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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. |
The species Ar, `K^(+) and Ca^(2+)` contain the same number of electrons. In which order do their radii increase ?A. `Ca^(2+) lt K^(+) lt Ar`B. `K^(+) lt Ar lt Ca^(2+)`C. `Ar lt K^(+) lt Ca^(2+)`D. `Ca^(2+) lt Ar lt K^(+)` |
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Answer» Correct Answer - A They are isoelectronic species but their nuclear charges are Ca = 20, K = 19, Ar = 18. Greater the nuclear charge of the isoelectronic species, greater is the pull on the electrons and smaller is the size |
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| 252. |
How many unpaired electrons are present in the atomic mercury `(Z = 80)`?A. oneB. zeroC. twoD. three |
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Answer» Correct Answer - B Mercury has `80` electrons. It takes `(2+8+8+18 +18) = 54` electrons to complete the fifth-period elements. Strating with the sixth period, we need `2` electrons for the `6s` subshell, `14` electrons for the `4f` subshell, and `10` more for the `5d` subshell. The total number of electrons adds up to `80`, thus, the electrons configuration of `Hg` is as follows: `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(6)4p^(10)4f^(14)5s^(2)5p^(6) 5d^(10)6s^(2)` or simply `[Xe] 4f^(14) 5d^(10) 6s^(2)` It has only paired electrons, i.e. it has zero unpaired electrons. |
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| 253. |
Statement-1. The radii of corresponding orbitals in all H-like particles are equal Statement-2 All H-like particles contain only one electron.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-4B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-4C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - D Correct statement-1. The radii of the corresponding orbits of H-like particles depend upon the atomic number (Z) and are not equal |
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| 254. |
Why do two electrons in an orbital have opposite spin ? |
| Answer» Because the electrons with parallel or same spin will repel one another. | |
| 255. |
Which d-orbital does not have four lobes? |
| Answer» `d_(z^(2))` orbital has only two lobes. | |
| 256. |
Statement-1. The opposite lobes of a p-orbital have opposite sign whereas opposite lobes of d-orbital have the same sign Statement-2 The opposite lobes of a p-orbital have opposite charge whereas opposite of lobes of d-orbital have the same chargeA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-3B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-3C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - C Correct statement-2. The + and -sign represent the sign of the wave function and nor the sign of charges. |
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| 257. |
Which d-orbtial has its four lobes along the axis? |
| Answer» `d_(x^(2)-y^(2))` orbital has four lobes along the x and y axes. | |
| 258. |
Statement 1. 4s orbital has lower energy than 3d orbital. Statement 2. s-orbital is spherically symmetrical and remains closer to the nucleus and has lower energy than d-orbital.A. Statement-1 is true , Statement-2 is also true. Statement-2 is the correct explanation of Statement-1B. Statement-1 is true , Statement-2 is also true. Statement-2 is not correct explanation of Statement-1C. Statement -1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - B Correct explanation. Because of shielding effect, 3d orbital has higher energy than 4s orbital. |
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| 259. |
The frequency of radiations emitted when electron falls from `n = 4` to `n = 1` in `H-"atom"` would be (Given `E_1` for `H = 2.18 xx 10^-18 J "atom"^-1` and `h = 6.625 xx 10^-34 Js`.)A. `1.54xx10^(15)s^(-1)`B. `1.03xx10^(15)s^(-1)`C. `3.08xx10^(15)s^(-1)`D. `2.0xx10^(15)s^(-1)` |
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Answer» Correct Answer - C According to Bohr,s theory, `DeltaE = Z^(2)E_(1) ((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))` where `Z^(2)E_(1) =-` Ionization enegry Since `DeltaE = hv`, we have `v = (DeltaE)/(h) = (Z^(2)E_(1))/(h) ((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))` `v = (2.18xx10^(-18)J)/(6.625xx10^(-34)Js) ((1)/(1^(2))-(1)/(4^(2)))` `= 0.329 xx 10^(16)s^(-1) ((15)/(16))` `= 3.08 xx 10^(15) s^(-1)` Note that we use positive sign of enegry because `v` has to be positive. |
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| 260. |
Write the electronic configuration of chlorine atom ? Hint `:` Chlorine `( Z = 17)` |
| Answer» `1s^(2) 2s^(2)2p^(6) 3s^(2) 3p^(5)` | |
| 261. |
What is the maximum numaber of emission lines obtained when the excited electron of a H atom in `n=6` drops to the ground state ? |
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Answer» No. of lines produced when electron from nth shell drops to ground state `Sigma (n_(2)-n_(1))=Sigma (6-1)= = Sigma 5=5+4+3+2+1=15` There are produced due to following transitions : `{:(6 rarr 5,5 rarr 4,4 rarr 3,3 rarr 2,2 rarr 1),(6 rarr 4,5 rarr 3,4 rarr 2,3rarr1,),(6 rarr 3,5 rarr 2,4 rarr 1,,),(6 rarr 2,5 rarr 1,,,),(6 rarr 1,,,,),(("5 lines"),("4 lines"),("3 lines"),("2 lines"),"(1 lines)"):}` |
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| 262. |
The radius of the first Bohr orbit of hydrogen atom is `0.59 Å`. The radius of the third orbit of `He^(+)` will beA. `8.46 Å`B. `0.705 Å`C. `1.59 Å`D. `2.38 Å` |
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Answer» Correct Answer - D `r_(n) = (a_(0) n^(2))/(Z) = (0.59 B208Å xx 3^(2))/(2) = 2.38 B208Å` |
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| 263. |
Which of the following has zero electron density in xy plane ?A. `d_(z^(2))`B. `d_(x^(2)- y^(2))`C. `p_(z)`D. `d_(xy)` |
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Answer» Correct Answer - C `p_(x)` orbital has zero electron density in the xy plane. |
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| 264. |
For which of the following orbitals is the radial probaility density `R^(2)` the maximum at the nucleus and descrease sharply as the distance form the nucleus increases?A. `1s`B. `2s`C. `3s`D. All the `s` orbitals |
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Answer» Correct Answer - A For `1s` orbital, `R^(2)` is large near the nucleus `(r = 0)`, indicating that the `1s` electron in most likely to be found in this region. The value of `R^(2)` decrease rapidly as the distance form the nucleus increases, but `psi^(2)` never goes to exactly zero, although the probaility does become extremely small at large distances from the nucleus. On teh other hand, for `2s` orbital, `R^(2)`(radial probability density) first decreases sharply to zero but gain starts increasing. After reaching a small maximum, it decreases again and approaches zero as the distance form the nucleus increases. the region where the radial probability density `(R^(2))` reduces to zero is called radial node or sherical node. |
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| 265. |
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.A. `3.31xx10^(-22)kJ`B. `3.31xx10^(-22)J`C. `2.98xx10^(-21)J`D. `3.0xx10^(-21)kJ` |
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Answer» Correct Answer - B Step I : Wavelength `(lamda)` = 589 nm = `589xx10^(-9)m` Frequency (v) = `c/lamda=(3xx10^(8))/(589xx10^(-9)) =5.093xx10^(14)` cycles per sec Step II : Wavelength `(lamda)` = 589.6 nm = `589.6xx10^(-9)m therefore" "v=c/lamda=(3xx10^(8))/(589.6xx10^(-9))=5.088xx10^(14)` cycles per sec Energy difference between two excited states, `DeltaE=6.626xx10^(-34)(5.093-5.088)10^(14) =6.626xx10^(-34)xx5xx10^(-3)xx10^(14)=3.31xx10^(-22)J` |
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| 266. |
The longest wavelength doublet absorption is observed at `589` and `589.6 nm`. Calculate the frequency of each transition and energy difference between two excited states. |
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Answer» For `lambda_(1) = 589 nm` Frequency of transition `(v_(1)) = (c)/(lambda_(1))` `= (3.0 xx 10^(8)ms^(-1))/(589 xx 10^(-9)m)` Frequency of transition `(v_(1)) = 5.093 xx 10^(14) s^(-1)` Similarly, for `lambda_(2) = 589.6 nm` Frequency of transition `(v_(2)) = (c )/(lambda_(2))` `= (3.0 xx 10^(8) ms^(-1))/(589.6 xx 10^(-9) m)` Frequency of transition `(v_(2)) = 5.088 xx 10^(14) s^(-1)` Energy difference `(DeltaE)` between excited states `= E_(1) - E_(2)` where, `E_(2) =` energy associated with `lambda_(2)` `E_(1) =` energy associated with `lambda_(2)` `DeltaE = hv_(1) - hv_(2)` `= h(v_(1) - v_(2))` `= (6.626 xx 10^(-34) Js) (5.093 xx 10^(14) - 5.088 xx 10^(14))s^(-1)` `= (6.626 xx 10^(-34)J)(5.0 xx 10^(-3) xx 10^(14))` `DeltaE = 3.31 xx 10^(-22) J` |
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| 267. |
A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference of V esu. If e and m are charge and mass of the electron respectively, then the value of `h//lamda` (where `lamda` is the wavelength associated with the electron wave) is given byA. `me V`B. `2 me V`C. `sqrt(me V)`D. `sqrt(2 meV)` |
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Answer» Correct Answer - D `lamda = (h)/(p) or (h)/(lamda) = p = mv` `KE = (1)/(2)mv^(2)` so that `v = sqrt((2KE)/(m)) " But " KE = eV` `:. v = sqrt((2eV)/(m)) :. (h)/(lamda) = mv = m sqrt((2eV)/(m)) = sqrt(2meV)` |
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| 268. |
The energy of a photon is given as `3.03xx10^(-19)J`. The wavelength of the photon isA. 6.56 nmB. 65.6 nmC. 0.656 nmD. 656 nm |
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Answer» Correct Answer - D `E=(hc)/lamdaorlamda=(hc)/E lamda=(6.626xx10^(-34)Jsxx3xx10^(8)ms^(-1))/(3.03xx10^(-19)J) 6.56xx10^(-7)m or 656nm` |
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| 269. |
Which out of `Cu^(2+), Fe^(2+) and Cr^(3+)` has highest paramagnetism and why ? |
| Answer» `Cu^(2+)` has one, `Fe^(2+)` has four and `Cr^(3+)` has three unpaired electrons. Hence, `Fe^(2+)` has highest paramagnetism | |
| 270. |
Not considering the electronic spin, the degenracy of the second excited state of `H^(-)` is |
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Answer» Correct Answer - 3 In case of H-atom, the energies of the orbitals are in the order. `1s lt 2s = 2p lt 3s = 3p = 3d lt 4s = 4p = 4d = 4f lt`.... ,br. In case of multielectron atoms or ions (e.g. `H^(-)` viz `1s^(2)`), the order is `1s lt 2s lt 2p lt 3s lt 3p`.... (i.e., they follows `(n + l)` rule) Hence, in case of `H^(-) (1s^(2))`, the first excited state would be `1s^(1) 2s^(1)` Second excited state would be `1s^(1) 2s^(0) 2p^(1)`. Thus, degeneracy of the second excited state `(2p^(1))` is three viz. `2p_(x) 2p_(y) 2p_(z)`, i.e., number of degenerate orbitals = 3 [In case of H-atom `(1s^(1))`, 1st excited state would be 2s or 2p and 2nd excited state would be 3s, 3p or 3d, i.e., number of degenerate orbitals would be 1(3s), 2 (3p) and 5(3d), i.e., total = 9] |
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| 271. |
What is the maximum number of electrons that can be accommodated in the d-subshell ? |
| Answer» Correct Answer - Ten | |
| 272. |
What is the maximum number of electrons that can be placed in each subshell?A. `2l^(2)`B. `2(l +1 )`C. `2(2l +2)`D. `2(2l +1)` |
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Answer» Correct Answer - D Each subshell of quantum number l contains `(2l +1)` orbitals. For example, if `l = 1`, there are three `p` orbitals. No more than two electrons can be placed in each orbitals. Therefore, the maximum number of electrons is simply twice the number fo orbitals that are employed. |
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| 273. |
Which of the following statement are correct ?A. The electronic configuration of Cr is `[Ar] 3d^(5) 4s^(1)` (Atomic No. of cr = 24)B. The magnetic quantum number may have a negative valueC. In silver atom, 23 electrons a spin of one type and 24 of the opposite type (Atomic No. of Ag = 47)D. The oxidation state of nitrogen in `HN_(3) " is " -3` |
| Answer» Correct Answer - A::B | |
| 274. |
Calculate the velocity of an electron in the first Bohr orbit of a hydrogen atom |
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Answer» Assume `n = 1`, take `Z = 1`, and insert the values of the constants to obtain `V_(1) = ((2)(3.14)(1)(4.8xx10^(-10)esu)^(2))/((1)(6.63xx10^(-27)erg s))` `= 2.18 xx 10^(8)cm s^(-1)` `= 2.18 xx 10^(6) m s^(-1)` |
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| 275. |
The least stable in amongst the following is :A. `C^(-)`B. `B^(-)`C. `Be^(-)`D. `Li^(-)` |
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Answer» Correct Answer - C `Li(1s^(2)2s^(1))rarrLi^(-)(1s^(2)2s^(2))` `Be(1s^(2)2s^(2))rarr Be^(-) (1s^(2)2s^(2)2p^(1))` `B(1s^(2)2s^(2)1p^(1)) rarr B^(-)(1s^(2)2s^(2)2p_(x)^(1)2p_(y)^(1))` `C^(-)(1s^(2)2s^(2)2p^(2))rarr C^(-) (1s^(2)2s^(2)2p^(3))` `Be^(-)` is the least stable because in the Be atom, all the subshells are completely full. Thus, on addition of `1` electron, the stable configuration is destabilized. |
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| 276. |
The correct enegry value order isA. `ns, np, nd, (n - 1)f`B. `ns, np, (n -1)d, (n -2)f`C. `ns, np, (n - 1)d, (n - 1)f`D. `ns, (n - 1)d, np (n - 1)f` |
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Answer» Correct Answer - D Considering the minimum value of `n` for the existence of the `f` subshell, we have `(1) rArr 5s, 5p, 5d, 4f (2) rArr 6s, 6p, 4d, 4f` `(3) rArr 5s, 5p, 4d, 4f (4) rArr 5s, 4d, 5p, 4f` According to the `(n+l)` rule, lower `(n +l)` implies lower energy, and if `(n +l)` is same, then lower `n` implies lower enegry. Therefore, teh correct enegry value order is `5s lt 4d lt 5p lt 4f` |
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| 277. |
Which of the following lanthanoid ions is diamagnetic? (Atomic number of `Ce=58,Sm=62,Eu=63,Yb=70`]A. `Sm^(2+)`B. `Eu^(2+)`C. `Yb^(2+)`D. `Ce^(2+)` |
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Answer» Correct Answer - C Diamagnetic specied contains no unpaired electrons: `Ce: [Xe] 4f^(1)5d^(1) 6s^(2)rArr Ce^(2+: [Xe]4f^(1)5d^(1)`, two unpaired eletrons `Sm: [Xe] 4f^(6) 6s^(2) rArr Sm^(2+):[Xe] 4f^(6)`[ six unpaired electrons `Eu: [Xe] 4f^(7) 6s^(2) rArr Eu^(2+): [Xe] 4f^(7)`, seven unpaired electrons `Yb:[Xe] 4f^(14) 6s^(2) rArr Yb^(2+):[Xe]4f^(14)`, zero unpiared electrons Thus, `Yb^(2+)` ion is diamagnetic. |
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| 278. |
Calculate the enegry of an electron in the second Bohr orbit of an `H` atom. Strategy : Use Eq. and proper values of `n` and `Z`. Atomic number of the `H`-atom is `1`. |
| Answer» `E_(2) =((1)^(2)(-13.6eV at om^(-1)))/((2)^(2)) =- 3.4 eV atom^(-1)` | |
| 279. |
Calculate the enegry of an electron in the first Bohr orbit if a hydrogen atom. Strategy: Use Eq. to obtain directly the energy of the lowest stationary state (or ground state). |
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Answer» Assume `n = 1`, take `Z = 1`, and insert the values of the constants of obtain `E_(1) =- ((2)(3.14)^(2)(1)^(2)(4.8xx10^(-10)esu)^(4)(9.11xx10^(-18)g))/((1)^(2)(6.63xx10^(-27)erg s)^(2))` `=- 2.188 xx 10^(-11) "erg atom"^(-1)` `=- 2.188 xx 10^(-18) J "atom"^(-1)` `=- 13.6eV "atom"^(-1)` `=- 1312 kJ mol^(-1)` |
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| 280. |
Calculate the velocity of an electron in the second Bohr orbit of `Be^(3+)`. Strategy: Use Eq. with proper values of `n` and `Z`. Atomic number of the `Be` atom is `4`. |
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Answer» `V_(2) = (ZV_(1))/(n) = ((4)(2.18xx10^(6)ms^(-1)))/(2)` `4.36xx10^(6)ms^(-1)` |
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| 281. |
The electron in the hydrogen atom makes a transition from the `n = 2` energy level (or state) to the ground state (corresponding to `n = 1`). Find the frequency of the emitted photon. Strategy: Use Eq. directly to obtain `bar(v)`, with `n_(f) = 1` (ground state) and `n_(i) = 2`. Then find the frequncy. |
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Answer» Step `1: bar(v) = R_(H) |((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))| = R_(H)|((1)/(1^(2))-(1)/(2^(2)))|` `= (3)/(4)R_(H) = (3)/(4) (1.097 xx 10^(7)m^(-1))` `=0.8228 xx 10^(7)m^(-1) = 8.228 xx 10^(6)m^(-1)` This wavenumber lies in the ultraviolet region. Step `2:` Since `v = cbar(v)`, the frequency of the emitted photon is `v = cbar(v)` `(3.0xx10^(8)ms^(-1))(8.228xx10^(6)m^(-1))` `= 24.68 xx 10^(14) s^(-1) = 2.5xx10^(15)s^(-1) (or Hz)` |
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| 282. |
The energy absorbed by each molecule `(A_2)` of a substance is `4.4 xx 10^-19 J` and bond energy per molecule is `4.0 xx 10^-19 J`. The kinetic energy of the molecule per atom will be.A. `2.2 xx 10^(-19)J`B. `2.0 xx 10^(-19)J`C. `4.0 xx 10^(-20)J`D. `2.0 xx 10^(-20)J` |
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Answer» Correct Answer - D Extra enegry available si equally divided between the two departing atoms of the `A_(2)` molecule and appears as the kinetic enegry. Thus, kinetic enegry per atom `=(("Enegry absorbed pr molecule")-("Bond energy per moelcule"))/(2)` `=((4.4xx10^(-19)J)-(4.0xx10^(-19)J))/(2)` `= (0.4xx10^(-19)J)/(2) = 0.2 xx 10^(-19)J` `=2.0 xx 10^(-20)J` |
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| 283. |
The energy absorbed by each molecule `(A_(2))` of a substance is `4.4 xx 10^(-19)J` and bond energy per molecule is `4.0 xx 10^(-19) J`. The kinetic energy of the molecule per atom will beA. `2.2 xx 10^(-19) J`B. `2.0 xx 10^(-19) J`C. `4.0 xx 10^(-20) J`D. `2.0 xx 10^(-20) J` |
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Answer» Correct Answer - D K.E. per atom `= ((4.4 xx 10^(-19)) - (4.0 xx 10^(-19)))/(2)J` `= (0.4 xx 10^(-19))/(2) J = 2.0 xx 10^(-20) J` |
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| 284. |
Which of the following is not permissible arrangement of electrons in an atom ?A. `n = 5, l = 3, m = 0, s =+1//2`B. `n = 3, l = 2, m =- 3, =+1//2`C. `n = 3, l = 2, m =- 2, s =- 1//2`D. `n = 4, l = 0, m = 0, s=- 1//2` |
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Answer» Correct Answer - B For an orbital electron whose azimuthal quantum number is `l`, the magnetic quantum number `m_(l)` can have any intergral value form `-l` to `+l`, including zero. Thus, the value of `m_(l)` can never be greater than that of `l`. In the second set. `l = 2`, so `m_(l)` cannot be `-3`. |
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| 285. |
Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is `5.77xx10^(5) m s^(-1)`. |
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Answer» Correct Answer - `10^(-10)m` According to uncertainty principle, `Deltax.m Deltav=(h)/(4pi) or Deltax=(h)/(4 pim Delta v)` `Deltav=5.77xx10^(5)m s^(-1), m=9.1xx10^(-31) "kg" h=6.62xx10^(-34)"kg"m^(2)s^(-1) , pi=3.143` `Deltax=((6.626xx10^(-34)"kg"m^(2)s^(-1)))/(4xx3.143xx(9.1xx10^(-31)"kg")xx(5.77xx10^(5) ms^(-1)))=10^(-10)m`. |
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| 286. |
Calculate the uncertainity in the position of a cricket ball of mass 150 g if the uncertainity in its velocity is`3.53 xx 10^(-24 ) ms^(-1)` |
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Answer» Mass of the ball`= 150 g = 0.150 kg` Uncertainty in velocity `( Deltav) = 3.52 xx 10^(-24) ms^(-1)` `Deltax . m DeltaV ge ( h)/( 4pi)` `Deltax = ( 6.626 xx 10^(-34) kgm^(2) s^(-1))/( 4 xx 3.142 xx ( 0.150 kg ) xx ( 3.52 xx 10(-24) ms^(-1)))` `Deltax = 10^(-10)m` `Deltax = 1Å ( 1 Å = 10^(-10) m ) ` |
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| 287. |
Radial wave function for an electron in hydrogen atom is `Psi = (1)/(16 sqrtpi) ((1)/(a_(0))^(3//2)) [(x -1) (x^(2) - 8x + 12)] e^(-x//2)` where `x = 2r//a_(0), a_(0) =` radius of first Bohr orbit. Calculate the minimum and maximum positions of radial nodes in terms of `a_(0)` |
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Answer» `Psi = (1)/(16 sqrtpi) ((1)/(a_(0))^(3//2)) [(x -1) (x^(2) -8x - 12)] e^(-x//2)` At radial node, `Psi = 0`. Hence, `(x -1) = 0 or (x^(2) -8x - 12) = 0` If `x -1 = 0`, then `x = 1`, i.e., `(2r)/(a_(0)) = 1 or r = (a_(0))/(2)` If `x^(2) -8x + 12 = 0`, i.e.,` (x -6) (x -2) = 0`, then `(x -6) = 0 or (x -2) = 0` If `(x -2) = 0`, then `x = 2 or (2r)/(a_(0)) = 2 or r = a_(0)` If `(x - 6) = 0`, then x = 6 or `(2r)/(a_(0)) = 6 or r = 3 a_(0)` Thus, nodes exist at `(a_(0))/(2), a_(0) and 3a_(0)`. Minimum is at `(a_(0))/(2)` and maximum at `3a_(0)` |
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| 288. |
`Psi^(2) = 0` representsA. a nodeB. an orbitalC. angular wave functionD. wave function |
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Answer» Correct Answer - A `Psi^(2) = 0` means probability of finding the electron in the orbital is zero, i.e., it represents a node |
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| 289. |
Write the electronic configuration of the elements with Z=17 and predict (i) no. of p electrons (ii) no. of filled orbitals (iii) no. of half filled orbitals ? |
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Answer» The element with Z=17 is chlorine. Its electronic configuration is : `1s^(2)2s^(2)2p^(6)3s^(2)3p_(x)^(2)3p_(y)^(2)3p_(z)^(1)` (i) No. of p electrons=11 , (ii) No. of filled orbitals=8, (iii) No. of half filled orbitals =1. |
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| 290. |
Mark the incorect statement regarding the photoelectric effect.A. There is no time lag between the striking of light beam and the ejection of electrons from the metal surface.B. The number of electrons ejected is inversely proportional to the intensity of light.C. Photoelectric effect is not observed below threshold frequency.D. The kinetic energy of the electrons increases with increase in frequency of light used. |
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Answer» Correct Answer - B The number of electrons ejected is directly proportional to the intensity of light. |
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| 291. |
Give the possible value for the missing quantum number (s) in each of the following sets. (a) `n=3, l=0, m_(l)=?` , (b) `n=3, l=?, m_(l)=-1` (c) `n=?, l=1, m_(l)=+1` , (d) `n=?, l=2, m_(l)=?`. |
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Answer» (a) `n=3, l=0, m_(l)=0` , (b) `n=3, l=1, m_(l)=-1` (c) `n=2, l=1, m_(1) =+1` , (d) `n=3, l=2, m_(l) =0`. |
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| 292. |
The most probable radius (in pm) for finding the electron in `He^(+)` isA. `0.0`B. `52.9`C. `26.5`D. `105.8` |
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Answer» Correct Answer - C `r_(n) = (52.9 xx n^(2))/(Z)` pm `:.` For `H^(+), r_(1) = (52.9 xx1^(2))/(2) = 26.5` pm |
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| 293. |
Consider the figure of electromagnetic wave, and choose the correct information related to it.A. These components have same wavelength and speed.B. They vibrate in two mutually perpendicular planes.C. Electromagnetic waves do not require medium and can move in vacuum.D. All of these. |
| Answer» Correct Answer - D | |
| 294. |
Which of the following is not correct?A. A wave is a vibrating disturbance by which enegry is transmitted.B. The speed of a wave depends on the type of wave and the neture of the medium through which the wave is traveling (for example, air water, or vaccum).C. Wave from repeats itself at regular intervals.D. Waves having different wavelengths and frequencies will also have different wave amplitudes. |
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Answer» Correct Answer - D They can have same amplitude as it is the vertical distance from the centre line of a waves to the peak or trough. |
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| 295. |
Match the values of column II with column I and mark the appropriate choice. A. `(A)to(i), (B)to(ii), (C)to (iv), (D)to(iii)`B. `(A)to(iii), (B)to(i), (C)to (ii), (D)to(iv)`C. `(A)to(ii), (B)to(iii), (C)to (iv), (D)to(i)`D. `(A)to(i), (B)to(iii), (C)to (ii), (D)to(iv)` |
| Answer» Correct Answer - B | |
| 296. |
Match the column I with column II and mark the appropriate choice. A. `(A)to(iii), (B)to(iv), (C)to(i), (D)to(ii)`B. `(A)to(ii), (B)to(i), (C)to(iv), (D)to(iii)`C. `(A)to(iv), (B)to(iii), (C)to(i), (D)to(ii)`D. `(A)to(i), (B)to(ii), (C)to(iv), (D)to(iii)` |
| Answer» Correct Answer - B | |
| 297. |
Two atoms are said to be isobars isA. they have same atomic number but different mass number.B. they have same number of electrons but different number of neutrons.C. they have same number of neutrons but different number of electrons.D. Sum of the number of protons and neutrons is same but the number of protons is different. |
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Answer» Correct Answer - D lsobars have same mass number (sum of protons) and neutrons but different atomic numbers (different number of protons). |
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| 298. |
The region where probability density function reduces to zero is calledA. probability density regionB. nodal surfacesC. orientation surfacesD. wave function. |
| Answer» Correct Answer - B | |
| 299. |
The radiation having maximum wave length isA. ultravioletB. radiowaveC. X-rayD. infra-red |
| Answer» Correct Answer - B | |
| 300. |
Which of the following statement about proton is correct?A. It is a nucleus of deuteriumB. It is an ionised hydrogen moleculeC. It is an ionised hydrogen atomD. It is an `alpha`-particle. |
| Answer» Correct Answer - C | |