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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.
| 2101. |
Which jave the same number of s-electrons as the d-dlectrons in ` Fe^(2+)` ?A. `Li`B. ` Na`C. `N`D. `P` |
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Answer» Correct Answer - B ` P ( 1s^(2) , 2s^(2) 2p^(6) , 3s^(2) 3p^(3) )` has `(6)` electron in s-subshells as in d-shell fo `Fe^(2+)`. |
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| 2102. |
Why Bohr orbit are also know as energy levels ? |
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Answer» Correct Answer - A Because each orbit is associated with definite amount of energy |
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| 2103. |
According to the Bohr model, the energy levels of hydrogen atom can be found by making two assumptions. (i). The electrons move in circulr orbit and (ii) the angular momentum of the electron in the `n^(th)` energy level is quantized to have a value `n(h)/(2pi)`. The levels calculated with nuclear charge Ze, with a single electron in the orbit, are called Hydrogenic levels. Assume that the two electrons in the ground state of a helium atom occupy the corresponding lowest Hydrogenic level. Q. The minimum repulsive energy between the two electrons would then be:A. 3.4 eVB. 6.8 eVC. 13.6 eVD. 27.2 eV |
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Answer» Correct Answer - D For singly ionised He atom, ground state energy is `((-Ze^(2))/(2r))=-54.4eV` `(-2e^(2))/(2r)=-54.4eV` `(e^(2))/(2r)=27.2eV` `therefore` Minimum repulsive energy =27.2eV (Two electrons will lie diametrically opposite). |
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| 2104. |
According to the Bohr model, the energy levels of hydrogen atom can be found by making two assumptions. (i). The electrons move in circulr orbit and (ii) the angular momentum of the electron in the `n^(th)` energy level is quantized to have a value `n(h)/(2pi)`. The levels calculated with nuclear charge Ze, with a single electron in the orbit, are called Hydrogenic levels. Assume that the two electrons in the ground state of a helium atom occupy the corresponding lowest Hydrogenic level. Q. if the hydrogen atom ionizatioin temperature is T, the temperature at which He atoms ionise completely (both electron having left the atom) would be:A. 8TB. 4TC. 6TD. 2T |
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Answer» Correct Answer - C E=Ionisation energy of He-atom `=2xx54.4eV`-repulsive energy `=108.8eV-27.2eV` `=6xx`Ionization energy of hydrogen `therefore` Ionisation temperature of Helium `=6xx`Ionisation temperature of hydrogen ltBrgt `=6T` |
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| 2105. |
Which of the following sets of quantum numbers are correct? n l m n l mA. `{:(n,l, ml),(1,1,+2):}`B. `{:(n,l,ml),(2,2,-1):}`C. `{:(n,l,ml),(3,2,-2):}`D. `{:(n,l,ml),(3,4,-2):}` |
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Answer» Correct Answer - C l values `rArr 1,2,3,4`.. Upto `(n-1)` `m_(1)` values `rarr -l....O ...+l` |
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| 2106. |
How many unique spectral lines are emitted when `e^(-)` returns from `n_(2)^(th)` orbit to `n_(1)^(th)` orbit in a sample of H-like species ? |
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Answer» No. of unique lines/wavelengths/ frequencies `=((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2)` If `e^(-)` is returning to ground state `(i.e. n_(1)=1)` then `["where" n_(2)=n]` No. of unique lines `=((n-1)(n))/(2)` |
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| 2107. |
The spectral lines corresponding to the radiatio emitted by an electron jumping from higher orbits to first orbit belong toA. Paschen seriesB. Balmer seriesC. Lyman seriesD. none of these |
| Answer» Correct Answer - C | |
| 2108. |
The spectral lines corresponding to the radiation emitted by an electron jumping from 6th, 5 th and 4th orbits to second orbit belong to:A. Lyman seriesB. Balmer seriesC. Paschen seriesD. Pfund series. |
| Answer» Correct Answer - B | |
| 2109. |
The distance between `4th` and `3rd` Bohr orbits of `He^(+)` is `:`A. `2.645 xx 10^(-10)m`B. `1.322 xx 10^(-10)m`C. `1.851 xx 10^(-10)m`D. `6.8 xx 10^(-10)m` |
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Answer» Correct Answer - C `r_(n,z) = 0.529 xx (n^(2))/(Z)` |
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| 2110. |
The magnetic quantum number of an atom is releted to theA. Size of the orbitalB. Spin angular momentumC. Orbital angular momentumD. Orientation of the orbit in space |
| Answer» Correct Answer - D | |
| 2111. |
Which one of the following consitutes a group of the isoelectronic species?A. `C_(2)^(2),O_(2)^(-),CO,NO`B. `NO^(+),C_(2)^(2-),CN^(-),N_(2)`C. `CN^(-),N_(2),O_(2)^(2-),C_(2)^(2-)`D. `N_(2),O_(2)^(-)`,NO^(+),CO` |
| Answer» Correct Answer - B | |
| 2112. |
An electron of kinetic energy`E_(0)` is scattered by an atomic hydrogen sample in ground state. Find the minimum value of `E_(0)`,so that a photon of wavelength `656.3`nm may be emitted by H-atom.A. `12.09 eV`B. 13.6 eV`C. `14.6 eV`D. None of these |
| Answer» Correct Answer - A | |
| 2113. |
29 electrons are removed from Zn-atom (Z=30)by certain means. The minimum energy needed to remove the 30th electron, will beA. `12.24keV`B. `408 eV`C. `0.45 eV`D. None of these |
| Answer» Correct Answer - A | |
| 2114. |
Calculate wavelength : (a) of electron moving with veloctiy of `10^(3)m//sec` (b) of cricket ball moving with velocity of `10^(3)m//sec` having mass `0.5 kg` |
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Answer» (a) `lambda_(e) = (6.626xx10^(-34))/(9.1xx10^(-31)xx10^(3))=727nm` (b) `lambda_(b) = (6.626xx10^(-34))/(0.5xx10^(3))=1.33xx10^(-27)nm` Note : `therefore (lambda_(b))/(lambda_(e))=10^(-30)` `because` Cricket ball has much larger mass than `e^(-)` so its wavelength is very small and negligible. |
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| 2115. |
If the energy of electron in H atom is given by expression `- 1312 n^(2) kJ "mole"^(-1)` then the energy required to excited the elcxtron from ground state to second orbit isA. `328 kJ`B. `656 kJ`C. `984 kJ`D. `312 kJ` |
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Answer» Correct Answer - C `(-1321)/(4) - ((1312)/(1)) = 984 kJ` |
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| 2116. |
Photochemical dissociation of oxygen result in the production of two oxygen atoms one in the ground state and in the excited state `O_(2) overset(hv) to O + O^(6)` The maximum wavelenght (1) neded for this is `17.4` nm .If the exchation energy `O rarr O^(6) is 3.15 xx 10^(-19) J` haw much energy in kJ `mol^(-1)` is neded for the dissociation of `1` and of oxygen into normal atomic is the ground state |
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Answer» Energy required per molecule in the process `iO_(2) overset(hv) to O + O^(6) `may be gives as `E = (hc)/(lambda) = (6.626 xx 10^(-16) xx 3 xx 10^(8))/(174 xx 10^(-9))` `= 11.424 xx 10^(-19)`joule per molecule Energy for `O rarr O^(6) is 3.15 xx 10^(-19) J` This energy for `O_(2) rarr 20 `will be `11.424 xx 10^(-19) - 3.15 xx 10^(-19)= 8.274 xx 10^(-19) J` For one mole`= E xx N_(A) `(Avogadro number) `= 8.247 xx 10^(-19) xx 6.023 xx 10^(23)` `= 498.3 kJ mol ^(-1)` |
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| 2117. |
The hydrogen-like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one radial node. Upon absorbing light the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the hydrogen atom. Q. The state `S_(1)` is:A. 1sB. 2sC. 2pD. 3s |
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Answer» Correct Answer - B 2s is symmetrical havig one radial node. |
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| 2118. |
The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one radial node. Upon absorbing light the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal to the ground state energy of the hydrogen atom. Energy of the state `S_(1)` in units of, the hydrogen atom in ground state energy isA. `0.75`B. `1.50`C. `2.25`D. `4.50` |
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Answer» Correct Answer - C `(E_(S_(1)))/(E_(H)) =((-13.6xx 3^(2))/(2^(2)))/(-13.6) = (9)/(4) = 2.25` |
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| 2119. |
Spectral line would not ne seen for a `2p_x - 2p_z` transition p-orbitals are degenerate orbitals.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) No transition between `p_x - p_y` tansition, since all the three p-orbitals possess same enegry level, i.e., degenerate orbitals. |
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| 2120. |
Assertion (A) : A spectral line will be seen for `2p,-2p` transition Reason (R ) : Energy is raleased in the form of wave of light when the electron drops from `2p_(x)` , to `2p_(y)` orbital.A. 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» Correct Answer - Both eare incorrect | |
| 2121. |
Assertion (A) : The position of electron can be determined with the help of an electronic microscope. Reason (R) : The product of uncertainty in momentum and the uncertainty in the position of an electron cannot be less than a finite limit.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 - D | |
| 2122. |
The maximum number of electrons can have pricipal quantum number `n = 3` and spin quantum number `Mz =-(1)/(2)` is (2011) |
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Answer» Correct Answer - 9 `n = 3, l = 0,1,2` for `{:(l =0,m =0,mz=-(1)/(2)+(1)/(2)),(l=1,m=-1,mz=-(1)/(2)+(1)/(2)),(l=1,m=0,mz=-(1)/(2)+(1)/(2)),(l=1,m=0,mz=-(1)/(2)+(1)/(2)):}` `{:("for",l =2,m=-2,mz=+(1)/(2)-(1)/(2)),("for",l=2,m=-1,mz=+(1)/(2)-(1)/(2)),("for",l=2,m=0,mz=+(1)/(2)-(1)/(2)),("for",l=2,m=+1,mz=+(1)/(2)-(1)/(2)),("for",l=2,m=+2,mz=+(1)/(2)-(1)/(2)OR):}` `n = 3` number of electros `=2n^(3) = 2xx 3^(2) = 18` electrons with `(ms =- (1)/(2)) = (18)/(2) = 9` |
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| 2123. |
A photon having `lambda = 854 Å` cause the ionization of a nitrogen atom. Give the I.E. per mole of nitrogen in KJ |
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Answer» Correct Answer - `1403kJ//mol` `IE=(hc)/(lambda)=(1240)/(85.4)` `=(1240)/(85.4)xx96.368kJ//"mole"~~1399.25kJ//"mole"` |
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| 2124. |
`{:(List-I,List-II),((I)"Wave number",(a)ms^(-1)),((II)"Frequency",(b)nm),((III)"Wavelength",(c)s^(-)),((IV)"Velcity",(d)m^(-1)):}` The correct match isA. `{:(I,II,III,IV),(a,b,c,d):}`B. `{:(I,II,III,IV),(d,c,b,a):}`C. `{:(I,II,III,IV),(b,c,d,a):}`D. `{:(I,II,III,IV),(c,d,b,a):}` |
| Answer» Correct Answer - B | |
| 2125. |
The radial wave equation for hydrogen of radial nodes from nucleus are: `Psi_(1s)=(1)/(16sqrt(4))(1/a_(0))^(3//2) [("x"-1)("x"^(2)-8"x"+12)]e^(-x//2)` where, `x=2r//a_(0),a_(0)` = radius of first Bohr orbit The minimum and maximum position of radial nodes from nucleus are:A. `a_(0),3a_(0)`B. `(a_(0))/(2),3a_(0)`C. `(a_(0))/(2),a_(0)`D. `(a_(0))/(2),4a_(0)` |
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Answer» Correct Answer - B At radial node, `Phi=0` `therefore` From given equation `x-1=0 and x^(2)-8x+12=0` `x-1=0impliesx=1` i.e., `(2r)/(a_(0))=1,r=(a_(0))/(2)` (Minimum) `x^(2)-8x+12=0` `(x-6)(x-2)=0` when x-2=0 x=2 `(2r)/(a_(0))=2,` i.e., `r=a_(0)` (Middle value) when `x-6=0` `x=6` `(2r)/(a_(0))=6` `r=3a_(0)` (Maximum). |
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| 2126. |
Assertion(A): It is not possible to predict position and the velocity of an electron exactly and simultaneously Reason(R): Electron moving with high speed possesses both the particle nature and the wave natureA. 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 | |
| 2127. |
Assertion (A) : A spectral line will be seen for `2p_(x)`-`2p_(y)` transition Reason (R ) : Energy is raleased in the form of wave of light when the electron drops from `2p_(x)` , to `2p_(y)` orbital.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 - D | |
| 2128. |
`{:(List-I,List-II),((A)"Energy",(1)(2pize^(2))/(nh)),((B)"Velocity",(2)(-2pi^(2)mz^(2)e^(4))/(n^(2)h^(2))),((C)"Rydberg constant",(3)(2pi^(2)mz^(2)e^(4))/(h^(3)c)),((D)"Radius",(4)(n^(2)h^(2))/(4pi^(2)mze^(2))),(,(5)(-4pi^(2)mz^(2)e^(4))/(n^(2)h^(2))):}` The correct match isA. `A = 2 B = 4 C = 5 D = 1`B. `A = 2B = 1 C = 3 D = 4`C. `A = 3 B = 2 C = 1D = 4`D. `A =4 B = 3C = 1 D = 5` |
| Answer» Correct Answer - B | |
| 2129. |
The given diagram shows point `P_(1),P_(2)P_(3)&P_(4)` in 1s orbital. The correct oreder of increasing radial probability at these point areA. `P_(1)=P_(2)=P_(4)gtP_(3)`B. `P_(1)=P_(3)ltP_(2)=P_(4)`C. `P_(1)=P_(3)gtP_(2)=P_(4)`D. `P_(1)=P_(2)=P_(4)ltP_(3)` |
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Answer» Correct Answer - C `Psi^(2)(r)` 1 s orbital do not depend upon angular part ,`therefore p_(1)=p_(3)` and decrease with increase in distance from the nucleus , therefore `P_(2)ltP_(3)` |
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| 2130. |
Statement-I : There are two spherical nodes in 3s-orbital. Because Statement-II : There is no angular node in 3s-orbital.A. Statement-I is true, Statement-II is true , Statement-II is correct explanation for Statement-I.B. Statement-I is true, Statement-II is true , Statement-II is NOT a correct explanation for statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - B | |
| 2131. |
Assertion (A) : There are two spherical nodes in `3s` orbital Reason (R ) : There is no planqaqr nodes in `3s` orbital.A. 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 - B `3s` orbital has two spherical nodes, but d orbital has no planner node |
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| 2132. |
The number of radial nodes of `3s` and `2p` orbital are, respectivelyA. `1 2,0`B. ` 0,2`C. ` 1,2`D. ` 2,1` |
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Answer» Correct Answer - A Number of radial nodes ` = n -l-1` `{:["For 3s it is"],["For 2p it is 0"]:}`. |
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| 2133. |
Assertion (A) : There are two spherical nodes in `3s` orbital Reason (R ) : There is no planar nodes in `3s` orbital.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 - B | |
| 2134. |
`{:(List-I,List-II),((I)"Radial probability distribution curve of 3s orbital",(a)1.1A^(@)),((II)"Distance of maximum probability of 1s electron",(b)1s "orbital"),((III)"Radial node for a 2s electron",(c)"3peaks, 2 radial nodes"),((IV)"No spherical nodes",(d)0.53A^(@)):}` The correct match isA. `{:(I,II,III,IV),(a,b,c,d):}`B. `{:(I,II,III,IV),(c,d,a,b):}`C. `{:(I,II,III,IV),(b,a,d,a):}`D. `{:(I,II,III,IV),(d,a,b,c):}` |
| Answer» Correct Answer - B | |
| 2135. |
Assertion (A) : There are two spherical nodes in `3s` orbital Reason (R ) : There is no planar nodes in `3s` orbital.A. If (A) and (R) are both correct and (R) is the correct reason for (A).B. If (A) and (R) are both correct but (R) is not the correct reason for (A).C. If (A) is true but (R) is false.D. If both (A) and (R) false. |
| Answer» Correct Answer - B | |
| 2136. |
When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with kinetic energy of `1.68 xx 10^(5) J "ml"^(-1)`. What is the minimum energy needed to remove an electron from sodium ? What is the maximum wavelength that will cause a photoelectron to be emitted. |
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Answer» The energy (E) of a 300 nm photon is given by `hv = (hc)/(lambda) = (6.626 xx 10^(-34) Js xx 3.0 xx 10^(8)ms^(-1))/(300 xx 10^(-9)m)` The energy of one mole of photons `= 6.626 xx 10^(-19)J xx 6.022 xx 10^(23) mol^(-1) = 3.99 xx 10^(5) J mol^(-1)` The minimum energy needed to remove one mole of electron from sodium. `= (3.99 - 1.68) 10^(5)J mol^(-1) = 2.31 xx 10^(5) J mol^(-1)` The minimum energy for one electron `= (2.31 xx 10^(5) J mol^(-1))/(6.022 xx 10^(23) "electrons" mol^(-1)) = 3.84 xx 10^(-19)J` This corresponds of the wavelength `:. lambda = (hc)/(E) = (6.626 xx 10^(-34)Js xx 3.0 xx 10^(8)ms^(-1))/(3.84 xx 10^(-19)J) = 517 nm` This wavelength corresponds to green colour in visible spectrum. |
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| 2137. |
Calculate the wavelength ( in nanometer ) associated with a proton moving at ` 1.0xx 10^3` m/s (Mass of proton ` =1.67 xx 10^(-27) kg ` and ` h=6.63 xx 10^(-34) is)` :A. `0.032 nm`B. `2.5nm`C. `14.0nm`D. `0.4nm` |
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Answer» Correct Answer - D `lambda = (h)/(mv)` |
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| 2138. |
Calculate the wavelength (in nanometer) associated with a proton moving at `1.0xx10^(3)ms^(-1)`(Mass proton `=1.67xx10^(-27)kg` and `h=6.63xx10^(-34)Js`):-A. 0.032 mmB. 0.40 mmC. 2.5 mmD. 14.0 mm |
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Answer» Correct Answer - B As `lambda=h/"mv" = (6.63xx10^(-34))/(1.67xx10^(-27)xx1xx10^3) =3.97xx10^(-10)m` `=0.397xx10^(-9) m = ~0.40 `nm |
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| 2139. |
How many of the following are possible `1p,2s,3p,3f,3d`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `3` `1p `is not possibleg because `n = 1= 8` `3f `is not possibleg because `n = 1= 3` |
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| 2140. |
An electron has kinetic energy `2.8 xx 10^-23 J` de-Broglie wavelength will be nearly. `(m_e = 9.1 xx 10^-31 kg)`.A. `9.28 xx 10^-4 m`B. `9.28 xx 10^-7 m`C. `9.28 xx 10^-8 m`D. `9.28 xx 10^-10 m` |
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Answer» Correct Answer - C ( c) Formule for de-Broglie wavelength is `lamda = (h)/(p)` or` lamda =(h)/(mv) rArr eV = (1)/(2) mv^2` or `v = sqrt((2 eV)/(m))` `lamda = (h)/(sqrt(2 meV)) = (6.62 xx 10^-34)/(sqrt(2 xx 9.1 xx 10^-31 xx 2.8 xx 10^-23))` `lamda = 9.28 xx 10^-8 "meter"`. |
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| 2141. |
An electron beam energes from an accelerator with kinetic energy 100eV. What is its de-Broglie wavelength? `[m = 9.1 xx 10^(-31)kg, h = 6.6 xx 10^(-34)Js, 1eV = 1.6 xx 10^(-19)J]` |
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Answer» Kinetic energy of electron `= 100eV` `= 100 xx 1.6 xx 10^(-19) J = 1.6 xx 10^(-17)J` we know, `lambda = (h)/(sqrt(2KE.m))` `= (6.626 xx 10^(-34))/(sqrt(2xx 1.6 xx 10^(-17) xx 9.1 xx 10^(-31))) = 1.228 xx 10^(-10)m` |
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| 2142. |
in a multi- electron atom ,which of the following orbitals described by the three quantum numbers, which of the following will have nearly same energy? `{:(( P) ,n=1","l=0"," m=0,(q) , n=2","l=0"," m=0),((r ) , n=2"," l=1"," m=1",",(S ) , n=3"," l=2"," m=1),((t ), n=3","l=2","m=0,","","",):}`A. a & cB. b & cC. c & dD. d & e |
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Answer» Correct Answer - D Same `(n+1)` value indicates same energy level |
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| 2143. |
The atomic number of an element having the valency shell electronic configuration `4s^2 4p^6` isA. 35B. 36C. 37D. 38 |
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Answer» Correct Answer - B Complete E.C. =`[Ar]^18 3d^10 4s^2 4p^6` Hence No. of `e^-` = no. of protons =36 = Z |
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| 2144. |
Maximum numbers of electrons in a subshell is given by-A. `(2l + 1)`B. `2(2 l + 1)`C. `(2 l + 1)2`D. `2(2 l + 1) 2` |
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Answer» Correct Answer - B (b) `2(2 l + 1)`. |
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| 2145. |
Photoelectrons are liberated by ultra light of wavelength `2000 Å` from a metallic surface for which the photoelectric threshold is 4000Å. Calculate the de Broglie wavelenth of electrons emitted with maximum kinetic energy. |
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Answer» K.E. = Quantum Energy - Threshold energy `=(6.626xx10^(-34)xx3xx10^(8))/(2000xx10^(-10))-(6.626xx10^(-34)xx3xx10^(8))/(4000xx10^(-10))` `(6.626xx10^(-34)xx3xx10^(8))/10^(-10)((1)/(2000)-(1)/(4000))=4.969xx10^(-19)` Joule `(1)/(2) mv^(2)=4.969xx10^(-19)implies m^(2)v^(2)=2xx4.969xx10^(-19)xx9.1xx10^(-31)` `mv = 9.51xx10^(-25)implies lambda=(h)/(mv)=(6.626xx10^(-34))/(9.51xx10^(-25))=0.696xx10^(-9)m` |
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| 2146. |
Consider the following statements: (a) Electron density in the `XY` plane in `3d_(x^(2)-y^(2))` orbital is zero (b) Electron density in the `XY` plane in `3d_(z^(2))` orbital is zero. (c ) `2s` orbital has one nodel surface (d) for `2p_(z)` orbital, `XY` is the nodal plane. Which of these are incorrect statements :A. 2 and 3B. 1,2,3,4C. Only 2D. 1 & 3 |
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Answer» Correct Answer - A for `dx^(2) - y^(2)` orbiral lobes are on xy plane |
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| 2147. |
The electron density of `3d_(xy)` orbital in YZ plane isA. `50%`B. `95%`C. `33.33%`D. Zero |
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Answer» Correct Answer - D yz plane is nodal plane for 3dxy orbital |
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| 2148. |
The difference between the incident energy and threshold energy for an electron in a photoelectric effect experiment is `5 eV`. The de Broglie wavelength of the electron is-A. `(6.6 xx 10^-9)/(sqrt(1456)) m`B. `(6.6 xx 10^-9)/(sqrt(145.6)) m`C. `(6.6 xx 10^-9)/(sqrt(1664))`D. `(6.6 xx 10^-9)/(sqrt(166.4))m` |
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Answer» Correct Answer - B (b) `KE of overline (e)^(-)` ejected = Energy of incident quantum - threshold energy `= 5 eV` `lamda = (h)/(mV) = (h)/(sqrt(2 m K.E))` =`(6.6 xx 10^-34)/(sqrt(2 xx 9.1 xx 10^-31 xx 5 xx 1.6 xx 10^-19))` `m = (6.6)/(sqrt(145.6)) m`. |
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| 2149. |
The wave function for an orbital is H-atom is given as `psi=(sqrt(2)/(81sqrt(pi)))((1)/(a_(0)))^((3)/(2))(6-(r )/(a_(0)))((r )/(a_(0)))e^((r )/(3a_(0))). sinthetasinphi` The orbital is :A. `2s`B. `3p`C. `2p`D. `3d` |
| Answer» Correct Answer - A::B | |
| 2150. |
An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. It jumps back to the ground state in steps e.g., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state etc. Total number of lines obtained in the spectrum would beA. 3B. 6C. 9D. 12 |
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Answer» Correct Answer - B total number of lines produced when the electron returns to the ground state. `=(n(n-1))/2=(4(4-1))/2=6` |
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