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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.
| 701. |
Electromagentic radiation, which of the following has greater wavelength than visible lightA. U.V raysB. I.R raysC. Gamma raysD. X-rays |
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Answer» Correct Answer - B `gamma`-raus |X-rays|U.V rays|I.R|Ratio `rarr lambda uarr` |
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| 702. |
Photoelectric emission is observed from a surface for frequencies `v_(1) "and" v_(2)` of the incident radiation `(v_(1) gt v_(2))` . If maximum kinetic energies of the photo electrons in the two cases are in the ratio `1:K` , then the threshold frequency is given by:A. `(v_(2)-v_(1))/(K-1)`B. `(Kv_(1)-v_(2))/(K-1)`C. `(Kv_(2)-v_(2))/(K-1)`D. `(v_(2)-v_(2))/(K)` |
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Answer» Correct Answer - B `hv_(1) = hv_(0) +1 rArr h[v_(1) -v_(0)] = 1` ..........(1) `hv_(2) = hv_(0) +k rArr h[v_(2)-v_(0)] = k` .........(2) `((1))/((2)) =(1)/(K) = (v_(1)-v_(0))/(v_(2)-v_(0))` `v_(2) - v_(0) = Kv_(1) - Kv_(0), Kv_(0) - v_(0) = Kv_(1) - v_(2)` `v_(0) [K-1] = Kv_(1) - v_(2) v_(0) = (Kv_(1)-v_(2))/(K-1)` |
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| 703. |
The frequency corresponding to transition `n = 1` to `n = 2` in hydrogen atom is.A. `15.66 xx 10^10 Hz`B. `24.66 xx 10^14 Hz`C. `30.57 xx 10^14 Hz`D. `40.57 xx 10^24 Hz` |
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Answer» Correct Answer - B (b) `v = (1)/(lamda) R[(1)/(n_1^2)-(1)/(n_2^2)]` =`109678[(1)/(1) -(1)/(4)] = 82258.5` `lamda = 1.21567 xx 10^-5 cm` or `lamda = 12.1567 xx 10^-6 cm` =`12.1567 xx 10^-8 m` `v = (c)/(lamda) = (3 xx 10^8)/(12.567 xx 10^-8) = 24.66 xx 10^14 Hz`. |
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| 704. |
Photoelectric emission is observed from a surface for frequencies `v_(1) "and" v_(2)` of the incident radiation `(v_(1) gt v_(2))` . If maximum kinetic energies of the photo electrons in the two cases are in the ratio `1:K` , then the threshold frequency is given by:A. ` ( v_2 - v_1)/( k- 1) `B. ` ( v_2 - v_2)/( k- 1) `C. ` ( v_2 - v_1)/( k- 1) `D. ` ( v_2 - v_1)/( k- 1) ` |
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Answer» Correct Answer - B ` hv_1 = hv_0 + 1/2 mu_1^2` …(1) ` hv_2 = hv_0 + 1/2 mu_1&2` …(2) ` :. 1/2 mi_1^2 = 1/k = 1/k ( 1/2 mu_2^2)` `:.` From eq.(i) ` hv_1 = hv_0 + 1/( 2k) mu_2^2` …(iii) or ` 1/2 mu_2^2 = khv_1 - khv_0` ..(iv) By eqws . (ii) and (iv) ` hv_2 = hv_0 - khv_0 + khv_1` or ` v_0 ( 1 - k ) = v_2 - kv_1` or ` v_0 = ( kv_1 - v_2)/( k -1)`. |
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| 705. |
Photoelectric emission is observed from a surface for frequencies `v_1` and `v_2` of the incident radiation `(v_1 gt v_2)`. If the maximum kinetic energies of the photoelectrons in two cases are in ratio `1 : K` then the threshold frequency `v_0` is given by.A. `(v_2 - v_1)/(K - 1)`B. `(Kv_1 - v_2)/(K - 1)`C. `(K v_2 - v_1)/(K - 1)`D. `(v_2 - v_1)/(K)` |
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Answer» Correct Answer - B (b) `KE_1 = hv_1 - hv_0` `KE_2 = hv_2 - hv_0` `As = (KE_1)/(KE_2) = (1)/(K)` `(h v_1 - hv_0)/( h v_2 - hv_0) = (1)/(K)` `(K v_1 - v_2) = v_0 (K - 1)` `v_0 = (K v_1 - v_2)/(K - 1)`. |
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| 706. |
According to ThomsonA. Negative charge of an atom is uniformly distributed throughout the atom.B. the volume occupied by positive charge is less than that occupied by the negative charge.C. electrons are embebbed in the positive charge which is spread uniformly.D. None of the above |
| Answer» Correct Answer - C | |
| 707. |
Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is.A. `4.41 xx 10^(-16)J "atom"^(-1)`B. `-4.41 xx 10^(-17)J "atom"^(-1)`C. `-2.2 xx 10^(-15)J "atom"^(-1)`D. `-8.83 xx 10^(-17)J "atom"^(-1)` |
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Answer» Correct Answer - B `I.E =- E_(1), E_(n) prop (z^(2))/(n^(2)) rArr ((E_(1))_(He^(+)))/((E_(1))_(Li^(+2))) =((z_(He^(+)))^(2))/((z_(Li^(+2)))^(2))` |
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| 708. |
Elements having the same number of valence electrons in their atoms have similarA. combining capacitiesB. Chemical propertiesC. atomic sizesD. metallic characters |
| Answer» Correct Answer - B | |
| 709. |
Two elements A, B have 14 and 9 electrons in M and N shells respectively. Then the ratio of their atomic numbers isA. `2:3`B. `3:4`C. `3:2`D. `1:2` |
| Answer» Correct Answer - A::C | |
| 710. |
The frequency corresponding to transition `n = 1` to `n = 2` in hydrogen atom is.A. `15.66xx10^10` HzB. `24.66xx10^14` HzC. `30.57xx10^14` HzD. `40.57xx10^24` Hz |
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Answer» Correct Answer - B `barv=1/lambda=R[1/n_1^2-1/n_2^2]=109678[1/1-1/4]=82258.5` `lambda=1.21567xx10^(-5)` cm or `lambda=12.1567xx10^(-6)` cm `=12.1567xx10^(-8)` m `v=c/lambda=(3xx10^8)/(12.1567xx10^(-8))=24.66xx10^14` Hz. |
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| 711. |
The photoelectric emission from a surface starts only when the light incident upon the surface has certain minimum:A. IntensityB. WavelengthC. FrequencyD. Velocity |
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Answer» Correct Answer - C Minimum frequency is threshold prequency |
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| 712. |
The product of which of the following is equal to the velocity of lightA. Wave length and wave numberB. Wave length and frequencyC. Frequency and wave numberD. Wave length and amplitude |
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Answer» Correct Answer - B `v = (c)/(lambda) rArr c = v lambda` |
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| 713. |
The number of valence electrons in `""_(20)^(40)X` isA. 7B. 9C. 5D. 2 |
| Answer» Correct Answer - C::D | |
| 714. |
The splitting of line into groups under the effect of magnetic field is calledA. Zeeman effectB. Stark effectC. Photoelectric effectD. None of these |
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Answer» Correct Answer - A The splitting of spectral line by the magnetic field is called Zeeman effect |
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| 715. |
If a hydrogen atom emit a photon of energy `12.1 eV` , its orbital angular momentum changes by `Delta L. then`Delta L` equalsA. `(h)/(2pi)`B. `(3h)/(2pi)`C. `(h)/(pi)`D. `(2h)/(pi)` |
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Answer» Correct Answer - C `DeltaE = E_(3) - E_(1) = 12.1eV` Electron jumps from 3rd orbit to 1st orbit `mvr = (3h)/(2pi) - (h)/(2pi)` |
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| 716. |
The energy of an electron in the 3rd orbit of an atom is `-E`. The energy of an electron in the first orbit will beA. `-3E`B. `- E/3`C. `- E/9`D. `-9E` |
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Answer» Correct Answer - D `E_3=-E=-13.6xxZ^2/3^2` `rArr -E=E_0xxZ^2/9` [where `E_0`=-13.6] `rArr E_0 =(-9E)/Z^2` ...(i) According to question, `E_1=-13.6 xx Z^2 /1^2 =E_0xxZ^2` `=(-9E)/Z^2 xxZ^2 rArr E_1 = -9E` |
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| 717. |
The isotope with zero neutrons isA. protiumB. deutuiumC. tritiumD. None of these |
| Answer» Correct Answer - A::C | |
| 718. |
The number of valence electrons in `""_(4)X^(8)` atom isA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - B::C | |
| 719. |
Which of the following orbitals has the lowest energy?A. 4dB. 4fC. 5SD. 5P |
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Answer» Correct Answer - C `(n+l)` rule energy order `5s lt 4d lt 5p lt 4f` |
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| 720. |
The frequency of an electromagnetic radiation is `2xx10^6` Hz. What is its wavelength in metres (Velocity of light =`3xx10^8 ms^(-1)`)A. `6.0xx10^14`B. `1.5xx10^4`C. `1.5xx10^2`D. `0.66xx10^(-2)` |
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Answer» Correct Answer - C When `c=vxxlambda` than `lambda=c/v=(3xx10^8)/(2xx10^6)=1.5xx10^2` m |
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| 721. |
Among the following , the orbitals that has the lowest energy isA. 5fB. 4fC. 6sD. 6p |
| Answer» Correct Answer - C | |
| 722. |
The ratio of number of electrons in N shell of A and M shell of B atomic numbers 40 and 32 respectively isA. `5:3`B. `9:5`C. `5:9`D. `5:4` |
| Answer» Correct Answer - C | |
| 723. |
Two elements X and Y have 6 and 7 electrons in their N-shell and M-shell respectively . Find the ratio of atomic numbers of X and Y.A. `3:4`B. `1:2`C. `2:1`D. `6:7` |
| Answer» Correct Answer - C | |
| 724. |
The ratio of the energies of two different radiation whose frequenceies are `3xx10^(14)`Hz and `5xx10^(14)`Hz is `"____________"` |
| Answer» Correct Answer - `3:5` | |
| 725. |
Given is the electronic configuration of element `X` `{:(K,L,M,N,),(2,8,11,2,):}` The number of electrons present with `l = 2` in an atom of element `X` is.A. 3B. 6C. 5D. 4 |
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Answer» Correct Answer - A (a) `X_23 : 1 s^2 2 s^2 2 p^6 3 s^2 3 p^(6) 3 d^3 4 s^2` No. of electron with `l = 2` are `3 (3 d^3)`. |
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| 726. |
The mass of a particle is `1mg` and its velocity is `4.5xx10^(5)cm` per second. What should be the wavelength of this particle if `h=6.652xx10^(-27) "erg second"`.A. `1.4722xx10^(-24)cm`B. `1.4722xx10^(-29)cm`C. `1.4722xx10^(-32)cm`D. `1.4722xx10^(-34)cm` |
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Answer» Given that `m=1 mg =1 xx 10^(-3)g," " c=4.5xx10^(5)cm//sec., " "h=6.652 xx10^(-27)` erg sec. `therefore " "lambda=(h)/(mc)=(6.652xx10^(-27))/(1xx10^(-3)xx4.5xx10^(5))=(6.652xx10^(-29))/(4.5)cm=1.4722xx10^(-29)cm` |
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| 727. |
Consider the ground state `Cr` atom `(Z = 24)`. The number of electron with the azimuthal number `l = 1` and `2` ,respectively areA. 16 and 5B. 12 and 5C. 16 and 4D. 12 and 4 |
| Answer» Correct Answer - B | |
| 728. |
The valency of hydrogen is 1, magnesium, 2 aluminium 3 and carbon 4 . Can you see any connection between the valency of an element and the number of electrons it has in its outermost electron shell ? What would you predict the valencies of helium (He), phosphorus (P), sulphur (S) and neon (Ne) to be ? |
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Answer» The valency of an element depends upon the number of electrons in the outermost shell (valence shell) of an atom of the element. The valency of an element is either equal to the number of valence electrons in an atom of the element or to the number of electrons requried to complete an octet of 8 electrons in its valence shell. Valency of a metal=number of valence electrons. Valency of a nonmetal =8-number of valence electrons. Helium (He) : An atom of helium contains 2 electrons in its K shell (n=1) . This shell is also the outermost shell of helium which is completely filled with 2 electrons. Hence, valency of helium =0 . Phosphorus (P): The electronic configuration of the P atom is 2,8,5. Thus , it has 5 valence electrons and it is a nonmetal . Hence , valency of P =8-5=3. Sulphur (S) : The electronic configuration of an atom of S is 2,8,6. Sulphur is a nonmetal. Hence , valency of S=8-6=2 Neon (Ne): The electronic configuration of neon is 2,8. The outermost shell of neon is thus completely filled. Hence, valency of neon =0. |
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| 729. |
A particle moving with a velocity of `6.626 xx 10^(7)` m/s has a de Broglie wavelength of `1 "Å"` in a circular path of radius `0.529 "Å"` . The angular momentum of particle is (h = `6.626 xx 10^(-34) J xx "sec"`):A. `3.5 xx 10^(-34) kgm^(2) sec^(-1)`B. `3.5 xx 10^(-35) "kg"m^(2) sec^(-1)`C. `3.5 xx 10^(-31) "kg"m^(2) sec^(-1)`D. `1.053 xx 10^(-34) "kg" m^(2) sec^(-1)` |
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Answer» Correct Answer - A |
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| 730. |
A photosensitive metallic surfaces has work function `hv_(0)` . If photon of `2 hv_(0)` fall on the surface ,the electrons come out with a maxiumum velocity of `4 xx 10^(6)m//s` . If energy of photon is increased to `5hv_(0)` , the maximum velocity of photoelectrons will be :A. `2 xx 10^(7) m//s`B. `8 xx 10^(6) m//s`C. `2 xx 10^(6) m//s`D. `8 xx 10^(5) m//s` |
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Answer» Correct Answer - B |
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| 731. |
Which of the following pair of orbitals possess two nodal planesA. `p_"xy", d_(x^2-y^2)`B. `d_"xy", d_"zx"`C. `p_"xy", d_"zx"`D. `d_z^2 , d_(x^2-y^2)` |
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Answer» Correct Answer - B `d_"xy"` and `d_"zx"` has two nodal planes |
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| 732. |
Total numbr of nodal planes aresame in :A. `3s 4d`B. `4s,3p`C. `5s, 4s, 4p`D. `4s, 4p` |
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Answer» Correct Answer - D Total nodes is an orbit is `n-1` |
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| 733. |
Two elecrtrons A and B in an atom have the following set of quantum numbers . What is true for A and B ?A. A and B have same enrgyB. A has more enrgy than (B)C. B has more energy thah AD. (A) and (B) represent same electron. |
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Answer» Correct Answer - B For A : `(n + l) = 5` [ Thus lower is value of `n +l` For B : `n +l =3` lower is nergy of subshell .] |
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| 734. |
An electron in H-atom in its ground state absorbs ` 1.5` times as much energy as the minimum required for its escape ( i. e., 13 . 6 eV) from the atom . Calculate the wavelength of emitted electron. |
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Answer» ` E= hv + 1/2 m u^2` ` ( 13. 6 xx 3)/2 = 13. 6 + 1/2 m u^2` ` 1/2 m u^2 = 6.8 eV = 6. 8 xx 1. 602 xx 10^(-12) erg` ` :. U^2 = ( 6.8 xx 2 xx 1.602 xx 10^(-12))/( 9. 108 xx 10^(28))` `therefore u = 1. 54 xx 10^8 cm //sec `. ` therefore lambda = h/( m u)= ( 6.626 xx 10^(27))/( 9. 108 xx 10^(-28) xx 1. 54 xx 10^8)` ` = 4 . 72 xx 10^(-8) xx 10 ^(-8) cm = 4. 72 Å`. |
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| 735. |
If ` lambda_v, lambda_x` and ` lambda_m` represents the wavelength of visible light X-ray and microwave respectively then :A. `lambda _m gt lambda_v gt lambda_x`B. ` lambda_m gt lambda_xgt lambda_v`C. ` lambda_v gt lambda_m gt lambda_x`D. ` lambda_vgt lambda_xgt lambda_m` |
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Answer» Correct Answer - A ` lambda_m= 1 to 10^(-1) cm`, ` lambda_v = 10^(-4)` to `1-0^(-5) cm`, ` lambda_x = 10^(-6)` cm and above . |
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| 736. |
Identify the incorrectly matched set from the followingA. `{:(list I, listII),("Wavelength","Naometers"):}`B. `{:(list I, listII),("Frequency","Hertz"):}`C. `{:(list I, listII),("Wavenumber",m^(-1)):}`D. `{:(list I, listII),("Velocity","ergs"):}` |
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Answer» Correct Answer - D unit of velocity `m.sec^(-1)` |
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| 737. |
How many spectral linears are emitted by atomic hydrgen excited to the nth energy level ? |
| Answer» Correct Answer - ` 1/2 n*(n-1)` | |
| 738. |
`gamma` rays are electromagnetic radiation of wavelength of `10^(-6)" to" 10^(-5) cm` |
| Answer» `gamma`rays are electromagnetic radiation wavelength of `10^(-9)` to `10^(-10) cm` | |
| 739. |
The wave number of the first line in the Balmer series of hydrogen atom is `15200 cm^(-1)`. What is the wave number of the first line in the Balmer series of `Be^(3+)`?A. `2.432 xx 10^(5)cm^(-1)`B. `15200 cm^(-1)`C. `4xx 15200 cm^(-1)`D. `2 xx 15200 cm^(-1)` |
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Answer» Correct Answer - A `bar(v) = R_(H)z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` |
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| 740. |
Which of the following connot be obtained form the solution of Schrodinger wave equation ?A. Wave function of an electronB. Energy of an elecrtron in a ` 1-D` boxC. Energyof an electron in orbitalsD. Velocity of electron in circular orbits |
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Answer» Correct Answer - D Schrodinger wave equation is based on wave mechanics . |
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| 741. |
Which of the following reflectst the wave nature of light ?A. phopotelectric effectB. `E = mc^2`C. diffractionD. indterfeence |
| Answer» Correct Answer - C::D | |
| 742. |
The hydrogen atom in the ground state is excited by mass of monchromatic radiations of wavelength ` lambda Å ` . The resulting spectrum consists fo maximum `15` different lines . What is the value of ` lambda` ? (`R_H = 109737 cm^(-1))`. |
| Answer» Correct Answer - `937.3Å` | |
| 743. |
Which of the following shows an increasing value of `e//m`?A. `prop gt e^(-) ltp lt n`B. `n lt alpha lt p lt e^-`C. `n lt p lt alpha lt e^-`D. `e^(-) lt p lt n lt alpha` |
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Answer» Correct Answer - D (d) `m` of `alpha` is highest in comparing to other and `m` of electron is minimum in comparing to others. |
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| 744. |
Each hydrogen atom is excited by giving 10.2eV. The maximum number of spectral lines in the emission is equal toA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A `E_(2) -E_(1) = 10.2ev`, so electron is excited to second orbit, therefore no of spectral lines `=(n(n-1))/(2)` |
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| 745. |
The first atom with obncomplete d-shell is :A. `Sc`B. `Cu`C. `Fe`D. `Zn` |
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Answer» Correct Answer - A The Sc atom has ` 3d^2, 4s^2`, configuration . |
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| 746. |
If the uncertainty in the position of an electron is zero the nucertainty in its momentum beA. zeroB. `lt h//(4 pi)`C. `gt h//(4 pi)`D. infinite |
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Answer» Correct Answer - D (d) `Delta x xx Delta p ge (h)/(4)` If `Delta x = 0` then `Delta p = oo`. |
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| 747. |
The energy required to remove and elrctron from the surface fo sodium metal is ` 2. 3 eV`. What is the longest wavelength of radiation with which it can shown photoelectric effect ? |
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Answer» Energy required to shown photo emission ` = 2 . 3 eV = 2.3 xx 1.6 xx 10^(-19 J` Thus ` E= ( hC)/(lambda) "or" lambda = (hc)/E` ` =( 6.626 xx 10^(-34) xx 3xx 10^(8))/( 2.3 xx 1.6 xx 10^(-19)) = 5. 4 xx 10^(-7)m` Thus longest wavelength to shown photoelectric effect ` = 5. 4 xx 10 ^(-7) m`. |
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| 748. |
Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is.A. `4.41xx10^(-16)J atom^(-1)`B. `-4.41xx10^(-17) J atom^(-1)`C. `-2.2xx10^(-15)J atom^(-1)`D. `8.82xx10^(-17)J atom^(-1)` |
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Answer» Correct Answer - B `IE=-E_(1)` `E_(1) for He^(+)=-19.6xx10^(-18)Jat om ^(-1)` `(E_(1))_(He^(+))/((E_(1))_(Li^(2+)))=(Z_(He^(+)))^(2)/((Z_(Li^(2+)))^(2))rArr (-19.6xx10^(-18))/((E_(1))_(Li^(2+)))=4/9` `E_(1)(Li^(2+))=(-19.6xx9xx10^(-18))/4` `-4.41xx10^(-17)Jat om^(-1)` |
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| 749. |
Which of the following has the same mass as that of an electron ?A. PhotonB. NeutronC. PositronD. Proton |
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Answer» Correct Answer - C ( c) Positron `(+ 1e^0)` has the same mass as that of an electron `(-1 e^0)`. |
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| 750. |
The wavelength of the radiation emitted when the electron jumps from 4th shell to 2nd shell isA. `4862 Å`B. `2056Å`C. `5241 Å`D. `109700` cm |
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Answer» Correct Answer - A According to Balmer equation Wave number `(barv)=109677(1/n_(1)^(2)-1/n_(2)^(2))cm^(-1)` `barv=109677(1/((2)^(2))-1/((4)^(2)))cm^(-1)` `=(109677xx3)/16cm^(-1)` `lambda=1/v=16/(109677xx3)=4862xx10^(-8)`cm `=4862xx10^(-10)m=4862 Å` |
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