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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.
| 551. |
Maximum number of different spectral lines which will be obtained in visible region when in a sample of large number of H-atoms containing atoms in 2nd , 3rd and 5th excited state only is given by :A. 3B. 4C. 15D. depends on relative number of atoms in 2nd , 3rd and 5th excited state. |
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Answer» Correct Answer - B |
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| 552. |
With a certain radiation of a particular frequency v , to which hydrogen atoms are exposed , the maximum number of spectral lines obtained is 15. The upper most energy level to which the `e^(-)` is excited is n =A. 4B. 5C. 6D. 7 |
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Answer» Correct Answer - C |
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| 553. |
Calculate the frequency and wave number of a radiation having wavelength `600nm` |
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Answer» Wavelength `(lambda)` of the radition `= 600nm = 600 xx 10^(-9)m = 6.0 xx 10^(-7)m` Velocity of radition ( C) `= 3 xx 10^(6)m s^(-1)` `c = v lambda` `:. Frequency v = (c )/(lambda) = (3 xx 10^(6) ms^(-1))/(6 xx 10^(-7)m) = 5 xx 10^(14)s^(-1)` Wavelength `bar v = (1)/(lambda) = (1)/( 6 xx 10^(-7) m) = 1.67 xx 10^(6)m^(-1)` |
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| 554. |
If there are only two H-atoms each is in 3rd excited state then:A. maximum number of different photons emitted is 4B. maximum number of different photons emitted is 3C. minimum number of different photons emitted is 1D. minimum number of different photons emitted is 2 |
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Answer» Correct Answer - A::C |
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| 555. |
Average life time of a hydrogen atom excited to `n=2 ` state is `10^(-8) s.` Find the number of revolutions made by the electron on an average before it jumps to the ground state. If your answer in scientific notation is `x xx 10^(y), ` then find the value of y. |
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Answer» Correct Answer - 6 |
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| 556. |
When a metal is irradiated with light of frequency `4.0 xx 10^(16)s^(-1)` the photo electrons emitted has six times the K.E as the K.E of photo electron emitted when the metal was irradiated with light of frequency `2.0 xx 10^(16)s^(-1)`. Calculate the critical frequency of the metal.A. `2.0 xx 10^(16)s^(-1)`B. `1.6xx 10^(16)s^(-1)`C. `3.0xx 10^(16)s^(-1)`D. `4.2 xx 10^(16)s^(-1)` |
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Answer» Correct Answer - B `(K.E_(2))/(K.E_(1)) = (v_(2)-v_(0))/(v_(1)-v_(0))` |
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| 557. |
The following quantum numbers are possible for how many orbitals `(s) n = 3, l = 2, m = + 2` ?A. 3B. 2C. 1D. 4 |
| Answer» Correct Answer - C | |
| 558. |
Which of the following configuration is correct for ironA. `1s^2 2s^2 2p^6 3s^2 3p^6 3d^5`B. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5`C. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7`D. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6` |
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Answer» Correct Answer - D `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6` it shows electronic configuration of Iron |
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| 559. |
When electron in a Bohr atom is excited then which of the following increases?A. Potential energyB. Time period of revolutionC. Angular momentumD. All of the above |
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Answer» Correct Answer - D |
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| 560. |
The value of the energy for the first excited state of hydrogen will beA. `-13.6` eVB. `-3.40` eVC. `-1.51` eVD. `-0.85` eV |
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Answer» Correct Answer - B `E=(-13.6)n^2 eV=(-13.6)/2^2 =(-13.6)/4 =-3.40` eV |
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| 561. |
The following quantum numbers are possible for how many orbitals `(s) n = 3, l = 2, m = + 2` ?A. 1B. 3C. 2D. 4 |
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Answer» Correct Answer - A (a) Quantum numbers `n = 3, l = 2, m = + 2` represents an orbital with `s = +- (1)/(2) (3 d_(xy) or 3d_(x^2 - y^2))`. |
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| 562. |
The average life of a excited state of hydrogen atom is of the order of `10^(-8)`sec. the number of revolutions made by an electron when it returns from n=2 to n=1 is:A. `2.28xx10^(6)`B. `22.8xx10^(6)`C. `8.23xx10^(6)`D. `2.82xx10^(6)` |
| Answer» Correct Answer - C | |
| 563. |
Which one of the following frequency of radiation (in Hz) has a wavelength of 600 nmA. `2xx 10^(13)`B. `5 xx 10^(16)`C. `2xx 10^(14)`D. `5 xx 10^(14)` |
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Answer» Correct Answer - D `v = (c)/(lambda)` |
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| 564. |
A sodium street light gives off yellow light that has a wavelength of 600 nm. Then (For energy of a photon take `E=(12400eV-"Å")/(lambda("Å")):`A. frequency of this light is `7xx10^(14)s^(-1)`B. frequency of this light is `5xx10^(14)s^(-1)`C. wave number of the light is `3xx10^(6) m^(-1)`D. energy of the photon is approximately 2.07 eV |
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Answer» Correct Answer - B::D |
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| 565. |
Yellow light emitted from a sodium lamp has eavelenght (i) of `500 nm` .Find the frequency (v) wavelength `(barv)` of the yellow light |
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Answer» Wavelength `(lambda)` of the radition `= 580 nm = 580 xx 10^(-19)m = 5.80 xx 10^(-7)m` Velocity of radition `c= 3 xx 10^(8)m s^(-1)` `c = v lambda` :. Frequency `v = (c )/(lambda) = (3 xx 10^(8) ms^(-1))/(5.80 xx 10^(-7)m) = 5.17 xx 10^(14)s^(-1)` Wavelength `bar v = (1)/(lambda) = (1)/( 5.80 xx 10^(-7) m) = 1.72 xx 10^(6)m^(-1)` |
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| 566. |
The radius of electron in the first excited state of hydrogen atom isA. `a_0`B. `4a_0`C. `2a_0`D. `8a_0` |
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Answer» Correct Answer - B `r prop n^2` (excited state n=2) `r=4a^0` |
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| 567. |
The quantum number which may be designated by s,p,d and f instead of number isA. nB. lC. `m_1`D. `m_s` |
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Answer» Correct Answer - B l=0 is s,l=1 is p and l=2 is d and so on hence, s,p,d may be used instead of no |
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| 568. |
Which of the following statement(s) is (are) correctA. The electronic configuration of Cr is `[Ar] 3d^5 4s^1` (Atomic Number of Cr =24)B. The magnetic quantum number may have a negative valueC. In silver atom, 23 electrons have a spin of one type and 24 of the opposite type . (Atomic Number of Ag=47)D. The oxidation state of nitrogen in `HN_3` is -3 |
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Answer» Correct Answer - A::B::C `._24Cr=1s^2 , 2s^2 sp^6 , 3s^2 3p^6 3d^5, 4s^1 =[Ar] 3d^5 , 4s^1` For magnetic quantum number (m), negative values are possible For s-subshell, l=0 , hence m=0 For p-subshell, l=1 , hence m=-1,0,+1 `._47Ag=1s^2 ,2s^2 2p^6 , 3s^2 3p^6 3d^10 , 4s^2 4p^6 4d^10 , 5s^1` Hence 23 electrons have a spin of one type and 24 of the opposite type. oxidation state of N in `NH_3` is -1/3 |
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| 569. |
The velocity of electrons in the 3rd excited state of a hydrogen atom is [`a_(0)` is Bohr radius ]:A. `(h^(2))/(3pi^(2) ma_(0))`B. `(h^(2))/(3pi^(2) ma_(0)^(2))`C. `(8h)/(pima_(0))`D. `(h)/(8 pi ma_(0))` |
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Answer» Correct Answer - D |
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| 570. |
Which of the following is the energy of a possible excited state of hydrogen?A. `-3.4 eV `B. `+6.8` eVC. `+13.6` eVD. `-6.8` eV |
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Answer» Correct Answer - A |
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| 571. |
Identify the correct statement (s) regarding `3p_(z)` orbital.A. Angular part of wave function has `0=90^(@)`B. No. of maximum in `R^(2) (r) us (r) ` curve is 3.C. Total 3 nodal planes are possible.D. No. of radial nodes = 1 |
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Answer» Correct Answer - a d |
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| 572. |
There is no difference between a 2 p and 3p orbital regarding ?A. ShapeB. SizeC. EnergyD. Value of n |
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Answer» Correct Answer - A Both 2p and 3p-orbitals have dumb-bell shape |
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| 573. |
Which of the following statements regarding an orbital are correctA. An orbital is a definite trajectory around the nucleus in which electron can moveB. An orbital always has spherical trajectoryC. An orbital is the region around the nucleus where there is a `90 - 95%` probability of finding all the electrons of an atomD. An orbital is characterized by 3 quantum numbers n,1 andm |
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Answer» Correct Answer - C in an orbital `e^(-)` probability is 95% or more |
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| 574. |
Which of the following is/are correct?A. the energy of an electron depends only on the principal quantum number not on the other quantum numbersB. the energy of an electron depends only on the principal quantum number in case of hydrogen and hydrogen like atomsC. The different in potential energies of any two energy level is always more than the different in kinetic energies of these two levelsD. An electron in ground state can emit a photon |
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Answer» Correct Answer - B::C Quantum numbers concept `(P.E)/(K.E) =- 2` |
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| 575. |
Which of the following statements on the atomic wave function `psi` is not correct?A. `psi` may be a real valued wave functionB. `psi` may be in some cases be a complex functionC. `psi` has a mathematical significance onlyD. `psi` is proportional to the probability of finding an electron |
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Answer» Correct Answer - D probability `alpha|psi^(2)|` |
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| 576. |
Choose the correct statement(s):A. The shape of an atomic orbital depends upon azimuthal quantum numberB. The orientation of an atomic orbital depends upon the magnetic quantum numberC. The energy of an electron in an atomic orbital of multi-electron atom depends upon principal quantum number onlyD. The number of degenerate atomic orbitals of one type depends upon the value of azimuthal quantum number. |
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Answer» Correct Answer - A::B::D Quantum numbers concept |
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| 577. |
What is the full degeneracy of the `n =3` state of a H-atom in the absence of a magnetic field?A. 4B. 10C. 8D. 18 |
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Answer» Correct Answer - D no of orbitals `=n^(2)` and `e^(-1) = 2n^(2)` |
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| 578. |
The isoelectronic pair of `32` electrons isA. `BO_(3)^(3-) and CO_(3)^(2-)`B. `N_(2) and CO`C. `PO_(4)^(2-) and CO_(3)^(2-)`D. All of the above |
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Answer» `BO_(3)^(2-)` `CO_(2)^(2-)` `5 + 24 + 3 = 32` 6 + 24 + 2 = 32` |
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| 579. |
The structure of external most shell of inert gases inA. `s^2p^3`B. `s^2 p^6`C. `s^1 p^2`D. `d^10 s^2` |
| Answer» Correct Answer - B | |
| 580. |
Which of the following sets of quantum number is not possible for an electron?A. n=3, l=+2 , m=0 , `s=+1/2`B. n=3, l=0 , m=0 , `s=-1/2`C. n=3, l=0 , m=1 , `s=+1/2`D. n=3, l=1 , m=0 , `s=-1/2` |
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Answer» Correct Answer - C For s-subshell l=0 then should be m=0 |
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| 581. |
Which of the following sets is possible for quantum numbersA. n=4,l=3,m=-2, s=0B. n=4, l=4, m=+2, `s=-1/2`C. n=4, l=4, m=-2, `s=+1/2`D. n=4, l=3, m=-2, `s=+1/2` |
| Answer» Correct Answer - D | |
| 582. |
Select the correct statement(s):A. Radial distribution function indicates that there is a higher probability of finding the 3s electron close to the nucleus than in case of 3p and 3d electronsB. Energy of 3s orbital is less than for the 3p and 3d orbitalsC. At the node, the value of the radial function changes from positive to negativeD. The radial function depends upon the quantum numbers n and 1 |
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Answer» Correct Answer - A::B::C::D Orbital and Quantum numbers concept |
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| 583. |
Photo electric effect can be explained only by assuming that lightA. is a form of transverse wavesB. is a form of longitudinal wavesC. can be polarisedD. consists of quanta |
| Answer» Correct Answer - D | |
| 584. |
How many sets of four quantum numbers are possible for electron present in `He^(2-)` anionA. 2B. 4C. 5D. 7 |
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Answer» Correct Answer - B no of sets = no of electrons |
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| 585. |
The set of quantum numbers, n = 2, l = 2, `m_(l)` = 0 :A. Describes an electron in a 2s orbitalB. Describes one of the five orbitals of a similar typeC. Describes an electron in a 2p orbitalD. Is not allowed |
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Answer» Correct Answer - D possible values of `m = (2l+1)` |
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| 586. |
Photoelectric effect supports quantum nature fo light because :A. there is a minum freqwuency below which no photoelectrons are emittedB. the maximum kinetic erngy fo the photoelectorns depends only on the frequency fo light and not on its intensityC. eve when the metal surface si fiantyl illuminated the photoelectrons levae the surfae immeditelyD. electric charge fo thephotelcteons is quantixzwed |
| Answer» Correct Answer - A::B::C | |
| 587. |
If the energy different between the electronic stazte in `214.68 mol^(-1)` calculate the frequency of light emited when an electron drop form the hight to the lower state planks constant , `h = 39,79 xx 10^(-14)kJmol^(-1)` |
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Answer» Correct Answer - A::C::D `Delta E = E_(2) - E_(1) = hv` `:. 214.68 = 39.79 xx 10^(-14) xx v` ` or v = 5.39 xx 10^(14) xx cps` |
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| 588. |
Calculate the wavelength of radiation emited when an electron in a hydrogen atom makes a transition from an energy level with `n = 3` to a level with `n= 2` |
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Answer» Correct Answer - A::B::D `E_(n) = -(1312)/(n^(2)) kJ mol^(-1)` `Delta E = E_(2) - E_(1) = -1312 ((1)/(9) - (1)/(4)) ` `= 182.2 k J mol^(-1)` `Delta E "atom"^(-1) = (182.2)/(6.023 xx 10^(23)) k J "atom"^(-1)` `= 3.03 xx 10^(-19) J "atom"^(-1)` `E = hv,v = (c )/(lambda)`, `lambda = (hc)/(E ) = ((6.63 xx 10^(-34))(3 xx 10^(8)))/(3.03 xx 10^ (-19))` `= 6.65 xx 10^(-7) = 656 nm` |
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| 589. |
For an electron the product of velocity fo electron and pricipa quantum no.A. Energy of elecronB. Revolution numberC. Wavelength of electronD. principal quantum number |
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Answer» Correct Answer - D ` u prop 1/n ,` Thus ` u xx n prop 1/2 xx n,` i.e., independent of n. |
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| 590. |
Consider three electron jumps described below for the hydrogen atom `{:["X: n=3 to n=1"],["Y: n=4 to n=2"],["Z: n=5 to n=3"]:}` (a) The photon emited in which trasition ` X, Y` or (Z) will have shortest wavelength ? (b) For which transition will the electron experience the longest charge in robit radius ? |
| Answer» Correct Answer - (a) `X` (b) `Z` | |
| 591. |
The pair having identical velue fo ` e//m:`A. A proton and a neutronB. A proton and a deutrium ionC. An alpha-particel and a deuterium ionD. An elecrron and gamma-rays |
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Answer» Correct Answer - C `e//m` for alpha-particel ` = 2/4 = 1/2 , e//m` for deuterum ion ` 1//2` . |
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| 592. |
Which of the following statements is/are wrong?A. If the value of l=0, the electron distribution is spherical.B. The shape of the orbital is given by magnetic quantum no.C. Angular moment of 1s,2s,3s electrons are equal.D. In an atom, all electrons travel with the same velocity. |
| Answer» Correct Answer - C | |
| 593. |
The angular momentum of electron can have the value (s) :A. `(h)/(2pi)`B. `(h)/(pi)`C. `(2h)/(pi)`D. `(5)/(2)(h)/(2pi)` |
| Answer» Correct Answer - A::B::C | |
| 594. |
Statement-1: Photoelectric effect is easily pronounced by caesium metal. Statement-2: Photoelectric effect is easily pronounced by the metals having high ionization energy.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - C | |
| 595. |
Which of the following is/are correct?A. An electron in excited state cannot absorb a photonB. Energy of electrons depends ony on the principal quantum numbersC. Energy of electrons depends only on the principal quantum number for hydrogen atomD. Difference in potential energy of two shell is equal to the difference in kinetic energy of these shells |
| Answer» Correct Answer - A::C::D | |
| 596. |
Mr.Santa has to decode a number "ABCDEF" where each alphabet is respersented by a single digit . Suppose an orbital whose radial wave functoin id respresented as `Psi_(r)=K_(1).e^_r//k_(2) (r^(2)-5k_(3)r+k_(3)^(2))` From the following information given about each alphabet then write down the answer in the from of "ABCDEF" , for above orbital . Info A= Value of n where "n" is principle quantum number Info B= No .of angular nodes InfoC=Azimuthal quantum number of subshell to orbital belongs Info D= No. of subshells having energy between (n+5)s to (n+5)p where n id principle quantum number Info E=Orbital angular momentum of given orbital. Info F= Radial distance of the spherical node which is farthest from the nuclues (Assuming `K_(3)` =1) |
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Answer» Correct Answer - 30303 Generalized equation of `Psi(g)` `Psi(r)=4p K_(1)e^(_r//k_(2))xxr^(l)` `(polynomial )^(n-l-1)` compare the given equantion with the above equation we get `n-l-1=2" ".........(i)` `l=0" ".........(ii)` n=3 and slove accordingly |
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| 597. |
A particle of charge equal to that of electron and mass 208 times the mass of the electron moves in a circular orbit around a nucleus of charge `+3e.` Assuming that the Bohr model of the atom is applicable to this system find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit of the hydrogen atom. |
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Answer» Correct Answer - 25 |
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| 598. |
`barr,` average distance also described as expectation value of the distance of the electron from the nucleus is different from `r_(max).` For 1s orbital of H-atom, `r_(max)=a_(0), barr=3/2a_(0).` For 2s orbital of H-atom `r_(max) =0.77a_(0) "and" 5.23a_(0).` Find `barr"for"r_(max) = 5.23a_(0).` Also find `barr_(2)` for 1s orbital of `Li^(2+)` ion. Hence find the value of `((barr_(1)xxbarr_(2)))/(a_(0)^(2)).` |
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Answer» Correct Answer - 3 |
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| 599. |
The ionization energy of the ground state of hydrogen atom is `2.18 xx 10^(-8)J`. The energy of an electron in its second orbit would beA. `-1.09 xx 10^(-18)J`B. `-2.18 xx 10^(-18)J`C. `-4.36 xx 10^(-18)J`D. `-5.45 xx 10^(-19)J` |
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Answer» Correct Answer - D `E_(n) = (-E_(1))/(n^(2))` |
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| 600. |
To whom is the discovery of the nuclear atom attributed ?A. Neils BohrB. Louis de BroglieC. Robert MillikanD. Ernest Rutherford |
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Answer» Correct Answer - D |
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