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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.
| 2051. |
Assertion (A) : The ionisation energy of N is more than that of O Reason (R ) : Electronic configuration of N is more stable due to half - fillied `2p` orbitalsA. 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 (A) is incorrect but (R ) is correct |
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Answer» Correct Answer - A N has electronic configuration `= 1s^(2) 2s6(2)3s^(2)3p^(3)` O has electronic configuration `= 1s^(2) 2s6(2)3s^(2)3p^(6)` N is stable due to half -filled p obtained . |
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| 2052. |
It is teming to think that all possible transituion are permissible and that an atomic spectrum series from the transition of an electron from any intial orbital to any other .However this is not so because a photon a photon has as intrinsic spin angular momentum of `sqrt(2) h// 2pi` corresponding to `S = 1` although it has no charge and no rest mass On the other hand , an electron has got two typwe of agular momentum orbit angular momentum `L = [sqrt(l(l+1))] h//2pi`,and spin angular momentum `L_(1) = sqrt(s(s + 1)) h//2pi` arising from orbital motion and spin motion of the electronn during any electton transition must compentum for the angular momentum carried away by the photon .To salary this condition the different between the azisition quantum number of teh orbital witjhin which the transition `(l = 2)` cannot make a transition into as x-orbital `(l = 0)`because the photon cannot carry away enough angular momentum The maximum orbital angular momentum of an electon with `n = 5` isA. `sqrt(6) (h)/(2pi)`B. `sqrt(12) (h)/(2pi)`C. `sqrt(42) (h)/(2pi)`D. `sqrt(20) (h)/(2pi)` |
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Answer» Correct Answer - D Maximum allowed value of `l = 4` for ` n = 5`1 `:. L = sqrt(4(4 +1)) (h)/(2pi) = sqrt(20)(h)/(2pi)` |
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| 2053. |
It is teming to think that all possible transituion are permissible and that an atomic spectrum series from the transition of an electron from any intial orbital to any other .However this is not so because a photon a photon has as intrinsic spin angular momentum of `sqrt(2) h// 2pi` corresponding to `S = 1` although it has no charge and no rest mass On the other hand , an electron has got two typwe of agular momentum orbit angular momentum `L = [sqrt(l(l+1))] h//2pi`,and spin angular momentum `L_(1) = sqrt(s(s + 1)) h//2pi` arising from orbital motion and spin motion of the electronn during any electton transition must compentum for the angular momentum carried away by the photon .To salary this condition the different between the azisition quantum number of teh orbital witjhin which the transition `(l = 2)` cannot make a transition into as x-orbital `(l = 0)`because the photon cannot carry away enough angular momentum Electron transition from `4s` to `3s` orbital is forbiddeon meating that it cannot becauseA. There will be no change in the orbital angular momentum of electron athough the emitted photon has angular momentumB. There will be change in the orbital angular momentum whereas the emitte photon has to momentumC. `Delta m_(1) `valuee between `4s`1 and `3s` is not zero , which is an important selection slection rule for allowed transitionD. In `4s` and `3s` orbitals the wavelength of the electeron wave `n = 5` is |
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Answer» Correct Answer - A a. It does not involve any change in the value of l value of l should change by `+- 1` b. There will be no change in angular momentum because photon have angular momentum c. `Delta m` for `4s` and `3s` is zero d. Wavelength of electron wave are not the same because n value is difference |
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| 2054. |
The effect of splitting of spectral lines under the influence of magnetic field is calledA. photoelectric effectB. Zeeman effectC. Raman effectD. Stark effect |
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Answer» Correct Answer - B The splitting of spectral lines under the influence of magnetic field is called Zeeman effect. |
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| 2055. |
The velocity of `e^(-)` in a certain Bohr orbit of the hydrogen atom bears the ratio `1:275` to the velocity of light. What is the quantum no. "n" of the orbit and the wave no. of the radiation emitted for the transition form the quantum state `(n+1)` to the ground state. |
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Answer» Correct Answer - `2;9.75xx10^(4)cm^(-1)` `V_(n)=2.18xx10^(6)xx(Z)/(n)=(2.18xx10^(6))/(n)` `(2.18xx10^(6))/(n)=(1)/(275)` `(2.18xx10^(6))/(n)=(1)/(3xx10^(8))=(1)/(275)` `(2.18)/(n(300))=(1)/(275)" "(1)/(n)=(300)/(599.5)` `n=(599.5)/(300)=(1)/(275)" "(1)/(n)=(300)/(599.5)` `n=1.99~=2` |
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| 2056. |
It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of an electron from nay initial orbital to any other orbital. However, this is no so, because a photon has an intrinsic spin angular momentum of `sqrt(2)(h)/(2pi)` corresponding to `S = 1` although it has no charge and no rest mass. on the otherhand, an electron has got two types of angular momentum: Orbital angular momentum, `L = = sqrt(l(l+1))(h)/(2pi)` and spin angular momentum, `L_(s) (=sqrt(s(s+1))(h)/(2pi))` arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carried away by the photon. To satisfy this condition the difference between the azimuthal quantum numbers of the orbitals within which transition takes place must differ by one. Thus, an electron in a d-orbital `(l=2)` cannot make a transition into an s-orbital `(l=0)` because the photon cannot carry away enough angular momentum. An electron, possess four quantum numbers, n l, m and s. Out of these four l determines the magnitude of orbital angular momentum (mentioned above) while m determines its Z-component as `m((h)/(2pi))`. The permissible values of only integers right from `-l` to `+l`. While those for l are also integers starting from 0 to `(n-1)`. The values of l denotes the sub-shell. For `l = 0,1,2,3,4...` the sub-shells are denoted by the symbols s,p,d,f,g....respectively. The spin-only magnetic moment of a free ion is `sqrt(8)B.M`. The spin angular momentum of electron will beA. `sqrt(2)(h)/(2pi)`B. `sqrt(8)(h)/(2pi)`C. `sqrt(6)(h)/(2pi)`D. `sqrt((3)/(4))(h)/(2pi)` |
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Answer» Correct Answer - A `mu_(s) = sqrt(n(n+2)) BM`. Where `n = 2` spin angular momentum: `sqrt(s(s+1)).(h)/(2pi)` |
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| 2057. |
Assertion : Splitting of the spectral lines in the presence of magnetic field is known as stark effect. Reason : Line spectrum is simplest for hydrogen atom.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true |
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Answer» Correct Answer - D Assertion is false but reason is true. Splitting of the splitting of the spectral lines in the presence of a magnetic field is known as Zeeman effect or in electric field it is known as stark effect. The splitting of spectral lines is due to different orientations which the orbitals can have in the presence of magnetic field |
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| 2058. |
If each hydrogen atom in the ground state `1.0 mol of H ` atom are excited by axeited by absorbing photon of energy `8.4 eV, 12.09 eV and 15.0 eV` of energy, then number of spectral lines emitted is equal toA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `Delta E = 13.6 xx l^(2) [(1)/(2^(2)) - (1)/(33^(2))] = 12.09 eV` This will be only absorbed rest is useless (unabsorbed ) so , from `n_(2) = 3`"to" `n_(1) = 1` we have `lambda` corresponding to `3 rarr 2,2 rarr 1 ` and ` 3 rarr 1 ` So, total true spectral line |
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| 2059. |
Assertion (A) : Hydrogen has only one electron in its 1s orbital but it produces several spectral lines. Reason (R) : There are many excited energy levels available in H atoms.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 - A | |
| 2060. |
Assertion (A) : Hydrogen has only one electron in its 1s orbital but it produces several spectral lines. Reason (R) : There are many excited energy levels available in H atoms.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 - C Explanation is correct reason for statement . |
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| 2061. |
The electronic, identified by quantum numbers n and l, (i) `n = 4, l = 1`, (ii) `n = 4, l = 0`, (iii) `n = 3,l = 2`, (iv) `n = 3, l = 1` can be placed in order of increasing energy, from the lowest to highest, asA. `(iv) lt (ii) lt (iii) lt(i)`B. `(ii) lt (iv) lt (i) lt (iii)`C. `(i) lt (iii) lt (ii) lt (iv)`D. `(iii) lt (i) lt (iv) lt (ii)` |
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Answer» Correct Answer - A |
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| 2062. |
The energy of an electron in the first Bohr orbit of H atom is `-13.6 eV` The potential energy value (s) of excited state(s) for the electron in the Bohr orbit of hydrogen is(are)A. `-3.4 eV`B. `-4.2 eV`C. `-6.8 eV`D. `+6.8 eV` |
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Answer» Correct Answer - A |
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| 2063. |
How many electrons can fit in the orbital for which n = 3 and l = 1 ?A. 14B. 2C. 6D. 10 |
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Answer» Correct Answer - B For n=3 , l=1 3p orbital can have only 2 electron |
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| 2064. |
X-raysA. are deflected in a magnetic fieldB. are deflected in an electric fieldC. remain undeflected by both the fieldD. are deflected in both the field. |
| Answer» Correct Answer - C | |
| 2065. |
When X-rays pass through air they:A. produce light track in the airB. ionise the gasC. produce fumes in the airD. accelerate gas atoms |
| Answer» Correct Answer - A | |
| 2066. |
An excited state of H atom emits a photon of wavelength `lamda` and returns in the ground state. The principal quantum number of excited state is given by:A. `sqrt(((lambdaR-1))/(lambdaR))`B. `sqrt(lambdaR(lambdaR-1))`C. `sqrt((lambdaR)/((lambdaR-1)))`D. `sqrt(lambdaR(lambdaR+1))` |
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Answer» Correct Answer - C `(1)/(lambda) = R ((1)/(1)-(1)/(n_(2)^(2)))` `n_(2)^(2) = (R lambda)/(R lambda -1), n_(2) = sqrt((R lambda)/(R lambda -1))` |
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| 2067. |
Momentum of a photon of wavelength `lamda` isA. `h//lamda`B. zeroC. `hlamda//c^(2)`D. `hlamda//c` |
| Answer» Correct Answer - A | |
| 2068. |
Photoelectric effect supports quantum nature of light because (a) there is a minimum frequency of light below which no photo electrons are emitted (b) the maximum kinetic energy of photo electrons depends only on the frequency of light and not on its intensity (c ) even when the metal surface is faintly illuminated, the photo electrons leave the surface immediately (d) electric charge of the photo electrons is quantisedA. there is a minimum frequency of light below which no photoelectrons are emittedB. the maximum kientic energy of photoelectrons depends only on the frequency of light and not on its intensity.C. even when metal surface is faintly illuminated the photoelectrons leave the surfaec immediatelyD. electric charge of photoelectrons is quantised. |
| Answer» Correct Answer - A | |
| 2069. |
In spectral series of hydrogen , the series which does not come in infrared region isA. PfundB. BrackettC. paschenD. Lyman |
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Answer» Correct Answer - D Lyman series lies in the ultraviolet region |
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| 2070. |
The frequency of yellow light having wavelength 600 nm isA. `5.0xx10^14` HzB. `2.5xx10^7` HzC. `5.0xx10^7` HzD. `2.5xx10^14` Hz |
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Answer» Correct Answer - A `v=c/lambda=(3xx10^8 ms^(-1))/(600xx10^(-9)m)=5.0xx10^14` Hz |
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| 2071. |
A hydrogen atom with an elerctron in the first shell ( n = 1) absorbls ` UV` light of a wavelength ` 1. 03 xx 10^(-7) m` To what shell does the electron jumps ? |
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Answer» ` Delta E = (hc)/( lambda) = ( 6.626 xx 10^(-34) xx 3.0 xx 10^8)/( 1.03 xx 10^(-7))` ` = 1.93 xx 10^(-18) J` ` E_n -E_1 = 1 . 93 xx 10 ^(-18) J` ` E_1 -E_1 = 1. 93xx 10^(-18) J` ` E_1 - E_1/n^2 = 1 . 93 xx 10^(-18) J` ` or ( n^2 -1)/n^2 = ( 1.93 xx 10^(-18))/E_1` ` = (1. 93 xx 10^(-18))/( 13. 7 xx 1. 602 xx 10^(-19))` ` [E_1 = 13 . 7 eV = 13 . 7 xx 1. 602 xx 10^(-19) J]` ` ( n^2 -1)/n^2 =0. 88 therefore n= 2. 89 =3` . |
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| 2072. |
The energy of a photon is given as, `Delta E`/atom `= 3.03 xx 10^(-19)J "atom"^(-1)` then, the wavelength `(lambda)` of the photon isA. 6.56 nmB. 65.6 nmC. 656 nmD. 0.656 nm |
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Answer» Correct Answer - C (c) According to formula =, `E=(hc)/(lambda)(v=(c)/(lambda))` Energy E=hv `3.03xx10^(-19)=(hc)/(lambda)` `lambda=(6.63xx10^(-34)xx3.0xx10^(8))/(3.03xx10^(-19))` `=6.56xx10^(-7) m` `=6.56xx10^(-7)xx10^(9)nm` =`6.56xx10^(2) nm` `=656 nm` |
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| 2073. |
If wavelength of photon is `2.2 xx 10^(-11)m`, `h = 6.6 xx 10^(-34) Js`, then momentum of photons isA. `3xx10^(-23) "kg ms"^(-1)`B. `3.33xx10^(22) "kg ms"^(-1)`C. `1.452xx10^(-44) "kg ms"^(-1)`D. `6.89xx10^(43) "kg ms"^(-1)` |
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Answer» Correct Answer - A `p=h/lambda=(6.6xx10^(-34))/(2.2xx10^(-11))=3xx10^(-23) kgms^(-1)` |
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| 2074. |
The uncertainty in the position of a moving bullet of mass `10 g` is `10^-5 m`. Calculate the uncertainty in its velocity.A. `5.2xx10^(-28)` m/secB. `3.0xx10^(-28)` m/secC. `5.2xx10^(-22)` m/secD. `3xx10^(-22)` m/sec |
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Answer» Correct Answer - A Uncertainty of moving bullet velocity `Deltav=h/(4pixxmxxDeltax)=(6.625xx10^(-34))/(4xx3.14xx.01xx10^(-5))` `=5.2xx10^(-28)` m/sec |
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| 2075. |
The uncertainty in the position of a moving bullet of mass `10 g` is `10^-5 m`. Calculate the uncertainty in its velocity.A. `5.2 xx 10^-28 m//sec`B. `3.0 xx 10^-28 m//sec`C. `5.2 xx 10^-22 m//sec`D. `3 xx 10^-22 m//sec` |
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Answer» Correct Answer - A (a) Uncertainly of moving bullet velocity `Delta v = (h)/(4 pi xx mxx Delta v) = (6.625 xx 10^-34)/(4 xx 3.14 xx .01 xx 10^-5)` =`5.2 xx 10^-28 m//sec`. |
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| 2076. |
The mass of a photon with a wavelength equal to `1.54 xx 10^-8 cm` is.A. `0.8268 xx 10^-34 kg`B. `1.2876 xx 10^-33 kg`C. `1.4285 xx 10^-32 kg`D. `1.8884 xx 10^-32 kg` |
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Answer» Correct Answer - C ( c) We know that `lamda = (h)/(mv)` , `:. m = (h)/(m lamda)` The velocity of photon `(v) = 3 xx 10^8 m sec^-1` `lamda = 1.54 xx 10^-8 cm = 1.54 xx 10^-34 meter` `:.=(5.626 xx 10^-34 Js)/(1.54 xx 10^-10 m xx 3 xx 1068 m sec^-1)` =`1.4385 xx 10^-37 kg`. |
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| 2077. |
The mass of a photon of wavelength `1.54 Å` isA. `2.5xx10^(-32)` kgB. `1.42xx10^(32)` kgC. Both of theseD. None of these |
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Answer» Correct Answer - B `lambda =h/(mv)=h/(lambdav)` where `v=3xx10^(8)m//sh=6.625xx10^(-34)` Js `lambda =1.54xx10^(-10)`m `:. M=1.42xx10^(-32)` kg |
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| 2078. |
The uncertainty in the velocity of moving bullet of mass 10 g, when uncertainty in its position is `10^(-5)` m isA. `5.2xx10^(-28)m//s`B. `5.2xx10^(-22)m//s`C. `3xx10^(-28) m//s`D. `3xx10^(-22)m//s` |
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Answer» Correct Answer - A `Deltax.Deltap=h/(4pi)or DeltaxmDeltav=h/(4pi)` `Deltav=h/(4piDeltaxm)=((6.626xx10^(-34))js)/(4xx3.14xx10^(-5)mxx10^(-2)kg)` `=5.2xx10^(-28)m//s` |
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| 2079. |
The basis of quantum mechanical model of an atom isA. Angular momentum of electronB. Qantum numbersC. Dual nature of electronD. Black body radiation |
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Answer» Correct Answer - C Dual nature of electron |
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| 2080. |
Coloumn I `&` Column II contain data on Schrondinger Wave-Mechanical model, where symbols have their usual meanings. Match the column :- |
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Answer» Correct Answer - (A)p,(B) p,q,s, (C) p,r, (D) q,s s-orbital is spherical hence it is non-directional. |
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| 2081. |
`K` and `Cs` are used in photoelectric cells. `K` and `Cs` emit electrons on exposure to light.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 - A (a) It is due to photoelectric effect. |
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| 2082. |
Which of the following orbitals have no spherical nodes?A. 1sB. 2sC. 2pD. 3p |
| Answer» Correct Answer - A::C | |
| 2083. |
Which of the following orbitals are associated with angular nodes?A. fB. `d`C. `p`D. `s` |
| Answer» Correct Answer - A::B::C | |
| 2084. |
The correct statement(s) among the following is/are:A. All d-orbitals except `d_(z^(2))` have two angular nodes.B. `d_(x^(2)y^(2)), d_(z^(2))` lie on the axes.C. The degeneracy of p-orbitals remains unaffected in the presence of external magnetic field.D. d-orbitals have 3-fold degeneracy. |
| Answer» Correct Answer - A::B::C | |
| 2085. |
Which of the following statements are correct for an electron that has n = 4 and m = -2 ?A. The electron may be in a d-orbitalB. The electron is in the fourth principal electronic shell.C. The electron may be in a p-orbitalD. the electron must have the spin quantum number =+1/2. |
| Answer» Correct Answer - B::C | |
| 2086. |
Statement-1: Kinetic energy of photoelectrons increases with increases in the frequency of incident radiation Statement-2: The number of photoelectrons ejected increases with increases in intensity of incident radiation.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 - D | |
| 2087. |
For the energy levels in an ato which one of the following statements is/are correct?A. There are seven principal electron energy levels.B. The second principal energy level can have 4 subnergy levels and contain a maximum of 8 electrons.C. The M energy level can have a maximum of 32 electrons.D. The 4s subenergy level is at a lower energy than the 3d subenergy level. |
| Answer» Correct Answer - A::D | |
| 2088. |
When a light of frequency `v_(1)` is incident on a metal surface the photoelectrons emitted have twice the kinetic energy as did the photoelectron emitted when the same metal has irradiated with light of frequency `v_(2)`. What will be the value of threshold frequency?A. `v_(0)=v_(1)-v_(2)`B. `v_(0)=v_(1)-2v_(2)`C. `v_(0)=2v_(1)-v_(2)`D. `v_(0)=v_(1)+v_(2)`. |
| Answer» Correct Answer - C | |
| 2089. |
The value of `(n_(2)+n_(1))` and `(n_(2)^(2)-n_(1)^(2))` for `He^(+)` ion in atomic spectrum are 4 and 8 reaspectively . The wave length of emitted photon whwn electron jump from `n_(2)` to `n_(1)` isA. `(32)/(9R_(H))`B. `(9)/(32R_(H))`C. `(32)/(9)R_(H)`D. `(9)/(32)R_(H)` |
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Answer» Correct Answer - B `n_(2)+n_(1)=4` . . . (i) `n_(2)^(2)-n_(1)^(2)=8` i.e., `(n_(2)+n_(1))(n_(2)-n_(1))=8` `4(n_(2)-n_(1))=8` `(n_(2)-n_(1))=2` From eqns. (i) and (ii) `n_(2)=3,n_(1)=1` `(1)/(lamda)=R_(H)Z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `=R_(H)xx2^(2)[(1)/(1^(2))-(1)/(3^(2))]` `(1)/(lamda)=R_(H)xx(32)/(9)` ltbr. `thereforelamda=(9)/(32R_(H))`. |
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| 2090. |
The value of `(n_(2)+n_(1))` and `(n_(2)^(2)-n_(1)^(2))` for `He^(+)` ion in atomic spectrum are 4 and 8 reaspectively . The wave length of emitted photon whwn electron jump from `n_(2)` to `n_(1)` isA. `(32)/(9)R_(H)`B. `(9)/(32)R_(H)`C. `(9)/(32R_(H))`D. `(32)/(9R_(H))` |
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Answer» Correct Answer - C `n_(2)+n_(1)=4` `n_(2)^(2)-n_(1)^(2)=8` `n_(2)=3," " n_(1)=1` `lambda=R.4[(1)/(1^(1))-(1)/(3^(2)]]=(32R)/(9)rArr lambda=(9)/(32R)` |
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| 2091. |
Number of unpaired electrons in `Mn^(4+)` isA. 3B. 5C. 6D. 4 |
| Answer» Correct Answer - A | |
| 2092. |
Write down elerctronic fonfigureation of the following and reprot no. of unpaired electron in each . (a) ` Mn^(2+)` (b) ` Cr^(2+)`. |
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Answer» Electronic configuration No. of unpaired (e ) (a) `._(25 ) Mn^(4 +) : 1 s^2 2s^2 2p^6 3s^6 3d^3 3` (b) ` ._(24) Cr^(2+) : 1 s^2 2s^2 2p^6 3s^6 3d^4 4` Note : In case of writing electrinc configuration of cation first write configuration of neutral atom and then take out desired electrons from outer most shell , e.g., `._(25)Mn : 1s^2 , 2s^2 , 2p^2, 3s^2 3p^6 3d^5 , 4s^2` ` ._(25) Mn^+ : 1s^2 , 2s^2 , 2p^2, 3s^2 3p^6 3d^5 , 4s^1` . |
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| 2093. |
The maximum of electrons can have principal quantum number `n = 3` and spin quantum number `m_(s) = 1//2` is |
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Answer» `n = 3 rArr 1= 0,1,2` `{:(For l = 0 , m= 0 rArr m_(s) = - 1//2 and + 1//2),(For l = 1 , m= -1 rArr m_(s) = - 1//2 and + 1//2):}` `{:(0 rArr m_(s) = - 1//2 and + 1//2),(-1 rArr m_(s) = - 1//2 and + 1//2),("For l= 2,m= "-2 rArr m_(s) = - 1//2 and + 1//2),(-1 rArr m_(s) = - 1//2 and + 1//2),(0 rArr m_(s) = - 1//2 and + 1//2),(1 rArr m_(s) = - 1//2 and + 1//2),(2 rArr m_(s) = - 1//2 and + 1//2):}` Alermatively `n= 3` Number of electron `= 2n^(2) = 2 xx 3^(2) = 18 ` Number of electron with `m_(s) = - (1)/(2) = (18)/(2) = 9` |
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| 2094. |
Which of the following arrangements of electron is mostly likely to the stable ?A. B. C. D. |
| Answer» Correct Answer - A | |
| 2095. |
An electron beam can undergo defraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes `1.54` Å? |
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Answer» Correct Answer - C For `1.54 Å` wavelength, the momentum for electron would be given by the expression `p = (h)/(lambda) rArr v = (h)/(m lambda)` The potential to which a beam of electron be acceletated is given by the expression `eV = (1)/(mv^(2))` `i.e. V = (1)/(2me) ((h)/(lambda))^(2)` Substituting the value , we get `V = (1)/((2)(9.1 xx 10^(-31)kg) (1.6 xx 10^(-19)C))((6.626 xx 10^(-34) J s)/(1.54 xx 10^(-10)m))^(2)` `= 63.573 m s^(-1)` |
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| 2096. |
In centre-symmetrical system, the orbital angular momentum, a measure of the momentum of a particle travelling around the nucleus, is quantised. Its magnitude isA. `sqrt(l(l + 1)) (h)/(2 pi)`B. `sqrt(l(l - 1)) (h)/(2 pi)`C. `sqrt(s(s + 1)) (h)/(2 pi)`D. `sqrt(s(s - 1)) (h)/(2 pi)` |
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Answer» Correct Answer - A (a) `sqrt(l(l + 1)) (h)/(2 pi)`. |
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| 2097. |
The number of neutrons in deuterium isA. 2B. 3C. 5D. 1 |
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Answer» Correct Answer - D `N = A - Z` |
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| 2098. |
Which orbital has two angular nodal planes ?A. `s`B. `p`C. `d`D. `f` |
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Answer» Correct Answer - C Angular node ` = l`. For d-orbital ` l=2`. |
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| 2099. |
Energy of electron in the first Bohr orbit of H-ato is `-313.6` kcal `mol^(-1)`, then the energy in second Bohr orbit will be:A. `+313.6` kcal `mol^(-1)`B. `-78.4` kcal `mol^(-1)`C. `-34.64` kcal `mol^(-1)`D. `-12.5` kcal `mol^(-1)` |
| Answer» Correct Answer - B | |
| 2100. |
Why energy level are also know as stationary state ? |
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Answer» Correct Answer - A Bohr orbit are fixed path with definite energy in which el,,ectron revolve .Therefore , they are called stationary state |
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