " "` (iv) `n=1, l=0, m_(l)=+1, m_(s)=+1//2` (v) `n=3, l=3, m_(l)=-3, m_(s)=+1//2 " "` (vi)`n=3, l=1, m_(l)=0, m_(s)=+1//2` |
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Answer» (i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero. (ii) The set of quantum numbers is possible. (iii) The set of quantum numbers is not possible because, for n = 1 , l can not be equal to 1. It can have 0 value. (iv) The set of quantum numbers is not possible for l = 0 , `m_(l)` cannot be + 1. It must be zero. (v) The set of quantum numbers is not possible because, for n = 3, `l != 3`. (vi) The set of quantum numbers is possible. |
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| 568. |
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible. (a) `n=0, l=0, m_(l)=0, s=+1//2` , (ii) ` n=1, l=0, m_(l)=0, s=-1//2` (iii) ` n=1, l=1, m_(1)=0, s=+1//2` , (iv) `n=1, l=0, m_(l)=+1, s=+1//2` |
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Answer» (i) The set of quantum numbers in not possible because the minimum value of n can be 1 and not zero. (ii) The set of quantum numbers is possible. (iii) The set of quantum numbers is not possible because for `n=1, l` cannot be equal to 1. It can have 0 value. (iv) The set of quantum numbers is not possible because for `l=0,m` cannot be +1 . It must be zero. |
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| 569. |
Arrange the electrons represented by the following sets of quantum numbers in decreasing order of energy. (i)`n=4, l=0, m_(l)=0, s=+1//2`, (ii) `n=3, l=1, m_(l)=1, s=-1//2` (iii) ` n=3, l=2, m_(l)=0, s=+1//2`, (iv) `n=3, l=0, m_(l)=0, s=-1//2` |
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Answer» (i) represents 4s orbital , (ii) represents 3p orbital (iii) represents 3d orbital, (iv) represents 3s orbitals The decreasing order of energy , `3d gt 4s gt 3p gt 3s`. |
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| 570. |
Which of the following do and which donot make sense ? `2p^(5), 2d^(4), 4s^(2), 3 f^(7), 4p^(10)`. |
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Answer» `2p^(5)`: Does make sense `2d^(4)` : Does not make sense because second shell cannot have d sub-shell `4s^(2) `: Does make sense `3f^(7)` : Does not make sense because third shell cannot have f-sub-shell. `4p^(10)` : Does not make sense because p sub-shell cannot have more than six electrons. |
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| 571. |
Deuterium nucleus contains :A. `1p +1n`B. `2p +0n`C. `1p + pe^(-)`D. `2p +2n` |
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Answer» Correct Answer - A is the correct answer. |
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| 572. |
The number of neutrons in deuterium is |
| Answer» Correct Answer - 2 | |
| 573. |
Deuterium nucleus containsA. `1 p + 1 n`B. `2p + 0n`C. `1p + 1e^(-)`D. `2p + 2n` |
| Answer» Correct Answer - A | |
| 574. |
Which one of the following pairs constitutes isotones ?A. `._(6)C^(13) and ._(6)C^(14)`B. `._(6)C^(13) and ._(7)N^(14)`C. `._(7)N^(14) and ._(9)F^(19)`D. `._(7)N^(14) and ._(7)N^(15)` |
| Answer» Correct Answer - B | |
| 575. |
Whatis the electronic configuration of Argon ? |
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Answer» Argon has atomic number `18 ( Z = 18)` Electronic configuration `1s^(2) 2s^(2) 2p^(6) 3s^(2)3p^(6)` |
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| 576. |
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of `2` ns and the number of photons emitted during the pulse is `2.5xx10^(15)`, calculate the energy of the source. |
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Answer» Time duration (t) `=2 " ns "=2xx10^(-9)s` Frequency `(v)=(1)/(t)=(1)/(2xx10^(-9)s)=(10^(9))/(2)s^(-1)` Energy of one photon, E`=hv=(6.626xx10^(-34)Js)xx(10^(9)//2 s^(-1))=3.25xx10^(-25)J` No. of photons `= 2.5xx10^(5)` `:. " "` Energy of source `=3.3125xx10^(-25) J xx 2.5xx10^(15)=8.28xx10^(-10)J` |
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| 577. |
Which of the following sets of quantum numbers are correct? n l m n l mA. `{:(,"n",l,m_(l)),((a),1,1,+2):}`B. `{:(,"n",l,m_(l)),((b),2,1,+1):}`C. `{:(,"n",l,m_(l)),((c),3,2,-2):}`D. `{:(,"n",l,m_(l)),((d),3,4,-2):}` |
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Answer» Correct Answer - B::C (b, c) are both correct. (a) is not correct because `m_(l)`. Can not be+2 for l=1. (d)is not correct because l can not be 4 for n=3. |
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| 578. |
The longest wavelength doublet absorption is observed at `589` and `589.6 nm`. Caiculate the frequency of each transition and energy difference between two excited states. |
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Answer» `lambda_(1)=589 nm = 589xx10^(-9)m` `v_(1)=(c)/(lambda_(1))=(3xx10^(8)ms^(-1))/(589xx10^(-9)m)=(3000)/(589)xx10^(14)s^(-1) " " =5.0934xx10^(14)s^(-1)` `lambda_(2)=589.6 nm = 589.6xx10^(-9)m` `v_(2)=(c)/(lambda_(2))=(3xx10^(8)m s^(-1))/(589.6xx10^(-9)m)=(3000)/(589.6)xx10^(14)s^(-1)=5.0882xx10^(14)s^(-1)` `DeltaE = E_(1)-E_(2)=(hc)/(lambda_(1))-(hc)/(lambda_(2))=hc[(1)/(lambda_(1))-(1)/(lambda_(2))]` `=(6.626xx10^(-34)Js xx3xx10^(8) ms^(-1))[(1)/(589xx10^(-9)m)-(1)/(589.6xx10^(-9)m)]` `=(19.878xx10^(-34)xx10^(8))/(10^(-9))[(589.6-589)/(589.6xx589)]J` `=(19.878xx10^(-17)xx0.6)/(589.6xx589)J=3.43xx10^(-22)J` |
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| 579. |
In which of the following pairs, the ions are iso-electronic ?A. `Na^(+) ,Mg^(2+)`B. `Al^(3+) , O^(-)`C. `Na^(+), O^(2-)`D. `N^(3-),Cl^(-)` |
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Answer» Correct Answer - A::C The ions involved have 10 electrons `Na^(+)=(Z=11) 1s^(2)2s^(2)2p^(6)` `Mg^(2+) (Z=12) 1s^(2)2s^(2)2p^(6)` `O^(2-)(Z=8) 1s^(2)2s^(2)2p^(6)`. |
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| 580. |
`Be^(2+)` is iso electronic with which of the following ions ?A. `H^(+)`B. `Li^(+)`C. `Na^(+)`D. `Mg^(2+)` |
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Answer» Correct Answer - B Both `Be^(2+)` and `Li^(+)` ions have electronic configuration `: 1s^(2)`. `:.` these are isoelectronic in nature. |
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| 581. |
Match the following : `{:("(i) Photon","(a) Value of 4 for N shell"),("(ii) Electron","(b) Probability density"),((ii) Psi^(2),"(c) Always positive value"),("(Iv) Principla quantum number n","(d) Exhibits both momentum and wavelength"):}` |
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Answer» Correct Answer - (i)-(d), (ii)-(d), (iii)-(b), (c), (iv)-(a), (c) Photon has particle nature as well as wave nature. Electron also has particle nature as well as wave nature. `Psi^(2)` represents probability density and always has positive values. Principal quantum number n = 4 for N-shell and always has positive values |
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| 582. |
Which one of the following pairs of atoms/atom-ion have identical ground state configuration?A. `Li^(+)` and `He^(+)`B. `Cl^(-)` and `Ar`C. Na and KD. `F^(+)` and Ne |
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Answer» Correct Answer - B Both `Cl^(-)` and Ar have identical ground state configuration `(1s^(2)2s^(2)2p^(6)3s^(2)3p^(6))`. |
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| 583. |
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microcope is `1.6xx10^(6) m s^(-1)`. Calculate de Broglie wavelength associated with this electron. |
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Answer» `lambda=(h)/( m upsilon)=((6.626xx10^(34)Js))/((9.1xx10^(-31)"kg" ) xx (1.6xx10^(6) ms^(-1)))` `=0.455xx10^(-34+25)m=0.455nm=455 `pm. |
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| 584. |
The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one node. Upon absorbing light , the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal is to the ground state energy of the hydrogen atom. The orbital angular momentum quantum number of the state `S_(2)` isA. 0.75B. `1.50`C. 2.25D. `4.50` |
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Answer» Correct Answer - C The energy of `S_(1)`state (for 2s orbital)of `Li^(2+)ion=(-13.6xx(3)^(2))/((2)^(2))=-13.6xx(9)/(4)` `=-13.6xx2.25 eV` `=2.25 ` times of `E_(H)`. |
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| 585. |
The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one node. Upon absorbing light , the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal is to the ground state energy of the hydrogen atom. The sate `S_(1)` isA. 1 sB. 2 sC. `2 p`D. 3 s |
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Answer» Correct Answer - B The state `S_(1)` in case of `Li^(2+)`ion must be 2s (It has one radial mode). It must have jumped to 3p (`S_(2)` state) which has one radial node and its energy is equal to that of ground state of H atom `E_(H)(1 s)=-13.6 eV` `E_(Li^(2+))(3p)=(-13.6 Z^(2))/(n^(2))=-(13.6xx(3)^(2))/((3)^(2))` `=-13.6 eV` |
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| 586. |
What about degeneracy of 2p orbitals in a magnetic field?A. No degenerate orbitalsB. Three degenerate orbitalC. Two degenerate orbitalD. Cannot be statd any thing about degenracy |
| Answer» Correct Answer - 1 | |
| 587. |
For `Be^(3+)` ion, the degeneracy of the `3rd` enegry level is equal toA. nineB. sixC. threeD. seven |
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Answer» Correct Answer - A Degeneracy is equal to the number of orbitals that have the same enegry. Different subshells of a particular shell have different energues in case of multi-electron atoms. However, in `H` atom or `H-`like ion such as `Be^(3+)`, these have the same enegry. Thus,`3s, 3p_(x),3p_(y),3p_(z),3d_(xy),3d_(xz),3d_(yz),3d_(x^(2)-y^(2))`, and `3d_(z^(2))` all possess the same enrgy. |
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| 588. |
Which of the following relates to photon both as wave motion and as a stream of particles ?A. DiffractionB. `E = hv`C. `E = mc^(2)`D. Interference |
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Answer» Correct Answer - B `E` is the energy of a light particle (photon) and `v` is the frequency of the associated wave. Diffraction and interference are wave like properties. In `E = mc^(2), E` is the enegry of photon, `m` is the mass of photon, and `c` is the speed of photon. |
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| 589. |
The hydrogen like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one radial node. Upon absorbing light, the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the hydrogen atom Energy of the state `S_(1)` in units of the hydrogen atom ground state energy isA. 0.75B. `1.50`C. 2.25D. `4.50` |
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Answer» Correct Answer - C Energy in `S_(1)` state (i.e., 2s orbital) of `Li^(2+)` ion `= - (13.6 xx 3^(2))/(2^(2)) = - 13.6 xx 2.25 eV` = 2.25 times of `E_(H)` |
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| 590. |
The hydrogen like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one radial node. Upon absorbing light, the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the hydrogen atom The state `S_(1)` isA. 1stB. 2sC. 2pD. 3s, 4d |
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Answer» Correct Answer - B The state `S_(1)` must be 2s (as it has one radial node) and it must have jumped to 3p (which again has one radial node and energy equal to that of ground state H-atom as `E_(H (Is)) = - 13.6 eV` `E_(Li^(2+) (3p)) = (-13.6 Z^(2))/(n^(2)) = - (13.6 xx 3^(2))/(3^(2))` `= -13.6 eV` |
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| 591. |
An electron in hydrogen jumps to a shell in which four fold degeneracy is present , then correct about that shell isA. 3B. 4C. 2D. This type of transition is not position |
| Answer» Correct Answer - 3 | |
| 592. |
Which of the following quantum numbers relates to the average distance of an electron from the nucleus in a particular orbital?A. Electron spin quantum numberB. Magnetic quantum numberC. Angular momentum quantum numberD. Principal quantum number |
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Answer» Correct Answer - D The larger `n` is, the greater is the average distance of an electron in the orbital form the nucleus and, therefore, the large is the orbital (and the less stable is the orbital as the electorn will be located away from the nucleus). |
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| 593. |
An electron jumps to higher excited state of a orbital which is non directional and have 4 radial node then shell with which electron belong? |
| Answer» Correct Answer - 5 | |
| 594. |
The number of radial nodes and angular nodes for d-orbital can be represented asA. (n - 2) radial nodes + 1 angular node = (n - 1) total nodesB. (n - 1) radial nodes + 1 angular node = (n - 1) total nodesC. (n - 3) radial nodes + 2 angular node = (n - l - 1) total nodesD. (n - 3) radial nodes + 2 angular node = (n - 1) total nodes |
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Answer» Correct Answer - D Total number of nodes = n - 1 For d-orbital, radial nodes = n - l - 1 = n - 3 and there are 2 angular nodes. |
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| 595. |
The magnitude of velocity of the electron______with increase of positive charge on the nucleus and___with increase of pricipal quantum number.A. increases,increasesB. increases, decreasesC. decreases, increasesD. decreases, decreases |
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Answer» Correct Answer - B To overcome increased attraction towards the nucleus, the electron must revolves around the nucleus faster. |
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| 596. |
Assertion. The plots of probability density and radial probability function versus distance r from the nucleus for any particular orbital are not identical Reason. Probability density is `Psi^(2)` whereas radial probability function represents probability of finding the electron in a shell of thickness `dr`.A. If both assertion and reason are true, and reason is the true explanation of the arrertionB. If both assertion and reason are true, but reason is not the true explanation of the assertionalC. If assertion is true, but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - A The plots are not identical (see Fig. 2.27 and 2.44) The probability density `(Psi^(2))` is along the radial distance whereas radial probability function `(4pi r^(2) Psi^(2))` is the probability in a shell of thickness dr. Hence R is the correct explanation of A. |
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| 597. |
The radius of maximum probability for `1s` orbital of `H` atom isA. `52.4 nm`B. `52.9 mu m`C. `52.9 Å`D. `52.9pm` |
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Answer» Correct Answer - D The radial probability distribution function for the `1s` orbital of the `H` atom initially increases with increase in distance form the nucleus. It reaches a maximum at a distance very close to the nucleus and then decreases. The maixmum in the curve corresponds to the distance at which the probability of finding the electron is the maximum. This distance, called the radius of maximum probability, for `H` atom has a value of `52.9 pm`. For `2s` orbital, ot os `275 pm` and for `2p`, it is `210 pm`. |
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| 598. |
When electromagnetic radiaiton of wavelength `300 nm` falls on the surface of sodium electrons are emitted with a kinetic enegry of `1.68xx10^(5)Jmol^(-)`. What is the minimum enegry needed to remove an electorn from sodium? Strategy: The minimum enegry required to remove an electron from target metal is called work function `W_(0)` of the metal. It can be calculated from Eq., provided we know the energy of the incident photon and kinetic enegry of a single photoelectorn. |
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Answer» Step `1:` Calculating the enegry `(E)` of incident photon: `E = hv = (hc)/(lambda)` `=((6.6xx10^(-34)Js)(3.0xx10^(8)ms^(-1)))/((300xx10^(-9)m))` `= 6.6 xx 10^(-19)J` Step `2:` Calculating the kinetic enegry of a single photoelecton: `KE = ((1.68xx10^(5)Jmol^(-1)))/(6.022xx10^(23)e mol^(-1))` `0.279 xx 10^(18)Je^(-1)` `= 2.79 xx 10^(-19) Je^(-1)` Step `3:` Calculating the minimun energy needed to remove an electorn: `hv = hv_(0) +kE` `:. hv_(0) = hv - kE` `=(6.6 xx 10^(19)J)-(2.79 xx 10^(-19)J)` `= 3.81 xx 10^(-19) J` |
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| 599. |
What are the two longest wavelength lines (in nanometers) in the Lyman series of the hydrogen spectrum? Strategy: The Lyman series is given by the Balmer-Rydberg equation with `n = 1` and `mgt1`. Since the left side of Eq. is a fraction that has `lambda` in the denominator, the value of `lambda` (the wavelength) increases as the value of the term on the right side of the equation decreases. Since the value of `1//n^(2)` is now fixed and we need to subtract `1//m^(2)` from this, the wavelength `lambda` is the greatest when `1//m^(2)` is the largest or when `m` is the smallest, i..e., when `m = 2` and `m = 3`. |
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Answer» Solving Eq. first for `n =2` gives `(1)/(lambda) = R((1)/(1^(2))-(1)/(2^(2)))` `= 1.097xx10^(7)m^(-1)(1-(1)/(4))` `= 0.8228 xx 10^(7)m^(-1)` `lambda = (1)/(0.8228 xx 10^(7)m^(-1)) = 1.215xx10^(-7)m = 121.5nm` Solving next for `m = 3` gives `(1)/(lambda) = R ((1)/(1^(2))-(1)/(3^(2)))` `= 1.097xx10^(7)m^(-1) (1-(1)/(9))` `= 0.9751xx10^(7)m^(-1)` `lambda = (1)/(0.9751 xx 10^(7)m^(-1)) = 1.026xx10^(-7)m = 102.6 nm` |
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Gaseous metal ion `M^(2+)` has `5` unpaired electron. What is the atomic number?A. `26`B. `25`C. `27`D. `24` |
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Answer» Correct Answer - B Let us write the electronic configuraiton of `Z = 24`: `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)` or simply `[Ar] 3d^(5)4s^(1)` `(Z = 25):[Ar] 3d^(5)4s^(2)` `(Z = 26):[Ar]3d^(6)4s^(2)` `(Z = 27):[Ar]3d^(7)4s^(2)` If we take `2e^(-)s` from each of the above configurations, we notice that only `(Z =25)` leads to the formation of `M^(2+)` with `5` unpaired electrons. Shortcut method: The presence of `5` unpaired electrons can be justified by`d^(5)` subshell only. Thus, the electronic configuration of `M^(2+)`ion will be `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)` accounting for `23e^(-)s`. Thus, neutral `M` atom should have `23 +2 = 25e^(-)s`, i.e., the atomic number should ber `25`. |
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