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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.
| 2151. |
The first loinsatisation in electron volts of nitrogen and oxygen atoms are respectively , given byA. `14.6,13.6`B. `13.6,14.6`C. `13.6,13.6`D. `14.6,14.6` |
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Answer» Correct Answer - A Ionisation potential of nitrogen is more of oxygen l.This is because nitrogen has stable fully half filled p orbitals |
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| 2152. |
The orbital diagram in which the Aufhan principle is violated isA. B. C. D. |
| Answer» Correct Answer - B | |
| 2153. |
In a mixture of `He^(Theta)` gas H atom and `He^(Theta)` ions Are excited to three respective first excited subsepuenly , H atom transfers its total excitation energy to `He^(Theta)`ions by collision .Assuming that Bohr model of an atom is applicable , answer the following question The quantum number n of the statement finaly populated is `He^(Theta)` ion is .A. 1B. 2C. 4D. 6 |
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Answer» Correct Answer - C In `H_(1 rarr 2)` the excitation transfer their excitation energy `(10.2 eV)` to the electron in `He^(o+)`is further excited to say `n = n_(2)` thus` 10.2 = 13.6 xx 2^(2) xx (1//2^(2) - 1//n_(2)^(2))` therefore `n_(1) = 4` |
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| 2154. |
In a mixture of `He^(o+)` gas H atom and `He^(o+)` ions Are excited to three respective first excited subsepuenly , H atom transfers its total excitation energy to `He^(o+)`ions by collision .Assuming that Bohr model of an atom is applicable , answer the following questionIf each hydrogen atom in the ground state of `1.0 mol ` of H atom is excited by absorbing photon of energy `8.4 eV, 12.09 eV` and `15.0 eV` of energy then the number of spectral lines emitted is equal toA. 5B. 2C. 3D. 4 |
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Answer» Correct Answer - C Ground state electron in H atom can only be excited by energy greater than `0.2 eV` Thus, `15 eV` photon energy will ionic the atom and `8.4 eV`photon will not be able to higher palte ,New this will be excited to `n = 3` During de-excitation , corresponding to three transition wavelength will be emitted `{:(-0.85 eV .........n = 4),(-1.51 eV .........n = 3),(-3.4 eV .........n = 2),(-13.6 eV .........n = 1):}` |
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| 2155. |
In a mixture of `He^(o+)` gas H atom and `He^(o+)` ions Are excited to three respective first excited subsepuenly , H atom transfers its total excitation energy to `He^(o+)`ions by collision .Assuming that Bohr model of an atom is applicable , answer the following question When an electron of H jumps from a higher to lower energy state ,thereA. Its potential energy decreasesB. Its kinetic energy increasesC. Its angular momentum remain unchangedD. Wavelength of de Broglie wave associated with the electron decrease |
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Answer» Correct Answer - A::B::C As electron from higher to orbit `En darr` hence `KREn uarr [KE = En]` Similarly `Pen darr [Pen = 2En]` Angular momentum `(L) = nh//2pi` hence `1 darr` as `n darr` `lambda_(e) = h//mv_(n)[Ken uarr ` hence Vn darr]` therefore `lambda_(n) darr` |
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| 2156. |
For the electrons of oxygen atom, which of the following statemetns correct?A. `Z_(eff)` for an electron in a 2s orbital is the same as `Z_(eff)` for an electron in a 2pB. An electron in the 2s orbital has the same energy as an electron in the 2p orbitalC. `Z_(eff)` for an an electron in 1s orbital is the same as `Z_(eff)` for an electron in a 2s orbitalD. The two electrons present in the 2s orbital have spin quantum numbers `m_(s)` but if opposite sign |
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Answer» Correct Answer - D (a) Electron in 2s and 2p orbitals have different screen effect. Hence, their Zeff is different Zeff of 2s orbital `gt` Zeff of 2p orbital. Therefore, it is not correct. (b) Energy os 2s orbital `lt` energy of 2p orbital Hence, it is not correct. (c) Zeff of 1s orbital `ne` Zeff of 2s orbital Hence, it is incorrect. (d) For the two electrons of 2s orbital, the valuw of `m_(s)` is `+(1)/(2)` and `-(1)/(2)` Hence, it is correct. |
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| 2157. |
The sum of the number of neutrons and proton in the isotope of hydrogen isA. 6B. 5C. 4D. 3 |
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Answer» Correct Answer - D |
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| 2158. |
Many elements have non-integral atomic masses becauseA. they have isotopesB. their isotopes have non-integral massesC. their isotopes have different massesD. the constituentsm neutrons, protons and electrons, combine to given fractional masses |
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Answer» Correct Answer - A::C |
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| 2159. |
Atomic number of an element represents:A. number of neutrons in the nucleusB. atomic mass of an elementC. valency of an elementD. number of protons in the nucleus. |
| Answer» Correct Answer - D | |
| 2160. |
The radius of an atomic nucleus is of the order ofA. `10^(-10)cm`B. `10^(-13)cm`C. `10^(-15)cm`D. `10^(-8)cm` |
| Answer» Correct Answer - B | |
| 2161. |
When alpha particles are sent through a thin metal foil, most of them go straight through the foil because:A. alpha particles are much heavier than electronsB. alpha particles are positively chargedC. most part of the atom is empty spaceD. alpha particles move with very high velocity. |
| Answer» Correct Answer - C | |
| 2162. |
When alpha particle are sent through a thin metal foil ,most of them go straight through the foil becauseA. alpha particles are much heavier than electronsB. alpha particles are positively chargedC. most part of the atom in empty spaceD. alpha particles move with high velocity |
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Answer» Correct Answer - A::C |
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| 2163. |
When alpha particle are sent through a thin metal foil ,most of them go straight through the foil becauseA. `alpha-particels are mauch haveier ha elecronsB. alpha-particels are positively cahrgedC. most part fo the atom is empty spaceD. apha-partcle mvoe eith higth velocity |
| Answer» Correct Answer - A::C | |
| 2164. |
Wavelength of Radiowaves is :A. `lt` MicrowavesB. `gt` MicrowavesC. `le` Infrared wavesD. `ge` UV rays |
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Answer» Correct Answer - B |
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| 2165. |
Which orbital is represented by the complete wave function , `Psi_(420)` ?A. 4sB. 4pC. 4dD. 4f |
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Answer» Correct Answer - C |
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| 2166. |
The wave function of an orbital is represented as `psi_(4,2,0)`. The azimuthal quantum number of that orbital is |
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Answer» Correct Answer - 2 `Psi_(n,l,m)` is the wave function of given orbital |
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| 2167. |
Statement : The `psi_(640)` represents an orbital . Explanation : The orbital may be ` 6g`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - B (b) `Psi` represents and orbital `Psi_(640)` means `n 6, l = 4, m = 0`, i.e., `6g` orbital. |
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| 2168. |
A hydrogen like species is in a spherical symmetrical orbital `S_(1)` having 3 radial nodes. It gets de-excited to another level `S_(2)` having no radial node. Energy of `S_(2)` orbital is 2.25 times energy of 1st Bohr robit of hydrogen atom. Identify the species involvedgtA. `He^(+)`B. `Be^(+3)`C. `Li^(+2)`D. `B^(+4)` |
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Answer» Correct Answer - c |
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| 2169. |
A hydrogen like species is in a spherical symmetrical orbital `S_(1)` having 3 radial nodes. It gets de-excited to another level `S_(2)` having no radial node. Energy of `S_(2)` orbital is 2.25 times energy of 1st Bohr robit of hydrogen atom. What is the orbital angular momentum quantum number of `S_(2)`?A. 1B. 0C. 2D. `h/(2pi)` |
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Answer» Correct Answer - a |
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| 2170. |
Statement : The `psi_(640)` represents an orbital . Explanation : The orbital may be ` 6g`.A. S is correct but E is wrongB. S is wrong but E is correct.C. Both S and E are correct and E is correct explanation of SD. Both S and E are correct but E is not correct explanation of S |
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Answer» Correct Answer - D `psi` represents an orbital `psi_(640)` means n=6, l=4, m=0, i.e., 6g orbital. |
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| 2171. |
The number of possible lines of Paschenc series when electron jumps from `7^(th)` excited state to ground state ( in hydrogen like atom ) is `:`A. 2B. 5C. 4D. 3 |
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Answer» Correct Answer - 2 Number of lines of Paschen series `=5(8 rarr 3,7 rarr 3,6rarr 3,5rarr 3,4 rarr 3)` |
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| 2172. |
The number of possible line of Paschen series when electron jumps from seventh excited state up to ground state (in hydrogen like atom) is :A. 2B. 5C. 4D. 3 |
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Answer» Correct Answer - B (b) Number of line of Paschen series =`5(8 rarr 3,7 rarr3,6 rarr 3,5 rarr 3,4 rarr 3)`. |
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| 2173. |
Which of the following is the energy of a possible excited state of hydrogen?A. `+13.6 eV`B. `-6.8 eV`C. `-3.4 eV`D. `+6.8 eV` |
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Answer» Correct Answer - C `(E_n)_H=-13.6 1^2/n^2 eV` `n=2 rArr E_2=-3.4 eV` |
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| 2174. |
Which of the following is the energy of a possible excited state of hydrogen?A. `+13.6 eV`B. `-6.8 eV`C. `+3.4 eV`D. `+6.8 ev` |
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Answer» Correct Answer - C |
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| 2175. |
Statement -I : Probability density of electron is zero at nucleus for all p & d orbitals. Statement -II : Probability density of electron at the nucleus depends upon azimuthal quantum number only.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 - C Probability density of electron at the nuclues depends upon principle quantum number as well as azimauthal quamtum number |
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| 2176. |
`P` is the probability of finding the Is electron of hydrogen atom in a spherical shell of infitesimal thickness, dr, at a distance `r` from the nucleus. The volume of this shell is `4pi r^(2)dr`. The qualitative sketch of the dependence of `P` on r isA. B. C. D. |
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Answer» Correct Answer - C |
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| 2177. |
What is the energy, momentum and wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n=2 to n=1? Given that ionization potential is 13.6 eV |
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Answer» Correct Answer - `16.32xx10^(19)J,5.44xx10^(-27)kg m//sec,` 1218Ã… `E_(1)=-13.6eV` `E_(2)=(-13.6)/(4)eV` `DeltaE=(3)/(4)xx13.6eV` `=0.75xx13.6xx1.6xx10^(-19)J=1.632xx10^(-18)J` `(hc)/(lamda)=1.632xx10^(-18)` `lamda=(6.626xx10^(-34)xx3xx10^(8))/(1.632xx10^(-18))=1218xx10^(-10)m=1218`Ã… `lamda=(h)/(p)` `thereforep=(h)/(lamda)=(6.626xx10^(-34))/(1218xx10^(-10))=5.44xx10^(-27)kg-m//sec`. |
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| 2178. |
An isostere isA. `NO_2^-` and `O_3`B. `NO_2^-` and `PO_4^(3-)`C. `CO_2, N_2O, NO_3^-`D. `CIO_4^-` and `OCN^-` |
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Answer» Correct Answer - A `NO_2^(-)` and `O_3` are isostere. The number of atoms in these (=3) and number of electrons (24) are same |
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| 2179. |
The nitride ion in lithium nitride is composed ofA. 7 protons + 10 electronsB. 10 protons + 10 electronsC. 7 protons + 7 protonsD. 10 protons + 7 electrons |
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Answer» Correct Answer - A In case of `N^(3-)` , p=7 and e=10 |
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| 2180. |
The nitride ion in lithium nitride is composed ofA. 7 protons `+7` electronsB. 10 protons `+7` electronC. 7 protons `+10` electronsD. 10 protons `+10` electrons |
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Answer» Correct Answer - C `N^(3-)` |
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| 2181. |
The transition from state `n = 4` to `n = 3` in a `He^(o+)` ion result in ultraviolet radition Intrated radiation will be abtained in the transiion fromA. `n = 2 rarr n = 1`B. `n = 3 rarr n = 2`C. `n = 5 rarr n = 4`D. `n = 8 rarr n = 6` |
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Answer» In case of H-spectrum UV spectrum appose from `n_(1) = 1.to n_(2) = 2,3….` Here in case of `He^(theta)` spectrum UV spectrum is from `n_(1) to n_(2) = 4` `v_(UV)= RZ^(2)(1/3^(2) -(1)/(4^(2))) = R xx 4((1)/(3^(2)) - (1)/(4^(2)))` in case of H-spectrum IR appose from `n_(1) = 3,n_(2) = 4,5....` `v_(m) for H_(2)^(o+) = R xx I^(2)((1)/(3^(2)) - (1)/(4^(2)))`.....(i) `v_(m) for H^(o+) = R xx 4((1)/((n)^(2)) - (1)/(n_(2)^(2))) = R((4)/(n_(1)^(2)) - (4)/(n_(2)^(2)))`......(iii) Comparing the coefficient of equation (i) and (ii) we get `(4)/(n_(1)^(2)) = (1)/(9) ,n_(1) =6` `(4)/(n_(2)^(2)) = (1)/(16) ,n_(2) =8` The transition for `He^(o+)` in IR is `n_(1) rarr 6` to`n_(2) rarr 8` |
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| 2182. |
The ultraviolet radiation has frequency `6xx10^(16)//s` The wavelength would beA. `2xx10^(6)` cmB. `5xx10^(-7)` cmC. `18xx10^(-26)` cmD. `2xx10^(7)` cm |
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Answer» Correct Answer - B `lambda=c/v=(3xx10^(8))/(6xx10^(16))=0.5xx10^(-8)m=5xx10^(-7)` cm |
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| 2183. |
If Aufbau rule is not followed in filling of suborbitals , then block of the element will change inA. `K(19)`B. `Sc(21)`C. `V(23)`D. `Ni(28)` |
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Answer» Correct Answer - A If Aufhan rule is not followed, electronic configuration is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(1)` Last electron is in `3d` (instead of `4s`) hence d block (but it is of s block) |
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| 2184. |
The increasing order of specific charge for electron `(e)` , proton `(p)` , neutron `(n)` and alpha particle`(a)` isA. `e,p,n, alpha`B. `n,p,e,alpha`C. `n,alpha,p,e`D. `n,p,alpha,e` |
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Answer» Correct Answer - C specific charge, `(e)/(m) =("charge")/("mass")` |
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| 2185. |
Calculate the shortest and longest wavelength in hydrogen spectrum of Lyman series. |
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Answer» For Lyman series `n_(1) = 1` For shortest wavelength in Lyman series (i.e, series limit), the energy difference in two states showing transition should be maximum, i.e., `n_(2) = oo` `(1)/(lambda) = R_(H) [(1)/(1^(2))-(1)/((oo)^(2))] = R_(H)` `lambda = (1)/(3xx 109678) = 9.11 xx 10^(-6) cm = 911. 7 A^(@)` For longest wavlength in Lyman series (i,e. first line the energy difference in two states showing transition should be minimum, i.e., `n_(2) = 2` `(1)/(lambda) = R_(H) [(1)/(1^(2))-(1)/((2)^(2))] = (3)/(4) R_(H)` `lambda =(4)/(3) (1)/(R_(H)) = (4)/(3 xx 109677)` `= 1215.7 xx 10^(-8) cm = 1215.7 A^(@)` |
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| 2186. |
If the diameter of a carbon atom is `0.15 nm`, calculate the number of carbon atom which can be placed side by side in a straight line length of scale of length `20 cm` long.A. `13.3 xx 10^(9)`B. `1.33 xx 10^(9)`C. `6.2 xx 10^(9)`D. `1.33 xx 10^(7)` |
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Answer» Correct Answer - B No of carbon items that can be placed in a straight line `= ("length")/("diameter")` |
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| 2187. |
Find out the energy of H atom in the first excitation state .The value of permittivity factor `4pi epsilon_(n) = 1.11264 xx 10^(-10) C^(2)N^(-1)m^(-1)` |
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Answer» In MKS system `E_(n) = (2pi^(2)Z^(2)me^(2))/((4 piepsilon_(0))^(2)n^(2)h^(2))` `= (2 xx (3.14)^(2) xx(1)^(2) xx 9.108 xx 10^(-31) xx (1.602 xx 10^(-19))^(4))/((1.11264 xx 10^(-10))^(2) xx (2)^(2) xx (6.625 xx 10^(-34))^(2))` `= 5.443 xx 10^(-19) J` |
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| 2188. |
When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had three twice the KE as did photoelectrons emitted when the same metal was irradited with light of frequency`2.0 xx 10^(16)s^(-1)` .Calculate the thereshold frequency of the metal |
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Answer» `KE = hv - hv_(0)` we have `hv_(1) hv_(0)=2( hv_(2)-hv_(0))` `v_(0) - 2v_(2) - v_(1)` `= 2(2.0 xx 10^(16)s^(-1)) - (3.2 xx 10^(16)s^(-1))` `= 0.8 xx 10^(16)s^(-1))` |
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| 2189. |
When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had twice the KE as did photoelectrons emitted when the same metal was irradiated with light of frequency`2.0 xx 10^(16)s^(-1)` . Calculate the thereshold frequency of the metal. |
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Answer» Correct Answer - `319.2kJ//mol` `hv_(1)=hv_(0)+2E_(1)" "hv_(2)=hv_(0)+E_(1)` `hv_(1)-w_(0)+2E_(1)" "hv_(2)-w_(0)+E_(1)` `2=(hv_(1)-w_(0))/(hv_(2)-w_(0))" "2 hv_(2)-2w_(0)=hv_(1)-w_(0)` `h[2v_(2)-v_(1)]=w_(0)` `w_(0)=6.62xx10^(-34)(2xx10^(15)-3.2xx10^(15))` `w_(0)=6.62xx10^(-34)xx0.8xx10^(15)` `w_(0)=5.29xx10^(-19)" "w_(0)=318.9kJ//mol` |
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| 2190. |
A certain when irradiated to light `(v = 3.2 xx 10^(16) Hz)` emit photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiation by light `( n = 2.0 xx 10^(16)Hz)` The `v_(0)` Threshold frequency ) of the metal isA. `12 xx 10^(14) Hz`B. `8 xx 10^(15) Hz`C. `1.2 xx 10^(16) Hz`D. `4 xx 10^(12) Hz` |
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Answer» Correct Answer - D `KE = hv_(0) - hv_(0)` `hv_(1) - hv_(0)= 2(hv_(2) - hv_(0))` `v_(0) = 2(v_(2) - v_(1))` `= 2(2.0 xx 10^(-16)) - (3.2 xx 10^(16))` `= 8 xx 10^(15) s^(-1) = 8 xx 10^(-15) Hz` |
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| 2191. |
Ionisation enrgy of a h-like ion (A) is greater than that of another H-like ion (B) Let r, u, e and (L) represent the radius fo the orbit speeed of the electron , energy of the atom and orbital angualr momentum of the electron repectively . In groun state :A. ` r_A gt r_B`B. `u_A gt u_B`C. `E_A lt E_B`D. `LA_A gt L_B` |
| Answer» Correct Answer - C::D | |
| 2192. |
Ionization energy of a hydrogen-like ion A is greater than that of another hydrogen like ion B . Let r, u, E and L represent the radius of the orbit , speed of the electron , total energy of the electron and angular momentum of the electron respectively (for the same n ). In ground state :A. `r_(A) gt r_(B)`B. `u_(A) gt u_(B)`C. `E_(A) gt E_(B)`D. `L_(A) gt L_(B)` |
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Answer» Correct Answer - B |
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| 2193. |
The electromagnetic radiations are, (a) Visible light (b) IR light UV light (d) Micro waves The correct order of increasing energy from lowest to highest isA. `a gt b gt c gt d`B. `a lt b lt c lt d`C. `d lt b lt a lt c`D. `b lt c lt d lt a` |
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Answer» Correct Answer - C `overset(U.V|"visible"||I.R|"micro")underset(Edarr)rarr` |
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| 2194. |
Which transition in `Li^(2 +)` would have the same wavelength as the `2 rarr 4` transition in `He^(+)` ion ?A. `4 to 2`B. `2 to 4`C. `3 to 6`D. `6 to 2` |
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Answer» Correct Answer - C |
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| 2195. |
Select the correct statement about electromagnetic waves.A. They have different wavelength velocity and frequency in vacuum.B. No medium is required for their propagation.C. perpendicular constant electric and magnetic field generates it .D. It consists of transverse vibrations produced from a combination of electric and magnetic fields perpendicular to each other . |
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Answer» Correct Answer - b d |
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| 2196. |
Whenever a hydrogen atom emits photon in the Balmer series:A. it may emit another photon in the Balmer seriesB. it may emit another photon in the Lyman seriesC. the second photon if emitted will have a wavelength of about 122 nmD. it may emit a second photon, but the wavelength of this photon cannot be predicted |
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Answer» Correct Answer - b c |
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| 2197. |
Radiatonal wavelength `(lambda)=124nm` falls on a metallic surface then the kinetic energy of the ejected photo electron(s) can be : [Given that threshold wavelength `(lambda_(0))=248nm]`A. 1 eVB. 2 eVC. 3 eVD. 5 eV |
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Answer» Correct Answer - a b c d |
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| 2198. |
`Psi_((r)) = ke^(-r//k_(1)) dot" r^(2) (r^(2) - k_(2)r + k_(3))` . If the orbital has no nodal plane , then , orbital can be :A. `5d_(xy)`B. `5d`C. `4d_(z^(2))`D. `4 p_(x)` |
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Answer» Correct Answer - A |
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| 2199. |
The potential energy of the electron present in the ground state of ` Li^(2+)` ion is represented by :A. `(3e^(2))/(8pi in_(0)r)`B. `-(3e^(2))/(8pi in_(0)r)`C. `(3e^(2))/(4pi in_(0)r)`D. `-(3e^(2))/(4pi in_(0)r)` |
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Answer» Correct Answer - A |
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| 2200. |
The potential energy of the electron present in the ground state of ` Li^(2+)` ion is represented by :A. `+(e^(2))/(pi in_(0)r)`B. `-(e)/(piin_(0)r)`C. `-(e^(2))/(pi in_(0)r^(2))`D. `-(e^(2))/(pi in_(0)r)` |
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Answer» Correct Answer - D |
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