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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.
| 1601. |
Write the numerical value of h and its unit |
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Answer» `h = 6.625 xx 10^(-27) erg s = 6.625 xx 10^(-34) Js` The unit of `h` is J s or erg s `( :. hv = E,:. h = (E )/(v) = (mg)/(s^(-1)))` |
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| 1602. |
Which of the following statement(s) is (are) correctA. The electronic configuration of Cr (at. No:24) is `[Ar] 3d^5 4s^1`B. The magnetic quantum number may have a negative valueC. In Ag (at. No. 47), 23 electrons have spins of one type and 24 electrons have spins of opposite typeD. The oxidation state of nitrogen in `HN_3` is -3 |
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Answer» Correct Answer - A::B::C Electronic configuration of Cr `to [Ar] 3d^5 4s^1` `to` Let I =1 , `m_1`=-1,0,+1 `to Ag to 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4d^10 5s^1` No. of electrons having `m_s=+1/2 "or" -1/2 =24` and rest 23 `to` oxidation number of N in `NH_3` is `-1/3` |
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| 1603. |
Which of the following is not possible?A. n=4, l=3, m=0B. n=4, l=2 , m=1C. n=4, l=4 , m=1D. n=4 , l=0 , m=0 |
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Answer» Correct Answer - C The highest value of l for a particular n is (n-1) . Hence , n=4, l=4 is not permissible |
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| 1604. |
Which of the following statements concerning light is false?A. it is a part of the electromagnetic spectrum.B. It travels with same velocity i.e., `3xx10^(10)cm//s`.C. It cannot be deflected by a magnet.D. It consists of photons of same energy. |
| Answer» Correct Answer - D | |
| 1605. |
An Eletromagnetic radition of wavelenght `242` nm is just sefficient to ionise a sodium atom .Calculate the ionisation energy of sodium in `KJ mol^(-1)`. |
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Answer» Energy associated with a photon of `242 nm` `= (6.625 xx 10^(-34) xx 3.0 xx 10^(8))/(242 xx 10^(-9))` `= 8.21 xx 10^(-19) J` Since one atom of requires `= 8.21 xx 10^(-19) J` for ionistion `6.023 xx 10^(23)` atoms of Na for ionisation required `8.21 xx 10^(-19) xx 6.023 xx 10^(-23) = 49.45 xx 10^(4)J = 494.5 kJ mol^(-1)` |
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| 1606. |
Which of the followng among the visible colours has the minimum wavelength?A. RedB. BlueC. GreenD. Violet |
| Answer» Correct Answer - D | |
| 1607. |
Of the following the radition having the maximum wavelength isA. UV raysB. RadiowavesC. X-raysD. IR rays |
| Answer» Correct Answer - Radiowaves | |
| 1608. |
Assertion(A): The energy of ultraviolet radiation is greater than the energy of infrared radiation Reason (R): The velocity of ultraviolet radition is greater than the velocity of infrared solution.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 | |
| 1609. |
The wave number which corresponds to electromagnetic radiations of 600 nm is equal to:A. `1.6xx10^(4)cm^(-1)`B. `0.16xx10^(4)cm^(-1)`C. `16xx10^(4)cm^(-1)`D. `160xx10^(4)cm^(-1)` |
| Answer» Correct Answer - A | |
| 1610. |
For n=2 the correct set of azimuthal and magnetic quantum numbers areA. l=2 , m=-2 , -1,0,+1,+2B. l=1 , m=-2 , -1,0,+1,+2C. l=0 , m=-1 , -1,0,+1D. l=1, m=-1 ,0,+1 |
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Answer» Correct Answer - D For n=2, l can be 0 and 1 For l=0 , m=0 and for l=1 , m=-1 , 0,+1 |
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| 1611. |
An element forms diatomic molecule with a triple bond. The configuration of the element may beA. `1s^2 2s^2 2p^5`B. `1s^2 2s^2 2p^6`C. `1s^2 2s^2 2p^3`D. `1s^2 2s^2 2p^4` |
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Answer» Correct Answer - C The element is option (c ) is nitrogen which forms triple bond. |
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| 1612. |
The radius of an orbit in hydrogen atom is equal toA. `n^(2)h^(2)//4pi^(2)mZe^(2)`B. `2pi Ze^(2)//nh`C. `2pi^(2)mZ^(2)e^(4)//n^(2)h^(2)`D. `-2pi^(2)mZ^(2)e^(2)//n^(2)h^(2)` |
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Answer» Correct Answer - A radius `alpha n^(2), alpha(1)/(z^(2))` |
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| 1613. |
The total energy of the electron in any orbit of one electron containing species is given by the expressionA. `-e^(2)//r^(2)`B. `-n^(2)h^(2)//2pi^(2)Z^(2)e^(4)m`C. `-2pi^(2)mZ^(2)e^(4)//n^(2)h^(2)`D. `nh//2pi` |
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Answer» Correct Answer - C Energy `alpha - z^(2), alpha -(1)/(n^(2))` |
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| 1614. |
Spin multiplicity of Nitrogen atom is |
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Answer» Correct Answer - 4 Spin multiplcity `= 2S +1` |
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| 1615. |
Which statement is not true, regarding 2s orbital?A. Number of radial nodes is greater than zeroB. Angular nodes is equal ot zero.C. `Psi(0,phi)= "constant".`D. Probability density is zero at nucleus. |
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Answer» Correct Answer - D |
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| 1616. |
The energy of the elkectron in the `3d` orbital is less than that in the `4s` orbital in the hydrogen atom |
| Answer» The energy of the electron in the `3d` orbital is more than that in the `4s` orbital in the hydrogen atom | |
| 1617. |
The spein multipltcity for the orbital enryron si ` 2s +1`where (s) is total enrctrons pin. The spin multiplicity for stage (I) , (II) and (III)are respectively .A. `1,1,1`B. `1,2,3`C. `1,13`D. `1,3,1` |
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Answer» Correct Answer - C For (I) (Ground state ) Total sjpins ` s = + 1/2 - 1/2 =0` ` :.` Spin multiplicity `= ( 2 xx 0) +1=1` For (II) (singlet excited state ) Total spin ` S= + 1/2 = 1/2 =1` `:.` Spin multiplicity ` = (2 xx 1) = 1 =3`. |
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| 1618. |
How many moles of photon would contain sufficient energy to raise the temperature of 225 g of water `21^(@)C` to `96^(@)C`? Specific heat of water is `4.18Jg^(-1)K^(-1)` and frequency of light radiation used is `2.45xx10^(9)s^(-1)`. |
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Answer» Energy associated with one mole of photons `=N_(0)xxhxxv` `=6.02xx10^(23)xx6.626xx10^(-34)xx2.45xx10^(9)` `=97.727xx10^(-2)Jmol^(-1)`. Energy required to raise the temperature of 225 g of water by `75^(@)C=mxxsxxt=225xx4.18xx75=70537.5J` Hence, number of moles of photons required `=(mst)/(N_(0)hv)=(70537.5)/(97.727xx10^(-2))=7.22xx10^(-4)`mol |
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| 1619. |
The electrons in a poly -electronic atom are filled one by one in roder fo increasing energy level . The enrgy fo subsgells and orientation depeds upon the values of three quantum numbers `9 i.e., n. l and m respectively ) derived from Schrodinger wave equation . The different orbitals fo a subsshells however possess same energy level and are called degenerater orbitals but their enrgy level charges in presenct fo magentic field and the orbitals are non-degenerate . (A) spectral line is noticed it an electron jups form one level to tohere . the paramgnetic nature of eelment sis due to the presence of unpaired electron . The umner of upaired electrons in cr atoms is :A. `2`B. `3`C. `5`D. `6` |
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Answer» Correct Answer - D `Cr (3d^5 , 4s^1)` has six unpaired electron. |
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| 1620. |
The electrons in a poly -electronic atom are filled one by one in roder fo increasing energy level . The enrgy fo subsgells and orientation depeds upon the values of three quantum numbers 9 i.e., n. l and m respectively derived from Schrodinger wave equation . The different orbitals fo a subsshells however possess same energy level and are called degenerater orbitals but their enrgy level charges in presenct fo magentic field and the orbitals are non-degenerate . (A) spectral line is noticed it an electron jups form one level to tohere . the paramgnetic nature of eelment sis due to the presence of unpaired electron . Which is each pair is most stable ion? ` Cu^+` or ` Cu^(2+) ` and ` Fe^(2+)` or ` Fe^(=)` .A. `Cu^+ , Fe^(3+)`B. `Cu^(2+) , Fe^(3+)`C. `Cu^(2+) , Fe^(3+)`D. `Cu^+ , Fe^(3+)` |
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Answer» Correct Answer - B `Cu^+ (3d^(10))` shows disproprotionation and thus less stable than ` cu^(2+)`. |
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| 1621. |
During photosynthesis, chlorophyll absorbs light of wavelength 440 nm and emits light of wavelength 670 nm. What is the energy available for photosynthesis from the absorption-emission of a mole of photons? |
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Answer» `DeltaE=[(Nhc)/(lamda)]_("absorbed")-[(Nhc)/(lamda)]_("evolved")` `=Nhc[(1)/(lamda_("absorbed"))-(1)/(lamda_("evolved"))]` `=6.023xx10^(23)xx6.626xx10^(-34)xx3xx10^(8)[(1)/(440xx10^(-9))-(1)/(670xx10^(-9))]` `=0.1197[2.272xx10^(6)-1.492xx10^(6)]` `=0.0933xx10^(6)J//mol=93.3kJ//mol` |
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| 1622. |
A certain dye absorbs light of `lamda = 4000 Å` and then fluresces light of `5000 Å`. Assuming that under given conditions `50 %` of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of number of quanta emitted out to the number of quanta absorbed.A. `(5)/(8)`B. `(8)/(5)`C. `(3)/(8)`D. `(8)/(3)` |
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Answer» Correct Answer - A (a) `E_(emit ted)= (50)/(100) xx E_(ab so rbed)` No. of emitted photons `xx` Energy of emitted photon `= (50)/(100)` `xx` No . Of absorbed photon xx Energy of absorbed photon. `:. n_e xx (12400)/(5000) = (50)/(100) xx n_a xx (12400)/(4000)` `:. (n_e)/(n_a) = (5)/(8)`. |
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| 1623. |
A certain dye absorbs light of `lamda = 4000 Å` and then fluresces light of `5000 Å`. Assuming that under given conditions `50 %` of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of number of quanta emitted out to the number of quanta absorbed. |
| Answer» Correct Answer - `0. 527` | |
| 1624. |
Cground state element conifiguration of nitrogen atom can be represented asA. B. C. D. |
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Answer» Correct Answer - A::B In single filled orbital electrons must align in one direction or they all must be spin up `(uarr)` or spin-down `(darr)` |
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| 1625. |
The set of quantum numbers for the outermost electron for copper in its ground state is ….A. 4,1,1,+1/2B. 3,2,2,+1/2C. 4,0,0,+1/2D. 4,2,2,+1/2 |
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Answer» Correct Answer - C Electronic configuration of Cu `._29Cu to 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^1` Outermost electron is in 4s sub-shell For 4s n=4, l=0, m=0, s=+1/2 or -1/2 |
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| 1626. |
How many times larger is the spacing between the energy levels with n=3 and n=4,then the spacing between the energy levels with n=8 and n=9 for a hydrogen like atom or ion?A. `0.71`B. `0.41`C. `2.43`D. `14.82` |
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Answer» Correct Answer - B `(r_(4)-r_(3))/(r_(9)-r_(8))=(r_(1)times4^(2)-r_(1)times3^(2))/(r_(1)times9^(2)-r_(1)times8^(2))=(16-9)/(81-64)=0.41` |
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| 1627. |
A certain dye absorbs light of `lambda = 4000 "Å"` and then flurescences light of `lambda = 5000"Å"` . Assuming that under given conditions 50 % of the absorbed energy is re-emitted out as fluorescence , calculate the ratio of the number of quanta emitted out to the number of quanta absorbed :A. `(5)/(8)`B. `(8)/(5)`C. `(3)/(8)`D. `(8)/(3)` |
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Answer» Correct Answer - A |
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| 1628. |
The wave length of a electron with mass `9.1 xx 10^(-31)kg` and kinetic energy `3.0 xx 10^(-25)J` isA. `89.67nm`B. `8.96nm`C. `456.7nm`D. `896.7nm` |
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Answer» Correct Answer - D `lambda = (h)/(sqrt(2mkE))` |
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| 1629. |
The number of orbitals present in the shell with n=4 isA. 16B. 8C. 18D. 32 |
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Answer» Correct Answer - A Number of orbitals =`n^2=4^2=16` |
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| 1630. |
A cricket ball of `0.5kg` moving with a velocity of `100 m s^(-1)`. The wavelength associated with its motion isA. `1//100 m`B. `6.6 xx 10^(-34)m`C. `1.32 xx 10^(-35)m`D. `6.6 xx 10^(-28)m` |
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Answer» Correct Answer - C `lambda = (h)/(mv)` |
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| 1631. |
Arrange the electrons represented by the following sets of quantum in decreasing orbit order of energy sA. `n = 4, l = 0, m = 0, s=+1//2`B. `n = 3, l = 1, m = 1, s=+1//2`C. `n = 3, l = 2, m = 0, s=+1//2`D. `n = 3, l = 0, m = 0, s=+1//2` |
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Answer» a. Represents `4s` electron b. Represents `3p` electron c. Represents `3d` electron d. Represents `3s` electron |
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| 1632. |
Which of the following sets of quantum numbers represents an impossible arrangement?A. `n=3 ,l=2 ,m= -2,s=1//2 `B. `n=4 ,l=0 ,m= 0,s=1//2 `C. `n=3 ,l=2 ,m= -3,s=1//2 `D. `n=5 ,l=3 ,m= 0,s=1//2 ` |
| Answer» For `l = 2, m = -2,m = -2, -10, +1,+2` | |
| 1633. |
Whichis are correct statement.A. The difference in angular momentum associated with the electron present in consecutive orbits of H-aom is `(n-1)(h)/(2pi)`B. Energy difference between energy levels will be changed If ,P.E. at infinty assigend value othe rthan zero.C. Frequncey of spectral line in a H-atom is in the order of`(2rarr1)lt(3rarr1)lt(4rarr1)`D. On moving away from the nucleus , kinetic energy of electron decreases. |
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Answer» Correct Answer - C::D Difference in angular momenture of consective orbits is=`(h)/(2pi)` Energy difference between two orbits its indendent of reference frequcecy of spectral line decrease on decreasing the difference in energy levels. On moving away speed & KE decreases as V `propto(1)/(r)` |
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| 1634. |
A base ball of mass `200 g` is moving with velocity of `3 xx 10^(3) cm s^(-1)`. If we can locate the base ball with an error equal to the magnitude of the wavelength of the light used `(5000 Å)`. How will the uncertainty in momentum be used with the total momentum of the base ball? |
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Answer» Correct Answer - `1.75xx10^(-29)` `DeltaV =30xx10^(2)cm//sec` `lambda=5000Å" "m=200g` `lambda=(h)/(mV)" "500=(h)/(mxxV)` `P=mV=(500)/(6.626xx10^(-26))=30xx10^(2)xx200` `=1.75xx10^(-29)` |
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| 1635. |
Out of the following options in which case in the velocity of electron greater than `1.5xx10^(6)`m//s. {hc=1240 eV-nm}A. An electron moving in the first Bhor radius of hydrogen atomB. The electron which is accelerated from rest positiove by an potenital of `3.4 eV`.C. The electron having deBroglie wavelenght of `2pixx0.529Å`.D. The electron with maximum kinetic energy emitted from a mental whose threshlod wavelenght wavelenght is 400 nm and is subjected to eleectromagnetic radiations having wavelenght 124nm. |
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Answer» Correct Answer - A::C::D (A) `V_(1)=2.188xx10^(6)xx(Z)/(n)m//sec.` `V_(1)(ngt1)=2.188xx10^(6)m//sec.` (B) `3.4xx1.6xx10^(-19)=(1)/(2)xx9.1xx10^(-31)xxV^(2)` `V=(2xx3.4xx1.6)/(9.1)=10^(+12)` `=1.09xx10^(6)m//sec` (c) `2pir=nlambda` `2pi(0.529xxn^(2)/Z)-nxx2pixx0.529` `n=1` `V_(1)=2.18 xx10^(6) m//sec` (D) energy of electron `= 6.9 eVrArr(1)/(2)mv^(2)` `v=1.55xx10^(6)m//sec`] |
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| 1636. |
Wavelength of electron waves in two Bohr orbits is in ratio3:5 the ratio of kinetic energy of electron is 25 : x, hence x is : |
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Answer» Correct Answer - 9 |
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| 1637. |
For two Bohr orbits `n_(1) "and"n_(2)` which follow the relationships, `(n_(2)^(2)=n_(1)^(2))=21 and (n_(2) - n_(1))=3.` If an electron makes transition from `n_(2) "to" n_(1)` directly then :A. third longest wavelength of Balmer series is emittedB. longest frequency of Lyman series is amittedC. longest frequency of Balmer saries is emittedD. only one spectral line is emitted |
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Answer» Correct Answer - a d |
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| 1638. |
The hydrogen -like species ` Li^(2+)` is in a spherically symmetric state ` S_1` whth one radisal node . Upon absorbing light the ion undergoes transitoj ot a state ` S_2` has one radial node and its enrgy is equal to the groun sate energy of hhe hydrogen atom. Energy fo the `(S_1)` in units fo the hydrogen atom groun sate enrgy is :A. `0. 75`B. ` 1. 50`C. ` 2. 25`D. ` 4. 50` |
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Answer» Correct Answer - C ` (E_(s_(1)))/(E_(H (ground)) )= ( -13.6 xx 9)/(4xx(-13 .6))0 = 2.25`. |
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| 1639. |
The hydrogen -like species ` Li^(2+)` is in a spherically symmetric state ` S_1` whth one radisal node . Upon absorbing light the ion undergoes transitoj ot a state ` S_2` has one radial node and its enrgy is equal to the groun sate energy of hhe hydrogen atom. The orbital angular momentum quantum number of the state ` s_2` is : |
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Answer» Correct Answer - B Azimuthal quantum number for ` S-2 = l =1` |
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| 1640. |
Isobars do not differ in the number of _______ . |
| Answer» protons+neutrons | |
| 1641. |
Why does chlorine have fractional atomic mass ? |
| Answer» Existence of isotopes | |
| 1642. |
The frequency of one of the lines in Paschen series of hydrogen atom is `2.340 xx 10^11 Hz`. The quantum number `n_2` Which produces this transition is.A. 6B. 5C. 4D. 3 |
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Answer» Correct Answer - B (b) `overline v = (1)/(lamda) = R_H [(1)/(n_1^2) -(1)/(n_2^2)]` =`(1)/(lamda) = R_H [(1)/(3^2) -(1)/(n_2^2)] = n_1 = 3` for Paschen series. |
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| 1643. |
The number of electrons in the atom which has 20 protons in the nucleusA. 20B. 10C. 30D. 40 |
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Answer» Correct Answer - A Nucleus of 20 protons atom have 20 electrons |
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| 1644. |
The hydrogen -like species ` Li^(2+)` is in a spherically symmetric state ` S_1` whth one radisal node . Upon absorbing light the ion undergoes transitoj ot a state ` S_2` has one radial node and its enrgy is equal to the groun sate energy of hhe hydrogen atom. The orbital angular momentum quantum number of the state ` s_2` is :A. 2B. 1C. 0D. 3 |
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Answer» Correct Answer - B For state `S_2` No. of radial node =1-n-l-1 …(i) Energy of `S_2` state=energy of `e^-` in lowest state of H-atom =-13.6 eV/atom =`-13.6 (3^2/n^2)` eV/atom n=3 Put in equation (i), l =1 so , orbital `rArr` 3p (for `S_2` state). |
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| 1645. |
The mass of an atom is constituted mainly byA. Neutron and neutrinoB. Neutron and electronC. Neutron and protonD. proton and electron |
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Answer» Correct Answer - C Mass of an atom is mainly due to nucleus (neutron + proton ) |
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| 1646. |
An electron beam in accelectrated by a potential difference of `1000 K` what is the wavelength of the w ave associated with the electron beam ? (Mass of electron`= 9.11 xx 10^(-31) kg)` |
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Answer» Kinetic energy of the electron is `1000 eV = 1000 xx 1.60 xx 10^(-19) J = 1.60 xx 10^(-16)J` `KE = (1)/(2) mv^(2)` Velocity of electron `V = ((2KE)/(m))^(1//2) = (2 xx 1.6 xx 10^(-16) Kg m^(2)s^(-2))/(9.11 xx 10^(-31) kg)` `= 5.9 xx 10^(-7) m s^(-1)` From de Brogle relation `lambda = (h)/(mv) = (6.63 xx 10^(-34)"kg" m^(2)s^(-1))/(9.11 xx 10^(-31) kg xx5.9 xx 10^(7) m s^(-1))` `= 0.123 xx 10^(-16) m = 0.123 Å` The wavelength of the electron beam is `0.125 Å` |
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| 1647. |
Number of unpaired electrons in the ground state of beryllium atom isA. 2B. 1C. 0D. All of above |
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Answer» Correct Answer - C `Be_4=1s^2 , 2s^2` =(Ground state) Number of unpaired electrons in the ground state of Beryllium atom is zero |
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| 1648. |
The no of electrons, protons and neutron in a species are equal to 10,11,12 respectively. Assign proper symbol to the species. |
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Answer» No. of protons =11, hence atomic no =11 so the element is Na. It has one electron less than the no. of protons, hence it has a unit +ve charge. No of neutrons =12 Mass number =no of protons +no of neutrons `=11 +12 = 13` Therefore the symbol of that species is `._(11)^(23)Na^(+)` |
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| 1649. |
Which will be the most stble among ` Cu^(+) , Fe^(+) Fe^(2+)` and `Fe^(3+)`.A. `Fe^(2+)`B. `Fe^(+)`C. ` Cu^(+)`D. `Fe^(3+)` |
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Answer» Correct Answer - D `Fe^(3+)` has `3d^(5)` configuration , i. e,. A half fillled one and thuss mor stable ` cu^(+)` no doubt has ` 3d^(10)` configuration but less stable because nuclear cahrge ( 29 protons ) in `Cu` is not sufficient enough to hold a core fo `18` electrons . |
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| 1650. |
The number of electrons, neutrons and protons in a species are equal to 10,8 and 8 respectively. The proper symbol of the species isA. `.^16O_8`B. `.^18O_8`C. `.^18Ne_10`D. `.^16O_8^(2-)` |
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Answer» Correct Answer - D From the given data, it is clear that the atomic no. Z, of the species is 8 (no. of protons). Since the no. of electrons are two more than the no.of protons, hence, it is a binegative species. Thus, the species is `._16O_8^(2-)` |
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