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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. |
Which of the following set of quantum numbers belong to highest energyA. n=4,l=0,m=0,`s=+ 1/2`B. n=3,l=0,m=0,`s=+ 1/2`C. n=3,l=1,m=1,`s=+ 1/2`D. n=3,l=2,m=1,`s=+ 1/2` |
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Answer» Correct Answer - D Orbitals are 4s, 3s ,3p and 3d . Out of these 3d has highest energy |
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| 252. |
To which electronic transtion between Bohr orbits in hydrogen, the second line in the Balmer series belongs ?A. `3 rarr 2`B. `4 rarr 2`C. `5 rarr 2`D. `6 rarr 2` |
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Answer» Correct Answer - B (b) For Balmer series `n_1 = 2 , n_2 = 3,4, 5….` for second line `n_2 = 4` so `4 rarr 2`. |
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| 253. |
Gives an isobar,isotone , and isotope of `_(6)C^(14)` |
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Answer» `"Isotone of" `_6C^(14)` is `_8O_(16)` `"Isotope of" `._(6)C^(14)` is `_(6)O^(16)` `"Isotope of " `_6C^(14) ` is `_6C^(13)` |
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| 254. |
Give an isobar,isotone , and isotope of `._(6)C^(14)` |
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Answer» Correct Answer - A Isobar `._(7)^(14)N` Isotone `._(8)^(16)O` Isotope `._(6)^(13)C` |
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| 255. |
By a sample of ground state atomic hydrogen ,UV light of energy `(13.6xx48)/(49) (eV)/("quanta")` is absorbed. How many different wavelengths will be observed in Balmer region of hydrogen spectrum? |
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Answer» Correct Answer - 5 |
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| 256. |
Find the number of quanta of radiations of frequency `4.67xx10^(13)s^(-1)`, that must be absorbed in order to melt 5 g of ice. The energy required to melt 1 g of ice is 333J. |
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Answer» Energy requried to melt 5g of ice `=5xx333=1665J` Energy associated with one quantum. `=hv=(6.62xx10^(-34))xx(4.67xx10^(13))` `=30.91xx10^(-21)J` Number of quanta requried to melt 5 g of ice `=(1665)/(30.91xx10^(-21))=53.8xx10^(21)=5.38xx10^(22)`. |
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| 257. |
An iodine dissociates into atom after absorting light of wave length `4500Å` If quantum of radition is absorbed by each molecule calculate the kinetic energy of iodine (Bood energy of `I_(2) is 240 kJ`(mol)) |
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Answer» Energy gives to iodine molecule `(hc)/(lambda) = (6.625 xx 10^(-34) xx 3 xx 10^(8))/(4500 xx 10^(-10))` `4.417 xx 10^(-19) J` Energy used for breaking up `I_(2)` moleculem is ` = (240 xx 10^(5))/(6.025 xx 10^(23)) = 3.894 xx 10^(-19) J` Therefore energy used in impairting kinetic energy to two atoms is `(4.417 - 3.984) xx 10^(-19) J` KE of iodine atom `= ((4.417 - 3.489))/(2) xx 10^(-19)` `= 0.216 xx 10^(-19) J` |
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| 258. |
Photochromic sunglasses, which darken when exposed to light, contain a small amount of colourless AgCl(s) embedded in the glass. When irradiated with light, metallic silver atoms are produced and the lass darkens. `AgCl(s)toAg(s)+Cl` Escape of chlorine atoms is prevented by the rigid structure of the glass and the reaction therefore, reverses as soon as the light is removed. if 310 kJ/mol of energy is required to make the reaction proceed, what wavelength of light is necessary.? |
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Answer» Energy per mole=Energy of one Einstein, i.e., energy of one mole quanta `=(Nhc)/(lamda)` `therefore310xx1000=(6.023xx10^(23)xx6.626xx10^(-34)xx3xx10^(8))/(lamda)` `lamda=3.862xx10^(-7)m=3862xx10^(-10)m=`3862Ã… |
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| 259. |
A bulb of 40 W is producing a light of wavelength 620 nm with `80%` of efficiency , then the number of photons emitted by the bulb in 20 seconds are : (`1eV = 1.6 xx 10^(-19) J , hc = 12400 eV`)A. `2 xx 10^(18)`B. `10^(18)`C. `10^(21)`D. `2 xx 10^(21)` |
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Answer» Correct Answer - D |
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| 260. |
A bulb emits light of wavelength `4500Å` .The bilb is nrated as `150` and `8%` of the energy is emmitted as light .How amny photon are emitted by the bulb per second? |
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Answer» A `150 W` bulb emits `150 J` of energy per second .The energy emitted by the bulb as light is `150 xx (8)/(100) = 12 J` Let the bulb emit n potons per second `E = nhv = (nhv)/(lambda)` `:. N = (E xx lambda)/(c xx h)` `= (12 J xx 45000 xx 10^(-19)m)/(3 xx 10^(8) ms^(-1) xx 6.63 xx 10^(-34) Js)` `= 2.715 xx 10^(19)` |
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| 261. |
The frequency of the rations having wave number `10m^(-1)` is:A. `10s^(-1)`B. `3xx10^(7)s^(-1)`C. `3xx10^(11)s^(-1)`D. `3xx10^(9)s^(-1)` |
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Answer» Correct Answer - D `overline(v)=(1)/(lamda)` `v=coverline(v)=(c)/(lamda)=3xx10^(8)xx10=3xx10^(9)s^(-1)` |
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| 262. |
Find the equivalent current due to motion of electron in first orbit of H-atom.A. `0.7times10^(-3) A`B. `9times10^(-3) A`C. `10^(-3) A`D. None of these |
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Answer» Correct Answer - A `l=theta/T=theta/(2pi//omega)=(thetaomega)/(2pi)=(thetav)/(2pir)` Here,`v=theta^(2)/(2varepsilon_(0)h)=(1.6times10^(-19))^(2)/(2times8.85times10^(-12)times6.63times10^(-34))` =`2.18times10^(6)m//s` `r=n^(2)r_(0)=(1)^(2)times0.53 A=0.53times10^(-10)` `thereforel=(16times10^(-19)times2.18times10^(6))/(3times3.14times0.53times10^(-10))=0.7times10^(-10) A` |
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| 263. |
With increasing member, the energy difference between adjacent levels in atoms.A. decreasesB. increasesC. remains constantD. decreases for low `Z` and increases for high `Z` |
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Answer» Correct Answer - A (a) `E prop (1)/(n^2)` |
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| 264. |
The distance oif closest approach of an alpha-particle fired towards a nucleus with momentum p is r. What will be the distance of closest approach when the momentum of alpha-particle is `2p`?A. `2r`B. `4r`C. `r//2`D. `r//4` |
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Answer» Correct Answer - D KE=PE or`p^(2)/(2m)=1/(4pivarepsilon_(0))cdot(q1q2)/rRightarrowrpropto1/p^(2)` Hence, momentum of alpha-particle will be the closest approach of `r//4`. |
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| 265. |
The energy of a photon of radiation having wavelength 300 nm is,A. `6.63xx10^(29)J`B. `6.63xx10^(-19)J`C. `6.63xx10^(-28)J`D. `6.63xx10^(-17)J` |
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Answer» Correct Answer - B `E=(hv)/(lamda)=(6.626xx10^(--34)xx3xx10^(8))/(300xx10^(-9))=6.63xx10^(-19)J` |
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| 266. |
A `100` W bnulb is emmiting light of wavelength `300` nm .Calculate the number of photon emitted by the bulb in `1` min? |
| Answer» `100 W` bulb is emmiting `100 J` energy per second so energy emitted in `1` min is `100 xx 60 = 6000 J` | |
| 267. |
Which of the following statements (regarding an atom of H ) are correct?A. Kinetic energy of the electron is maximum in the first orbitB. Potential energy of the electron is maximum in the first orbitC. Radius of the second orbit is four times the first orbitD. Various energy levels are equally spaced on energy scale |
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Answer» Correct Answer - a c |
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| 268. |
Assertion (A) : Each electron in an atom has two spin quantum number Reason (R ) : Spin quantum numbers are obtained by solving schrodinger wave equationA. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If both (A) and (R ) are incorrect |
| Answer» Schrodinger wave equation gives only n,l, and m | |
| 269. |
At the closest approach, the distance between Mars and Earth is found to be `58` million `km`. When planets are at this closest distance, how long would it take to send a radio message from a space probe of mars to earth? |
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Answer» Distance between mars and earth `=58` million `km=58xx10^(6)km` Time taken to send a radio message to earth `=("Distance")/("Velocity of light") = ((58xx10^(9)m))/((3xx10^(8)ms^(-1)))=193.3s` |
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| 270. |
A substance absorbs electromagnetic radiations of wavelength 12.3 nm and then emits another electromagnetic radiations of wavelength 24.6 nm . If ratio of number of photons absorbed to number of photons emitted is `2:1` then ratio of energy absorbed to energy emitted will be :A. `2:1`B. `1:1`C. `4:1`D. `1:4` |
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Answer» Correct Answer - C |
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| 271. |
Find the energy of a photon that a corresponds in light frequency at `3 xx 10^(6) Hz` b. Has a wavelength of `300 nm` |
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Answer» Energy of photon `E = hv` `= 6.626 xx 10^(-34) Js xx 3 xx 10^(6)s^(-1)` `= 1.98 xx 10^(-27)J` b. Energy of photon `E = hv = (hc)/(lambda)` `= (6.626 xx 10^(-34) J s xx 3 xx 10^(8) m s^(-1))/(300 xx 10^(-8) m)` `= 6.626 xx 10^(-19) J` |
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| 272. |
The wave number of electromagnetic radiations emitted during the transition of electron in between two levels of `Li^(2+)` ion having sum of the principal quantum numbers 4 and difference is 2 ,will be: (`R_(H)` = Rydberg constant )A. `3.5 R_(H)`B. `4R_(H)`C. `8R_(H)`D. `(8)/(9) R_(H)` |
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Answer» Correct Answer - C |
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| 273. |
The velocity of electron in first orbit of H-atom as compared to the velocity of light isA. `(1)/(10)`thB. `(1)/(100)`thC. `(1)/(1000)`thD. Same |
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Answer» Correct Answer - B `C = 3 xx 10^(10)cm//sec, V = 2.18 xx 10^(8) cm//sec`, |
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| 274. |
Calculate and compare the energies of two radition one with a wavelength of `300 nm` and the other `600` nm |
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Answer» Energy of radiation `:. E_(1) = (hc)/(lambda_(1)) = (6.626 xx 10^(-34) Js xx 3xx 10^(8) m s^(-1))/(300 xx 10^(-9) m ) = 6.626 xx 10^(-19) J` and `E_(2) = (hc)/(lambda_(2)) = (6.626 xx 10^(-34) Js xx 3xx 10^(8) m s^(-1))/(600 xx 10^(-9) m ) = 3.313 xx 10^(-19) J` The ratio of `E_(1)` and `E_(2)` is `(E_(1))/(E_(2)) = (6.262 xx 10^(-19)J)/(3.313 xx 10^(-19)J) = 2` `:. E_(1) = 2E_(2)` |
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| 275. |
Compare the energies of two radiations on with `l = 600 nm` and other with 30 nm. |
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Answer» `(E_(1))/(E_(2)) = (lambda_(2))/(lambda_(1)) = (30)/(600) = (1)/(20)` The relation between two energies is `E_(2) = 20E_(1)` |
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| 276. |
The configureation is `1s^(2) 2s^(2) 2p^(5)` , `3s^(1)` shows :A. Groun state fo fluorineB. Exctred state fo fluoineC. Excite state of noen atomD. Excited state fo `O^(-)` ion |
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Answer» Correct Answer - C Ground state configuration of Ne is ` 1s^(2) 2s^(2) 2p^(6)`. |
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| 277. |
How many unpaired electrons are there in `Ni^(2+)`? |
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Answer» Correct Answer - B `Ni(Z= 28) or 3d^(6) 4S^(8)4S^(2)`ltBRgt `:. Ni^(2+) = d^(8)` |
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| 278. |
Calculate the wavelength in Angstroms of the photon that is emitted when an electron is Bohr orbit `n = 2` return to the orbit `n = 1` in the hydrogen atom .The ionisation potential of the ground state hydrogen atom is `2.17 xx 10^(-11)` ergs per atom |
| Answer» Correct Answer - A::B | |
| 279. |
The increasing order (lowest first) for the values of `e//m` (charge//mass) for electron `(e)`, proton `(p)`, neutron `(n)`, and alpha particle `(alpha)` isA. e,p,n,`alpha`B. n,p,e, `alpha`C. n,p,`alpha`,eD. n,`alpha`,p,e |
| Answer» Correct Answer - D | |
| 280. |
The increasing order (lowest first) for the values of `e//m` (charge//mass) for electron `(e)`, proton `(p)`, neutron `(n)`, and alpha particle `(alpha)` isA. `e,p,n,alpha`B. `n,p,e,alpha`C. `n,p,alpha,e`D. `n,alpha,p,e` |
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Answer» Correct Answer - D `e/m` for (i) neutron =`0/1`=0 (ii)`alpha`-particle=`2/4=0.5` (iii)proton=`1/1` =1 (iv)electron=`1/(1//1837)=1837` |
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| 281. |
The increasing order (lowest first) for the values of q/m (charge/mass) for electrons (e) , proton (p), neutron (n) and alpha particles (`alpha`) is :A. e,p,n, `alpha`B. n,p,e,`alpha`C. n,p, `alpha`, eD. n, `alpha`, p, e |
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Answer» Correct Answer - D |
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| 282. |
The increasing order (lowest first) for the value of `e//m ("charge" //"mass")` for electron(e), proton (p) neutron (n) and alpha particle `(alpha)` isA. ` e, p, n. alpha`B. `n, alpha, p, e`C. ` n, p, e, alpha`D. ` n, p, alpha, e` |
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Answer» Correct Answer - B Charge on neutons is zero and mass of electron is minimum . |
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| 283. |
How many unpaired electrons are there in `Ni^(2+)`?A. ZeroB. `2`C. `4`D. `8` |
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Answer» Correct Answer - B ` Ni^(2+) : 1 s^2 , 2s^2 2p^7 , 3p^6 3p^6 3d^8` , Removal of electron should be from outermost shell . |
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| 284. |
The azimuthal quantum number and the principal quantum number of the 17th electron areA. `l = 1, n = 3`B. `l = 3, n = 2`C. `l = 1, n = 17`D. `l = 2, n = 1` |
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Answer» Correct Answer - A 3p orbitals |
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| 285. |
What is the maximum number of electron that may be present in all the atomic orbitals with principal quantum number `3` and azimuthal quantum number` 2`? |
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Answer» Correct Answer - A::C The primncipal quantum number `3` and the Azimuthal quantum number `2` mean `3d` subshell A d subshell five orbitals, and each of them can have a maximum of two electrons , therefore , this sub-shell can have a maximum of `10` electron |
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| 286. |
Which electronic transition in atomic hydrogen corresponds to the emission of visible light ?A. `n = 5 to n = 2`B. `n = 1 to n = 2`C. `n = 3 to n = 4`D. `n = 3 to n = 1` |
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Answer» Correct Answer - A |
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| 287. |
When the atomic electron is at infinite distance from the nucleus, its energy isA. infinityB. zeroC. negativeD. positive |
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Answer» Correct Answer - B `E_(n) = (-2pi^(2)mZ^(2)e^(4))/(n^(2)h^(2))`, if `n = oo` then `E = 0` |
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| 288. |
What would be the maximum number of emission lines for atomic hydrogen that you would expert to see with the naked eye if the only electronic energy involved are those as shown in figure ? A. 4B. 6C. 5D. 15 |
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Answer» Correct Answer - A |
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| 289. |
During one of its ambitious project 'ISRP" gathered information through its stallites for a planet in another galaxy. As far as structure of atoms is considered , it is very similar to the structure of an atom in our planet except that energy of an orbit in which electron resides is `(E_(n))=(45xxZ)/(n=1) eV//"atom".` Calculate wavelength `("in""Å")` of photon emitted when electron makes transition from 7th excited state to 1st excited state in H-atom of that planet:A. `1240 "Å"`B. `12400 "Å"`C. `310 "Å"`D. `3100 "Å"` |
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Answer» Correct Answer - a |
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| 290. |
During one of its ambitious project 'ISRP" gathered information through its stallites for a planet in another galaxy. As far as structure of atoms is considered , it is very similar to the structure of an atom in our planet except that energy of an orbit in which electron resides is `(E_(n))=(45xxZ)/(n=1) eV//"atom".` If photon of wavelength 7nm strikes on an atom of hydrogen on that planet, then wavelength of ejected electron will be approximately:A. `0.1 "Å"`B. `1 "Å"`C. `5 "Å"`D. `11 "Å"` |
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Answer» Correct Answer - b |
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| 291. |
During one of its ambitious project 'ISRP" gathered information through its stallites for a planet in another galaxy. As far as structure of atoms is considered , it is very similar to the structure of an atom in our planet except that energy of an orbit in which electron resides is `(E_(n))=(45xxZ)/(n=1) eV//"atom".` If collection of H-atom and `He^(+)` ion makes transition from 2nd orbit and 5th orbit respectively, then total different spectral lines emitted will be :A. 10B. 11C. 6D. 9 |
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Answer» Correct Answer - a |
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| 292. |
Monochromatic radiation of specific wavelength is incident on H-atoms in ground state. H-atoms absorb energy and emit subsequently radiations of six different wavelength. Find wavelength of incident radiations:A. `9.75 nm`B. `50nm`C. `85.8nm`D. `97.25nm` |
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Answer» Correct Answer - D For six different spectral lines, requilred transitions is `4 rarr 1` `(1)/(lambda) = R_(H)z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` |
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| 293. |
Monochromatic radiation of wavelength `lambda` are incident on a hydrogen sample in ground state. Hydrogen atoms absorb the light and subsequently emit radiations of 10 different wavelength . The value of `lambda` is nearly :A. 203 nmB. 95 nmC. 80 nmD. 73 nm |
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Answer» Correct Answer - B |
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| 294. |
Monochromatic radiation of wavelength `lambda` is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiations of six different wavelength . Find the wavelength `lambda`.A. 97.5 nmB. 121.6 nmC. 110.3 nmD. None of these |
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Answer» Correct Answer - A |
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| 295. |
The second ionization potential of Be is `17.98 eV` if the electron in Be is assumed to move in a spherical orbit with a centeral field of effective nuclearcharge `(Z_(6H))` consisting of the nucless and otherelectron by haw many units of charge in the nucless shicided by other electrons? (the energy of electrons in first Bohr of H is `-13.6 eV)` If the extent of shielding by the if electron of Li atom is the same as you have calculated above , find the ionisation potential of Li |
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Answer» `IE = (13.6Z_(eff)^(2))/(n^(2)) = 17.98` `Z_(eff) = 2.3` Shieiding extent `= Z - Z_(eff) = 4 - 2.3 = 1.7 ` `Z_(eff)` for lithium `= 3 - 1.7 = 1.3 ` `:.IE of lithium = 13.6 xx (1.3)^(2) = 5.746 eV` |
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| 296. |
The ionization enthalpy of hydrogen atom is `1.312 xx 10^6 J mol^-1`. The energy required to excite the electron in the atom from `n= 1` to `n = 2` is :A. `8.51 xx 10^(5)J mol^(-1)`B. `6.56 xx 10^(5)J mol^(-1)`C. `7.56 xx 10^(5)J mol^(-1)`D. `9.84 xx 10^(5)J mol^(-1)` |
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Answer» Correct Answer - D `DeltaE = E_(2) - E_(1) [:. I.E. =- E_(1)]` `= (-1.312 xx 10^(6))/(2^(2))-((-1.312xx10^(6))/(1))` |
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| 297. |
A metal surface ejects electrons when hit by green light but none when hit by yellow light. The electrons will be ejected when the surface is hit byA. yesB. NoC. Yes, if the red bream is quite intenseD. Yes, if the red beam continues to fall upon |
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Answer» Correct Answer - B Red light has less frequency than green. So, it will not cause ejection of electrons |
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| 298. |
Which of the following violates the Pauli exclusion principle :-A. `uarrdarruarrdarr`B. `uarrdarruarrdarruarr`C. `uarruarruarruarruarr`D. `uarruarruarruarr` |
| Answer» Correct Answer - C | |
| 299. |
Assertion(A): Threshold frequency is a characteristic for a metal Reason(R): Threshold frequency is a maximum frequency required for the ejection of electron from the metal surface.A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - C | |
| 300. |
Which of the following gives neither emission spectrum nor absorption spectrum?A. `He^(+)`B. `H_(2)`C. `H^(+)`D. `He` |
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Answer» Correct Answer - C It has no electron |
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