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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.
| 1051. |
In a certain electronic transition in the hydrogen atoms from an initial state `(1)` to a final state `(2)`, the difference in the orbit radius `((r_(1)-r_(2))` is 24 times the first Bohr radius. Identify the transition-A. `5rarr1`B. `25rarr1`C. `8rarr3`D. `7rarr5` |
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Answer» Correct Answer - A Let state `(1) = n_(1)` state `(2) = n_(2)` `r_(1)-r_(2)=624 r_(0)` `0.529xx(n_(1)^(2))/(Z)-(0.529n_(2)^(2))/(Z)=624xx(0.529xx1)/(Z)` `n_(1)^(2)-n_(2)^(2)=624` `n_(1)=25` `n_(2)=1` `25rarr1` |
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| 1052. |
In a certain electronic transition in the hydrogen atoms from an initial state `(1)` to a final state `(2)`, the difference in the orbit radius `((r_(1)-r_(2))` is 24 times the first Bohr radius. Identify the transition-A. `5 rarr 1`B. `25 rarr 1`C. `8 rarr 3`D. `1 rarr 5` |
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Answer» Correct Answer - A `r_(1) - r_(2) = 0.529 (n_(1)^(2) -n_(2)^(2))` |
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| 1053. |
In a certain electronic transition in the hydrogen atoms from an initial state `(1)` to a final state `(2)`, the difference in the orbit radius `((r_(1)-r_(2))` is 24 times the first Bohr radius. Identify the transition-A. `5 to 1`B. `25 to 1`C. `8 to 3`D. `6 to 5` |
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Answer» Correct Answer - A |
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| 1054. |
if the value of azimuthal quantum number is 3, the possible values of the magnetic quantum number would beA. 0,1,2,3B. 0,-1,-2,-3C. `0,pm1, pm2,pm3`D. `pm1, pm2, pm3` |
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Answer» Correct Answer - C When l=3 then m=-3,-2,-1,0,+1,+2,+3. m=-1 to +1 including zero |
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| 1055. |
The configuration `1 s^2, 2 s^2 2 p^5, 3 s^1` showsA. Excited state of `O_2^-`B. Excited state of neonC. Excited state of fluorineD. Ground state of fluorine atom |
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Answer» Correct Answer - B (b) The ground state of neon is `1 s^2 2 s^2 2 p^2` on excitation an electron from `2 p` jumps to `3 s` orbital. The excited neon configuration us `1 s^2 2 s^2 2 p^5 3 s^1`. |
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| 1056. |
The four quantum number of the valence electron of potassium are.A. `4, 1, 0 and (1)/(2)`B. `4, 0, 1 and (1)/(2)`C. `4, 0, 0 and + (1)/(2)`D. `4, 1, 1 and (1)/(2)` |
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Answer» Correct Answer - C ( c) Atomic number potassium is `19` and hence electronic configuration will be `1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 4 s^1` Hence for `4 s^1` electron value of quantum number are Principle quantum numbert `n = 4` Azimuthal quantum number `l = 0` Magnetic quantum number `m = 0` Spin quantum number `s = + 1//2`. |
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| 1057. |
Calculate the wavelength of the first and the last line in the Balmer series of hydrogen spectrum? |
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Answer» For the first line in Balmer series `n_(1)=2` and `n_(2)=3` `(1)/(lambda_(1))=109,677((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=109677[(1)/((2)^(2))-(1)/((3)^(2))]` `=109677xx(5)/(36)cm^(-1)=15232.98cm^(-1)` `lambda_(1)=6.565xx10^(-5)cm=656.5nm` For the last line in Balmer series `n_(1) = 2` and `n_(2)=oo` `(1)/(lambda_(1))=109677((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=109677[(1)/((2)^(2))-(1)/(oo)]` `=109677xx(1)/(4)cm^(-1)=27419.3cm^(-1)` `lambda_(1)=3.647xx10^(-5)cm=364.7nm` |
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| 1058. |
Which of the following is electronic configuration of `Cu^(2+) (Z = 29)` ?A. `[Ar]4s^1 3d^8`B. `[Ar]4s^2 3d^10 4 p^1`C. `[Ar]4s^1 3d^10`D. `[Ar 3d^9` |
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Answer» Correct Answer - D (d) `Cu : 1 s^2 2 s^2 2 p^6 3 s^2 3 d^10 4 s^1` `:. Cu^(2+) : 1 s^2 2 s^2 2p^6 3 s^2 3 p^6 3 d^9 or [Ar] 3 d^9`. |
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| 1059. |
What will be the differents in the mass number the number is halved and the number of electrons is doubled in `_(8)O^(16)`?A. `25%` decreasesB. `90 %` increasesC. `150 %` increasesD. No difference |
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Answer» `._(8)O^(16)` initial weight final weight `{:("Protons",8p , rarr ,8p),("Neutrons",8n , rarr,4n),("Weight" , 16,rarr, 12):}` Thus decreases in mass number `= 25%` |
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| 1060. |
What is the common nodal plane of `d_(xy)` and `d_(zx)` |
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Answer» (i) shape of dzx and `d_(xy)` orbitals (ii) orienthtion of lobes of `d_(xy)` and `d_(xz)` orbitals (iii) planes where probability of finding an electron is zero |
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| 1061. |
Nuturally occurring boron consists of two isotopes whese atomic weight are `10.01` and `11.01`The atomic weight of nurons is `10.81` Calculate the percentage of each isotope is natural boron |
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Answer» Let the percentage of isotope with atomic weight `10.01` be x . So the percentage of isotope with atomic weight `11.01` will be `100 - x`. Average atomic weight = `(m_(1)x_(1) + m_(2) x_(2))/(s_(1) + s_(2))` `= (s xx 10.01 + (100 - s) xx 11.01)/(100)` `10.81 = (s xx 10.01 + (100 - s) xx 11.01)/(100) = 20` `%` of isotope with atomic weight `10.01 = 20` `%` of isotope with atomic weight `11.01 = 100 - x = 80` |
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| 1062. |
Naturally occureing boron consists of two insotopes whose atomic weight are `10.01 and 11.01 ` The atomic weight of the natural boron is `10.81 ` Calkculate the precentage of e ach isotopes in natURAL BORON |
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Answer» Let the percentage of the isotope be a percentage of the second isotope will be `100 - a ` `10.81 = ((a xx 10.01) + (100 - a) xx 11.01)/(100)` On solving we get `a = 20,(100 - a) = 80` |
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| 1063. |
In the following transition which statement is correct ? A. `E_(3-1) = E_(3-2) - E_(2-1)`B. `lambda_(3) = lambda_(1) + lambda_(2)`C. `v_(3) = v_(2) + v_(1)`D. All of these |
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Answer» Correct Answer - C |
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| 1064. |
A photon of wavelength `5000 lambda` strikes a metal surface with work function `2.20 eV` calculate aTHe energy of the photon in eV b. The kinetic energy of the emitted photo electron c. The velocity of the photoelectron |
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Answer» `E = hv= (hc)/(lambda)` `= ((6.6 xx 10^(-34)) J s(3 xx 10^(8) m s^(-1)))/(5 xx 10^(-3) m) = 3.96 xx 10^(-19) J` `1 eV = 1.6 xx 10^(-19)J` `:. E = (3.96 xx 10^(-19) J)/(1.6 xx 10^(-19) J eV^(-2)) = 2.475 eV` Kinetic energy of the emitted photoelectron work function `= 2.20 eV` Therefore` KE= 2.475 - 2.20 = 0.275 eV = 4.4 xx 10^(-20) J` Velocity of the photoelectron `KE = (1)/(2) mv^(2) = 4.4 xx 10^(-20) J` Therefore velocity (v) `sqrt((2 xx 4.4 xx 10^(-20))/(9.1 xx 10^(-31))) = 3.11 xx 10^(5) m s^(-1)` |
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| 1065. |
If a metal is exposed with light of wavelength `lambda,` the maximum kinetic energy produced is found to be 2 eV. When the same metal is exposed to a wavelength `(lambda)/(5)` the maximum kinetic energy was found to be 14 eV. Find the value of work function (in eV). |
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Answer» Correct Answer - 1 |
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| 1066. |
If `n_(1)+n_(2) = 4 "and" n_(2)^(2)-n_(1)^(2)= 8,` then calculate maximum value of wavelength emitted in transition form `n_(2) to n_(1) "for"Li^(2+) "in nm[Given"R_(H) = 10^(7) m^(-1)].` |
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Answer» Correct Answer - 80 |
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| 1067. |
Calculate the wavelength emitted during the transition of an electron in between two level of `Li^(2+)` ion whose sum is `4` and difference is `2` |
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Answer» Let the transition occure vbetween levels `n_(1) and n_(2) and n_(2)gt n_(1)`1 Given that `n_(1) + n_(2) = 4 and n_(2) - n_(1) = 2` `:. N_(1) = 1 and n_(2) = 3` `:. (1)/(lambda) = R_(H) xx Z^(2)[(1)/((1)^(2)) - (1)/((3)^(2))]` `= 109.678 xx (3)^(2)[(8)/(9)]` `:. lambda = 1.14 xx 10^(-4) cm` |
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| 1068. |
Medical experts generally consider a lead of `30 mu g Pb` per `(dL)` of blood to pose a significant health risk `(Pb = 208)`. Number of lead atoms per `cm^3` of blood is.A. `8.64 xx 10^10`B. `8.86 xx 10^16`C. `8.67 xx 10^12`D. `8.68 xx 10^14` |
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Answer» Correct Answer - D (d) `1 dL = (1)/(10) L = 0.1 L = 100 cm^3` `100 cm^3` of blood has `= 30 mg Pb = 30 xx 10^-6 g Pb` =`(3 xx 10^-6)/(208) mol Pb = (30 xx 10^-6 xx N_0)/(208)` atoms `Pb = 8.6827 xx 10^16 "atoms" Pb` `1 cm^3` f blood has `= 8.627 xx 10^(+14) "atoms" Pb`. |
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| 1069. |
A beam of specific kind of particles of velocity `2.1 xx 10^7 m//s` is scattered by a gold `(Z = 79)` nuclei. Find out specific charge (charge/mass) of this particle if the distance of closest approach is `2.5 xx 10^-14 m`.A. `4.84 xx 10^7 C//g`B. `4.84 xx 10^-7 C//g`C. `2.42 xx 10^7 C//g`D. `3 xx 10^-12 C//g` |
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Answer» Correct Answer - A (a) `(1)/(2) mv^2 = (k(Zq_1)q_2)/( r)` `rArr (q_2)/(m) = (r.v^2)/(2k.q_1.Z)` `(q_2)/(m) = (2.5 xx 10^-14 xx (2.1 xx 10^7)^2)/(2xx 9 xx 10^9 xx 70 xx 1.6 xx 10^-19)` `4.84 xx 10^7 C//g`. |
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| 1070. |
The total allowed values of l for an given value of n are equal to ………….. |
| Answer» Correct Answer - n | |
| 1071. |
One otomic mass unit is quivvalent to …………. Energy |
| Answer» Correct Answer - `931 MeV` | |
| 1072. |
A beam of `aplha` particle moves with a velocity of `3.28 xx 10^(3) m s^(-1)` Calculate the wavelength of the `alpha` particles. |
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Answer» Correct Answer - A::B::C::D `0.3 Å` Hint `lambda = (h)/(m v)` |
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| 1073. |
What is the wavelength associated with welectron traviling at one throusmath the speed of light ? |
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Answer» Correct Answer - A::B::D `4.42 nm` Hint : Speed (v) =` (3 xx 10^(8) m s^(-1))/(1000)` `m_(e) = 9.1 xx 10^(-27) g` `lambda = (h)/(mv)` |
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| 1074. |
The uncertainty in the position of a buller weight `20 g` is `+- 10^(-4) m ` .Calculate the uncertainty in its velocity |
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Answer» Correct Answer - A::B::C::D `2.63 xx 10^(29) m s^(-1)` Hint: `Delta x = (h)/(4pi -m-Delta x) ` |
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| 1075. |
One requiored enrgy ` R_(n)` to remove mucleon and an energy ` E_e` to remove an electron form the orbit of an atom , then :A. `E_n = E_e`B. `E_n lt E_e`C. `E_ngtEe`D. `E_n leE_e` |
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Answer» Correct Answer - C ` E_(n) gt E_(e)` . |
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| 1076. |
The ammount of enrgy requred to remove the electron from a `Li^(2+)` ion in its grojnd state is , how many times greater than the amount of enrgy reuired to remove the elrcton firm an H-atom in its grojn state ?A. `9`B. `2`C. `3`D. `5` |
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Answer» Correct Answer - A `E_(Li^(2+)) = E_(H) xx Z^(2)` ` thereforeE_(1 Li^(2+))/E_(1H) = Z^(2) = 3^(2) =9`. |
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| 1077. |
Statement -I : If the potential difference applied to an electron is made 4 times, the de-Broglie wavelenght assoiated is halved. (Initial kinetic energy of the electron was zero ). Statement -II : On making potential difference 4 times velocity is doubled and hence according to De -broglie hypothesis`lambda` is halved.A. If both Statement -I & Statement -II are True & the Statement -II is a correct explanation of the Statement I.B. If both Statement -I & Statement -II are True but the Statement -II is not a correct explanation of the Statement- I.C. If Statement -I is True but the Statement -II is False.D. If Statement -I is True but the Statement -II is True . |
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Answer» Correct Answer - A `lambda_(1)=sqrt((150)/(V))Å`, `lambda_(2)sqrt((150)/(4V))Å`=`(1)/(2)sqrt((150)/(V))Å`=`(lambda_(1))/(2)` `(1)/(2)m_(e)u_(1)^(2)=eV" "(1)/(2)m_(e)u_(2)^(2)=4eV` `rArru_(1)sqrt((2eV)/(m_(e)))" "rArru_(2)=2xxsqrt((2eV)/(m_(e)))` `lambda_(1)=(h)/("mu"_(1))" "2mu_(1)` `lambda_(2)=(h)/("mu"_(2))`=`(h)/("2mu"_(1))`=`lambda_(1)/(2)` |
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| 1078. |
What are the speed and de broglie wavelength of an electron that has been accelerated by a potent5ial difference of `500 V`? |
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Answer» The KE of the electron under a PD of `500 V `is `1//2 mu^(2) = eV` `u = ((2eV)/(m))^(1//2)` `((2 xx (1.602 xx 10^(-19) C) (500V))/(9.1 xx 10^(-31)kg ))^(1//2)` ` = 1.326 xx 10^(7) m s^(-1)` Using de Broglie equation `lambda = (h)/(mu)` `lambda = (6.26 xx 10^(-34)J s)/((9.1 xx 10^(-31)kg) (1.326 xx 10^(7)ms^(-1))) = 5.5 xx 10^(-11)m` |
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| 1079. |
If `a= (h)/(4 pi^(2)me^(2))` then correct expression for calculate of the first orbit of hydrogen atom isA. `sqrt(4h^(2)) pi a`B. `2 pi r`C. `sqrt(4) pi ha`D. a and c are correct |
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Answer» Circumference `= 2pi r` `2 xx pi xx (n^(2)h^(2))/(4pi^(2)mZe^(2)) ,n = 1,Z = 1 and (h)/(4pi^(2)me^(2)) = a` `Thus 2 xx pi xx h xx a or sqrt(4) pi h a or sqrt(4h^(2)) pi a` |
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| 1080. |
The wavelength of the first line in the hbalmer series is `656 nm ` .Calculate the wavelength of the second line and the limeting line in the Ralmerseries |
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Answer» Wavelength of the second line According to Rydbery formula `(1)/(lambda) = R_(H)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` For the first line `n_(1) = 2,n_(2)= 3 and lambda = 656 nm` `(1)/(656) = R_(H) ((1)/(2^(2)) - (1)/(3^(2)))` ` R_(H) = ((1)/(4) - (1)/(9)) = R_(H) xx (5)/(36) = (5R_(H))/(36)`....(i) For the second line `n_(1) = 2,n_(2) = 4` `(1)/(lambda) = R_(H) ((1)/(2^(2)) - (1)/(4^(2))) = R_(H)((1)/(4) - (1)/(16))` `=R_(H) xx(3)/(16) = (3R_(H))/(16)`.....(ii) Dividing equation (i) by equation (ii),we get `(lambda)/(656) = (5)/(36) xx (16)/(3)`M `or lambda = (656 xx 5xx 16)/(36 xx 3) = (52450)/(108) = 485.9 m` IN balmer series the limiting line coreesponding to transition from `n = oo "to" n = 2` therefore `n_(1) = 2 and n_(2) = oo` `therefore (1)/(lambda) = R_(H)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `(1)/(lambda) = 109677 [(1)/(2^(2)) - (1)/(oo^(2))] = 109677 xx (1)/(4) cm^(-1) = 27419.25 cm^(-1)` Wavelength lambda = `(1)/(27419.25) = 3.47 xx 10^(-5) = 3647 Å` |
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| 1081. |
Calculate the wavelength of the first line in the Balmer series of hydrogen spectrum. |
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Answer» Correct Answer - A::B::D Balmer series `(1)/(lambda) = R [(1)/(2^(2)) - (1)/(n^(2))] ,n = 3,4,5……`when `n = 3` (first number of balmer series) `(1)/(lambda) = 1.09678 xx 10^(7)((1)/(2^(2)) - (1)/(3^(2))) = 1.09678 xx 10^(7) xx (5)/(36)` `:. Lambda = 6564 xx 10^(-10) m = 6564 Å` |
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| 1082. |
For each of the following pairs of the hydrogen orbits indicate which is higher is energyA. `1S ,2S`B. `2p,3p`C. `3d_(xy),3d_(yz)`D. `3s,3d` |
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Answer» For the electron presents in the hydrogen orbital the energy depens only upon the principal quantum number (n) keping this in mind , the orbit of light energy in the given parts is a. `2s` b. `3p` c. Same energy d. `5s` |
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| 1083. |
Calculate the IE a one `Li^(2+)` ion b one mole of `Li^(2+)` ion.Given Rydherg constant `= 1.0974 xx 10^(7)m^(-1)` |
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Answer» For ioniasion `n_(1) = 1,n_(2) = alpha` :. Ifd is given by a. `Delta E = Z^(2)Rhc = (9)(1.974 xx 10^(7)m^(-1))` `xx (6.626 xx 10^(-34) Js)(3 xx 10^(8) ms^(-1))` ` = 19.638 xx 10^(-18) J` b.`Delta E = N_(A) Delta E = (6.023 xx 10^(23)mol^(-1))` ` 1.118 xx 10^(7) J mol^(-1)` `= 1.118 xx 10^(4) KJ mol^(-1)` |
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| 1084. |
The shortest and longest wave number is H spectrum of Lyman series is (R = Rydherg constant)A. `(3)/(4)R,R`B. `(1)/(R ),(4)/(3)R`C. `R,(4)/(3) R`D. `R (3)/(4) R` |
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Answer» Correct Answer - A Shortest`bar v` means shortest E and vice versa When `n = 1,n_(1) = 2` `bar v = R ((1)/(1^(2))- (1)/(2^(2))) = (3)/(4) R` Longest `bar v` means longest E When `n_(1) = 1,n_(1) = oo` `bar v = R ((1)/(1^(2))- (1)/(oo^(2))) = R` |
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| 1085. |
Calculate the IE of (a) one `Li^(2+)` ion (b) one mole of `Li^(2+)` ions. Given Rydberg constant `= 1.0974 xx 10^(7)m^(-1)` |
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Answer» (a). `19.638xx10^(-18)J` (b). `1.118xx10^(4)kJ" "mol^(-1)`. |
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| 1086. |
A spectral line in the spectrum of H atom has a wavelength of `15222.22 cm^(-1)` .The transition responsible for this rediation is (Rydherg constant `R = 10977 cm^(-1)`A. `2 rarr 1`B. `4 rarr 2`C. `5 rarr 2`D. `2 rarr 3` |
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Answer» `lambda = (1)/(bar v) = (1)/(15222.32) = 6.569 xx 10^(-5)cm = 6569 Å`(Visible light wavelength) Clearly it lies in the visible regain i.e. in balmer series Hence `n_(1) = 2` Using the relation for wavelength foir H atom `bar v = (1)/(lambda) = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))` `15222.22=109677(1/2^(2)-1/n_(2)^(2))` `n_(1) = 3` THe required transition is `3 rarr 2` gtHence c is correct |
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| 1087. |
The ionisation energy for the Hydrogen atom in the ground state is `2.18 xx 10^(-18)J "atom"^(-1)`. The energy required for the following process `He^(+)(g)rarr He^(2+) (g) +e^(-)` isA. `8.72 xx 10^(-18)J atom^(-1)`B. `8.72 xx 10^(-19)J atom^(-1)`C. `4.35 xx 10^(-18)J atom^(-1)`D. `2.62 xx 10^(-19)J atom^(-1)` |
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Answer» Correct Answer - A `[IE]_(z) = [IE]_(H) (z^(2))/(n^(2))` |
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| 1088. |
what are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n =2 state in the hydrogen atom? |
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Answer» Since `n_(i) = 5` and `n_(f) = 2`, this transition gives rise to a special line in the visible region of the Bamer series. `DeltaE = 2.18 xx 10^(-18)J [(1)/(5^(2))-(1)/(2^(2))] =- 4.58 xx 10^(-19)J` It is an emission energy. The frequency of the photon (taking energy in terms of magnitude ) is given by `v = (Delta E)/(h) = (4.58 xx 10^(-19)J)/(6.626 xx 10^(-34)Js) = 6.91 xx 10^(14)Hz` `lambda = (c)/(v) = (3.0 xx 10^(8)ms^(-1))/(6.91 xx 10^(14)Hz) = 434 nm` |
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| 1089. |
Calculate the wavelength and energy of radiation for the elctronic transition form infinity to ground state for one H-atom . Given ` e_1 =- 13. 6 eV ( 1 1 eV = 1,. 6 xx 10^(-19) J)`. |
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Answer» Given : `n_1 = 1, n_2 = infty` ` therefore Delta E = E_(infty) - E_1 =0- ( - 13.6 ) = 13.6eV` ` = 13.6 xx 1.602 xx 10^(-19) J = 217.9 xx 10^(-20)J` Also `Delta E = (hc)/( lambda)` ` lambda= ( 6.626 xx 10^(-34 )xx 3.0 xx 10^8)/( 217 .9xx 10^(-20)) = 9. 12 xx 10^(-8) m` `= 912xx10^(-10) m= 912Å`. |
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| 1090. |
How much energy is needed to obtain to H-atom in first excted state from ground state ? |
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Answer» The excitation to first excited state form ground state means that electron in H-atom is boosted upto 2 nd shell from 1st shell . thus,` DeltaE = E_2 -E_1 ( therefore E_n =- ( 13.7)/(n^2) eV //"atom" )` ` = [ ( 13.7)/(4 + 13.7 )] = 10 . 275 eV //"atom"` . |
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| 1091. |
An electronic transition in hydrogen atom result in the formation of `Halpha` line of Hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in `kcal mol^(-1)`) areA. `-313.6,-34.84`B. `-313.6,-78.4`C. `-78.4,-34.84`D. `-78.4,-19.6` |
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Answer» Correct Answer - B `E_(n) = (-313.52)/(n^(2)) xx Z^(2)` |
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| 1092. |
Calcultte the enrgy required for the process , ` He^+ (g) rarr He^(2+) (g) + e` The ionization energy for the H-atom in the grounds state is ` 2. 18 xx 10 ^(-18) J "atom"^(-1)`. |
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Answer» Correct Answer - A `E_(1)` of H atom `= - IE = - 2.18 xx 10^(-18) J` `E_(1)` of H^(o+) = `E_(1)(H) xx Z^(2) = - 2.18 xx 10^(-18) xx 2^(2)J` `=8.72 xx 10^(-18) J` The given process represents ionisation of `He^(o+)` The energy required for ionisation of `He^(o+) = - E_(1) (He) = 8.72 xx 10^(-18) J` |
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| 1093. |
The most probable radius (in pm) for finding the electron in `He^(+)` is.A. `0.0`B. `52.9`C. `26.5`D. `105.8` |
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Answer» Correct Answer - C Most probable radius =`a_0//Z` where `a_0=52.9` pm. For helium ion, Z=2 `r_"mp"=52.9/2=26.45` pm |
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| 1094. |
For azimuthal quantum number l=3, the maximum number of electrons will beA. 2B. 6C. 0D. 14 |
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Answer» Correct Answer - D (d) When azimuthal quantum number is 3 `m=(2l+1)` ` l=3` `m=(2xx3+1)=7` orbitals Then total values of `m=(2xx3+1)=7` orbitals, We know that, one orbital contains two electrons. Hence, total number of electrons `=7xx2=14`. Alternative Total number of electrons `= 4l+2` `=4xx3+2=12+2=14` electrons |
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| 1095. |
An ion has 18 electrons in the outermost shell it isA. `Cu^(+)`B. `Th^(4+)`C. `Cs^(+)`D. `K^(+)` |
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Answer» Correct Answer - A (a) The electronic configuration of Cu, Th, Cs and K are Cu(29) = 2, 8,, 18, 1 `Cu^(+) = 2,8,18` Th(90)= 2, 8, 18, 32, 18, 10, 2 `Th^(4+)=2, 8, 18, 32, 18, 8 Cs(55) = 2, 8, 18, 18, 8,1 `Cs^(+)=2, 8, 18, 18, 8` K(19)=2, 8, 8, 1 `K^(+)=2, 8, 8` Thus, `Cu^(+)` has 18 electrons in the outermost shell. |
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| 1096. |
The atomic number of an element is 35 and mass number is 81.The number of electrons in the outer most shell isA. 7B. 6C. 5D. 3 |
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Answer» Correct Answer - A Bromine consists of outer most electronic configuration `[Ar]3d^10 4s^2 4p^5` |
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| 1097. |
With what velocity should an alpha `(alpha)`- particle travel towards the nucleus of a copper atom arrive at a distance of `10^(-13)m` from the nucleus of the copper atom ? |
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Answer» Correct Answer - `6.3 xx 10^(6)` |
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| 1098. |
For principle quantum number n=4,the total number of orbitals having l=3 isA. 3B. 7C. 5D. 9 |
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Answer» Correct Answer - B `n=4 to 1s^2, 2s^2 , 2p^6, 3s^2, 3p^6, 3d^10 , 4s^2 , 4p^6 , 4d^10 , 4f^14` So l=(n-1)=4-1=3 which is f subshell contain 7 orbital |
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| 1099. |
To what effective potential a proton beam be subjected to give its protons a wavelength of `1xx10^(-10)m`. |
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Answer» Correct Answer - `0.0826` volts `1xx1^(-10)=6.6xx10^(-34)` `=sqrt(2xx1.67xx10^(-27)xx1.6xx10^(-19)xxV)` `therefore1=6.6xx1^(-24)=sqrt(5.344xx10^(-8))eV` `therefore 1 =6.6 xx10^(-20)=sqrt(5.344xxV)` `therefore sqrt(5.344xxV)=6.6xx10^(-20)` |
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| 1100. |
Assertion : The transition of electrons `n_3 to n_2` in H atom will emit greater energy than `n_4 to n_3` Reason : `n_3` and `n_2` are closer to nucleus than `n_4`A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
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Answer» Correct Answer - B Both assertion and reason are true but reason is not the correct explanation of assertion. The difference between the energies of adjacent energy levels decreases as we move away from the nucleus. Thus in H atom `E_2-E_1 gt E_3 -E_2 gt E_4 -E_3`.... |
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