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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.
| 351. |
Electromegnetic radiation of wavelength `500` nm is just safficient to ionic a sodium atom .Calculate the energy corresponding to this wavelength the ionisation potential of Na |
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Answer» Wavelength of radiation `= 300 xx 10^(-9)m` Energy of radiation `E = hv = (hc)/(lambda) ` `= (6.626 xx 10^(-34)Js xx 3 xx 10^(8) ms^(-1))/(300 xx 10^(-9)m)` `= 6.626 xx 10^(-19) J` Ionisation potential of Na `N_(A) xx E ("where" N_(A) = "Avogadro number") = 6.023 xx 10^(23) xx 6.621 xx 10^(-19) J mol^(-1) = 398.7 kJ mol^(-1)` |
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| 352. |
Statement : The plot of atomic number ( y -axis ) versus number of neutrons ( x -axis ) for stable nuclei shows a curvature towards x-axis fron the line of ` 45^@` slope as the atomic number is increased . Explanation : proton -proton electrostatic repulsions begin to overcome attracive forces involving protons and neutrons in heavier nuclides.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 falseD. Statement 1 is false , statement 2 is true |
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Answer» Correct Answer - C In heavier nuclides, attractive forces between protons & neutrons overcome proton-proton electrostatic repulsions. |
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| 353. |
Calculate the uncertainty in the position of a dust particle with mass equal to `1 mg` if the uncertiainty in its velocity is `5.5 xx 10^(-20)ms^(-1)` |
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Answer» According to Herisenberg uncertainty principle `Delta x m Delta V_(x) = (h)/(4pi)` `or Delta x = (h)/(4pi m Delta V_(x))` `Delta v_(x) = 5.5 xx 10^(-20 m s^(-1)` ` m = 1 mg = 10^(-6) kg , pi = 3.143 ` `h = 6.626 xx 10^(-34) m^(2)s^(-1)` `:. Delta x = (6.626 xx 10^(-34) kg m^(2)s^(-1))/(4 xx 3.143 xx 10^(-6) kg xx 5.5 xx 10^(-20) ms^(-1))` `= 9.58 xx 10^(-10)m` |
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| 354. |
A certain transition emits `6.37 xx 10^(15)` quats per second per square .Calculate the power out put in joule equare metre per second .Given `lambda = 632.8 nm` |
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Answer» Energy failing per aqure metre per second is .Number of quanta falling per metre second xx Energy of one question `= 6.37 xx 10^(15) xx (hc)/(lambda)` `= 6.37 xx 10^(12) xx (6.25 xx 10^(-34) xx 3 xx 10^(8))/(632.6 xx 10^(-6))` `= 2 xx 10^(-3) J m^(-1)s^(-1)` |
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| 355. |
Assertion: The first ionisation energy of `Be` is greater than that of `B`. Reason: 2p-orbital is lower in energy than 2s-orbital.A. Both Statement I and Statement II are correct Statement II is the correct explanation of Statement IB. Both Statement I and Statement II are correct, Statement II is not the correct explanation of Statement IC. Statement I is correct, Statement II is incorrectD. Statement I is incorrect, Statement II is correct |
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Answer» Correct Answer - C |
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| 356. |
Match the entries in Column I with the correctly related quantum number (s) in Column II. `{:(Column I,Column II),((A)"Orbital angular momentum of the electron",(P)"Principal quantum number in a hydrogen-like atomic orbital"),((B)"A hydrogen-like one-electron wave function",(Q)"Azimuthal quantum number obeying Pauli principle"),((C)"Shape size and orientation of",(R)"Magnetic quantum number hydrogen-like atomic orbitals"),((D)"Probability density at the nucleus in hydrogen like-atom",(S)"Electron spin of electron quantum number"):}` |
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Answer» Correct Answer - A-q, B-s, C-p,q,r, D-p,q Orbital concept |
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| 357. |
Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate your answer by darkening the appropriate bubbles of the `4xx4` matrix given in the ORS. `{:(,"Column I",,"Column II",),((A),"Orbital angular momentum of the",(P),"Principal quantum number",),(,"electron in a hydrogen-like atomic orbital",,,),((B),"A hydrogen-like one-electron wave",(Q),"Azimuthal equation number",),(,"function obeying Pauli principle",,,),((C),"Shape.size and orientation of hydrogen like atomic orbitals" ,(R),"Magnetic quantum number",),((D),"Probability density of electron at the nucleus" ,(S),"Electron spin quantum number",),(,"in hydrogen-like atom",,,):}` |
| Answer» Correct Answer - (A) QR(B) P,Q,R,S(C ) P,Q,R(D) P,Q | |
| 358. |
The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one 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 is to the ground state energy of the hydrogen atom. The orbital angular momentum quantum number of the state `S_(2)` isA. 1sB. 2sC. 2pD. 3s |
| Answer» Correct Answer - B | |
| 359. |
The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one 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 is to the ground state energy of the hydrogen atom. The orbital angular momentum quantum number of the state `S_(2)` is |
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Answer» Correct Answer - B |
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| 360. |
The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one 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 is to the ground state energy of the hydrogen atom. The orbital angular momentum quantum number of the state `S_(2)` isA. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - B Orbital angular momentum quantum number of 3p subshell, i.e., `l=1` `underset(2s)(S_(1))overset("Transition")tounderset(3p)(S_(2))` |
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| 361. |
For an orbital in `B^(+4)` radial function is : `R(r ) = (1)/(9sqrt(6))((z)/(a_(0)))^((3)/(4))(4-sigma)sigma e^(-sigma//2` where `sigma = (Zr)/(a_(0))` and `a_(0)=0.529Å,Z` = atomic number, `r=` radial distance from nucleus. The radial node of orbital is at distance from nucleous.A. `0.529Å`B. `2.12Å`C. `1.06Å`D. `0.423Å` |
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Answer» For radial node `[R(r )]=0` `sigma =4 = (Zr)/(a_(0))" "implies(4a_(0))/(Z)=(4xx0.529Å)/(5)=0.423Å` |
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| 362. |
The wave function of 3s and `3p_(z)` orbitals are given by : `Psi_(3s) = 1/(9sqrt3) ((1)/(4pi))^(1//2) ((Z)/(sigma_(0)))^(3//2)(6=6sigma+sigma)e^(-sigma//2)` `Psi_(3s_(z))=1/(9sqrt6)((3)/(4pi))^(1//2)((Z)/(sigma_(0)))^(3//2)(4-sigma)sigmae^(-sigma//2)cos0,` `sigma=(2Zr)/(nalpha_(0))` where`alpha_(0)=1st` Bohr radius , Z= charge number of nucleus, r= distance from nucleus. From this we can conclude:A. Total number of nodal surface is same for 3s and `3p_(x)` orbitalsB. The angular nodal surface of `3p_(z)` orbital occur at `0=(pi)/2`C. The radial nodal surface of 3s and `3p_(z)` orbitals are at equal distance from nucleus.D. 3s electrons have greater penetrating power into the nucleus compared to `3p_(z)` electron. |
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Answer» Correct Answer - a b d |
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| 363. |
Question : Is the orbital of hydrogen atom `3p_(x)` ? STATE 1 : The radial function of the orbital is `R(r ) = (1)/(9sqrt(6)a_(0)^(3//2))(4-sigma)sigma e^(-sigma//2),sigma = (r )/(2)` STATE 2 : The orbital has 1 radial node `& 0` angular node.(a). Statement `(1)` alone is sufficient.(b). Statement `(2)` alone is sufficient(c). Both together is sufficient(d). Neither is sufficientA. Statement (1) alone is sufficient.B. Statement (2) alone is sufficient.C. Both togther is sufficient.D. Neither is sufficient. |
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Answer» Correct Answer - B |
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| 364. |
Question : Is the orbital of hydrogen atom `3p_(x)` ? STATE 1 : The radial function of the orbital is `R(r ) = (1)/(9sqrt(6)a_(0)^(3//2))(4-sigma)sigma e^(-sigma//2),sigma = (r )/(2)` STATE 2 : The orbital has 1 radial node `& 0` angular node.A. Statement `(1)` alone is sufficient.B. Statement `(2)` alone is sufficientC. Both together is sufficientD. Neither is sufficient |
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Answer» Correct Answer - B `x rarr y + ._(2)^(4) H_(2)` `y rarr O^(18) + ._(1)H^(1)` `y = 9^(y^(19^(8)))` `x=11^(X^(23))` Hence `x=Na` Na present in `3^(rd)` period No of neutron `=23-11implies12` mole of `Na = (4.6)/(23)implies 0.2` Mole of neutron `implies 0.2 xx12 implies 2.4` |
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| 365. |
Calculate the wavelength of the radiation that would cause photo dissociation of chlorine molecule if the `Cl-Cl` bond energy is `243kJ//mol`. |
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Answer» Correct Answer - `4.9xx10^(-7)m` `DeltaE=(hc)/(lambda)` (put value of `DeltaE` in eV/atoms) `lambda=(hc)/(DeltaE)=4.9xx10^(-7)m` |
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| 366. |
The valence electronic configuration of an atom is `6s^2` If d-orbital of the penultimate shell contains two unpaired electrons, calculate the atomic number |
| Answer» Since the valence electronic configuration of the atom is `6s^2` and cl-orbital of the penultinute shell contains two unpaired electrons, the electromc configuration is `[Xe]6s^24f^(14)5d^2`Therefore, the atomic number of the elen1ent is 72. | |
| 367. |
The ionisation energy of `He^(o+)` is `19.6 xx 10^(-18) J atom^(-1)` .The energy of the first stationary state of `Li^(2+)` will beA. `84.2 xx 10^(-18) J" atom" ^(-1)`B. `84.10 xx 10^(-18) J "atom" ^(-1)`C. `63.2 xx 10^(-18) J "atom" ^(-1)`D. `21.2 xx 10^(-18 J) "atom" ^(-1)` |
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Answer» `E_(1)` "for"` Li^(2+) = E_(1) for H xx Z^(2)for H xx 4` E_(1)`" for" `He^(o+) = E_(1) xx Z_(He)^(2)= E_(1)"for" H xx 4` or `E_(1)" for" Li^(2+) = (9)/(4) E_(1) for He^(o+) = 19.6 xx 10^(-16) xx (9)/(4) = 44.10 xx 10^(-19) J" atom"^(-1)` |
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| 368. |
The ionisation energy of `He^(o+)` is `19.6 xx 10^(-18) J atom^(-1)` .The energy of the first stationary state of `Li^(2+)` will be |
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Answer» ` E_1 "for" He^+ = E_1 "for" H xx z^2 = E_1 xx 4` ` E_1 "for" Li^(2+) = E_1 "for" Hxx Z^2 =E_1 xx 9` ` therefore E_1 "for" Li^2+ = E_1 "for" He^+ xx 9//4` ` = 19 . 6 xx 10 ^(-18) xx 9//4` ` = 44.1xx 10 ^(-18) "J atom"^(-1)` . |
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| 369. |
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. `8.82xx10^(-17)J atom^(-1)`B. `4.41xx10^(-16)J atom^(-1)`C. `-4.41xx10^(-17)J atom^(-1)`D. `-2.2xx10^(-15)J atom^(-1)` |
| Answer» Correct Answer - C | |
| 370. |
The maximum probability of finding electron in the `d_(xy)` orbital is -A. Along with x-axisB. Along the y-axisC. At an angle of `45^(@)` from the X and Y axisD. At an angle of `90^(@)` from the x and y axis |
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Answer» Correct Answer - C in xy plane in between x and y axes |
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| 371. |
The radius of first Bohr orbit is x, then de-Broglie wavelength of electron in 3rd orbit is nearlyA. `2px`B. `6px`C. `9x`D. `x//3` |
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Answer» Correct Answer - B `r_(n) = n^(2)r_(1), mv_(n)r_(n) = (nh)/(2pi), lambda = (h)/(mv_(n))` |
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| 372. |
If the velocity of hydrogen molecule is `5 xx 10^4 cm sec^-1`, then its de-Broglie wavelength is.A. `2 Å`B. `4 Å`C. `8 Å`D. `100 Å` |
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Answer» Correct Answer - B (b) According to de-Broglie `lamda = (h)/(mv) = (6.62 xx 10^-20 erg.sec)/((2)/(6.023 xx 10^23) xx 5 xx 10^4 cm//sec)` =`(6.62 xx 10^-27 xx 6.023 xx 10^23)/(2 xx 5 xx 10^4) cm` =`4 xx 10^-8 cm = 4 Å`. |
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| 373. |
The frequency of a wave of light is `12xx10^14 s^(-1)` . The wave number associated with this light isA. `5xx10^(-7) m`B. `4xx10^(-8) cm^(-1)`C. `2xx10^(-7) m^(-1)`D. `4xx10^(4) cm^(-1)` |
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Answer» Correct Answer - D Frequency `v=12xx10^14 s^(-1)` and velocity of light `c=3xx10^10 cm s^(-1)` . We know that the wave number `barv=v/c =(12xx10^14)/(3xx10^10)=4xx10^4 cm^(-1)` |
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| 374. |
Probilty of finding a `d_(yz)` electron is zero along theA. x-axisB. y-axisC. z-axisD. all of these |
| Answer» Correct Answer - D | |
| 375. |
Which of the following orbitals has `//` have zero `-` probability of finding the electron in xy plane ?A. `p_x`B. `p_z`C. `d_(yz)`D. `d_(x^2-y^2)` |
| Answer» Correct Answer - C | |
| 376. |
The de-Broglie wavelength of a particle with mass `1 g` and velocity `100 m//sec` is.A. `6.63xx10^(-33)` mB. `6.63xx10^(-34)` mC. `6.63xx10^(-35)` mD. `6.65xx10^(-35)` m |
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Answer» Correct Answer - A `lambda=h/"mv"=(6.63xx10^(-34))/(10^(-3)xx100)=6.63xx10^(-33) m` |
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| 377. |
What is the de-Broglie wavelength associated with the hydrogen electron in its third orbitA. `9.96xx10^(-10)` cmB. `9.96xx10^(-8)` cmC. `9.96xx10^(4)` cmD. `9.96xx10^(8)` cm |
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Answer» Correct Answer - B Velocity of `e^-` in orbit =`Vn=2.188xx10^6 (Z/n)ms^(-1)` and `lambda=h/"mV"=(6.62xx10^(-34)) /(9.1xx10^(-31)xx[(2.188xx10^6)/3]) "as" n=3` on solving `lambda=9.96xx10^(-8)cm = 9.96xx10^(-10) `m |
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| 378. |
Which of the following orbitals will have zero probability of finding the electron in the yz planeA. `P_x`B. `P_y`C. `P_z`D. `d_"yz"` |
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Answer» Correct Answer - A The `P_x`-orbital lies along the x-axis and the probability of finding electron is zero in the xz-plane |
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| 379. |
If the wavelength of the first line of the Balmer series of hydrogen atom is `656.1` `nm ` the wavelngth of the second line of this series would beA. `218.7 nm`B. `328.0 nm`C. `486.0 nm`D. 640.0 nm |
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Answer» Correct Answer - C ( c) `(1)/(lamda_1) = R_H ((1)/(2^2) -(1)/(3^2)) = R_H ((5)/(36))` `R_H = (36)/(5 xx 656.1)` `(1)/(lamda_2)= R_H ((1)/(2^2) -(1)/(4^2))` `(1)/(lamda_2) = (36)/(5 xx 656.1) ((3)/(16))` `lamda = (5 xx 656.1 xx 16)/(3 xx 36) = 486.0 nm`. |
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| 380. |
The velocity of `alpha` particels increases then angle of deviationA. increasesB. decreasesC. remains sameD. Cannot be predicated |
| Answer» Correct Answer - B::C | |
| 381. |
If the wavelength of the first line of the Balmer series of hydrogen atom is `656.1 nm `the wavelngth of the second line of this series would beA. `218.7 nm`B. `328.0 nm`C. `486.nm`D. `640.0nm` |
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Answer» Correct Answer - C `bar v_(1) = (1)/(lambda_(1)) = R ((1)/(2^(2)) - (1)/(3^(2))) = (5R)/(36)` `lambda = (36)/(5R)` `bar v_(2) = (1)/(lambda_(2)) = R ((1)/(2^(2)) - (1)/(4^(2))) = (3R)/(16)` `lambda_(2) = (16)/(3R)` When `(36)/(5R) rarr 656.1 nm` ` (16)/(3R) rArr (656.1 xx 5R xx 16)/(36 xx 3R) = 486 nm` |
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| 382. |
If ` lambda_0` is the Threshold wavelength of a metal for photoelencron emisssion . If the metal is exposed to the light of wavelngth ` lambda` then the velocity of ejectred electgron will be ` sqrt ((2h)/m ( lambda_0 - lambda) K )`. The value fo (K) is :A. `1`B. `c/( lambda_0 lambda)`C. `1/( lambda lambda_0)`D. `(c. lambda)/( lambda_0)` |
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Answer» Correct Answer - B ` (hc)/(lambda) = (hc)/(lambda_0) + 1/2 mv^2` ` v= sqrt ( (2hc)/(m)) [(lambda_0 -lambda)/(lambdalambda_0)]` `:. K = c/(lambda lambda_0)`. |
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| 383. |
Two particels (A) and (B) are in motion . If the wavelngth wavelength associated with particel (B) if its momentum is half of (A). |
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Answer» ` lambda _A = h/(P_A) and lambda _B = h/(p_B)` ` ( lambda _A)/( lambda _B) = P_B/P_A ("Given " P_B = 1/ 2 P_A)` ` ( 5 xx 10^(-8))/( lambda _B) = 1/2` ` :. lambda _B = 1 xx 10^(-7) m`. |
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| 384. |
The longest wavelength of ` He^(+)` in paschen series is "m", then shortest wavelenght of `Be^(+3)` in Pacchen series is( in terms of m):A. `(7)/(64)m`B. `(5)/(36)m`C. mD. `(53)/(8)m` |
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Answer» Correct Answer - A `(1)/(lamda_(He^(+)))=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `(1)/(m)=Rxx4[(1)/(9)-(1)/(16)]=(7R)/(36)`. . . .(i) `(1)/(lamda_(Be^(3+)))=Rxx16[(1)/(9)-(1)/(infty)]=(16R)/(9)` .. . .(ii) Dividing eq. (i) by (ii) `(lamdaBe^(3+)))/(m)=(7xx9)/(16xx36)` `lamda_(Be^(3+))=(7)/(64)m` |
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| 385. |
The longest wavelength of ` He^(+)` in paschen series is "m", then shortest wavelenght of `Be^(+3)` in Pacchen series is( in terms of m):A. `(5)/(36)m`B. `(64)/(7)m`C. `(53)/(8)m`D. `(7)/(64)m` |
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Answer» Correct Answer - D `(1)/(lambda_("longest"))`=`(1)/(m_(e))`=`RZ^(2)((1)/(3^(2))-(1)/(4^2))`=`(7Rxx4)/(16xx9)=(7)/(36)R` `(1)/(lambda_("longest"))`=`RZ^(2)((1)/(3^(2)))`=`(4R)/(9)` `(lambda_("longest"))`=`(7)/(64)m` |
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| 386. |
In photoelectric effect, the energy of the photon striking a metallic surface is `5.6 xx 10^(-19)J`. The kinetic energy of the ejected electrons is `12.0 xx 10^(-20)J`. The work function is:A. `6.4 xx 10^(-19)J`B. `6.8 xx 10^(-19)J`C. `4.4 xx 10^(-19)J`D. `6.4 xx 10^(-20)J` |
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Answer» Correct Answer - C `E = w_(0) +K.E` |
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| 387. |
In a hydrogen atom , in transition of electron a photon of energy 2.55 eV is emitted , then the change in wavelenght of the eletron isA. `3.32Å`B. `6.64Å`C. `9.97Å`D. None of these |
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Answer» Correct Answer - B `2pir_(n)=nlambda` `4 to 2` `lambda_(4)=(2pir_(4))/(4)" "lambda_(2)=(2pir_(2))/(2)` `lambda_(4)lambda_(2P)=2pi[(0.529(4)^(2))/(1xx4)-(0.529(2)^(2))/(2)]` =`2pi(0.529)(2)` `=4pir_(1)`=`6.64 Å` |
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| 388. |
For which of the following transitions would a hydrogen atom absorb a photon with longest wavelenght ?A. n = 1 to n = 2B. n = 3 to n = 2C. n = 5 to n = 6D. n = 7 to n = 6 |
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Answer» Correct Answer - C |
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| 389. |
A photon of wavelenght is ` 4.0 xx 10^(-7) m` strikes on a mnetal surfce , the work function of metal being ` 3. 4xx 10^(-19) J`. Select the correct statyements :A. The enrgy of pghoton is ` 4. 97 x 10 ^(19) J`B. The kinetic enrgy fo the emeission is ` 1. 57 xx 10^(-19) J`C. The kinetic enrgy fo the emission is ` 0.98 eV`D. The velocity fo photoelectron is ` 5. 87 xx 10^5 ms^(-1)`. |
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Answer» Correct Answer - A::B::C::D `E//"photon" = ( h. c)/(lambda) = ( 6.625 xx 10^(-34) xx 3.0 xx 10^8)/( 4 .0 xx 10^(-7))` ` = 4.97 xx 10^(-19) J` K.E, of the emission `= E _("photon") - W` ` ( :. W= "work function of metal" )` ` =4. 97 xx 10^(-19) - 3 . 4 xx 10^(-19)` ` = 1. 57 xx 10^(-19) J = 0. 98 eV ( :. 1 eV = 1 . 602 xx 10^(-19) J)` ` :. K.E`. of meitted photoelectron `= 1/2` `mu^2` ` :. u = sqrt (( 2 xx k.E.)/(m))` ` = sqrt(( 2 xx 1. 57 xx 10^(-19))/(9. 1 xx 10^(31))` ` u = 5. 87 xx 10^5 ms^(-1)`. |
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| 390. |
Electromagnetic radiation of wavelength `242` nm is just sufficient to ionise a sodium atom. Calculate the energy corresponding to this wavelength and the ionisation potential of Na. |
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Answer» Energy associated with a photon of 242nm =`(6.62xx10^(-34)xx3.0xx10^(8))/(242xx10^(-9)`=`8.21xx10^(-19)("joule")` `because` 1atom of Na for ionisation requires `=8.21xx10^(-19)J` `therefore6.023xx10^(23) ` atoms of Na for ionisation requires `=8.21xx10^(19)xx6.023xx10(23)` `=49.45xx10(4)J`=`494.5 KJ mol_(-1)` |
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| 391. |
An Electromagnetic radiation of wavelength `242` nm is just sufficient to ionise a sodium atom .Calculate the ionisation energy of sodium in `KJ mol^(-1)`. |
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Answer» `lamda=242nm=242xx10^(-9)m` `c=3xx10^(8)ms^(-1)` `E=hv=h.(c)/(lamda)=6.6256xx10^(-34)xx(3xx10^(8))/(242xx10^(-9))` `=0.082xx10^(-17)J=0.082xx10^(-20)kJ` Energy per mole for ionisation `=0.082xx10^(-20)xx6.02xx10^(23)` `=493.6kJmol^(-1)`. |
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| 392. |
de-Broglie proposed dual natrue for elecrton by putting his famous equation ` lambda = ( lambda) /( mu)`, Later on Heisenberg proposed uncertiainty principle as . `Deltap.Deltaxle(h)/(2)(h=(h)/(2pi))` . On the contray particel natrue of electron wa estabilshed on the basis fo photoelectric effect . When a photon strikes the metal suface it given up its energy to the electro . Part of this enrgy ( say W) is used by the electrons to escpe form the metal and the remaining imparts he kinetc enrgy `(1//2 mu^2)` to photoeleton . The potential applied on the surface to reduce the velocity fo photoelecton to zero is known as stoppong potential The element mlst commonly used in photoelectric cell is :A. `Na`B. `Ba`C. `Cs`D. `Ni` |
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Answer» Correct Answer - C Cs las lowest ioisation enrgy . |
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| 393. |
de-Broglie proposed dual natrue for elecrton by putting his famous equation ` lambda = ( lambda) /( mu)`, Later on Heisenberg proposed uncertiainty principle as . `Deltap.Deltaxle(h)/(2)(h=(h)/(2pi))` . On the contray particel natrue of electron wa estabilshed on the basis fo photoelectric effect . When a photon strikes the metal suface it given up its energy to the electro . Part of this enrgy ( say W) is used by the electrons to escpe form the metal and the remaining imparts he kinetc enrgy `(1//2 mu^2)` to photoeleton . The potential applied on the surface to reduce the velocity fo photoelecton to zero is known as stoppong potential .The wavelength fo a golf ball weightin `200 g` and moving at a speed fo `5 meter //hr is of the order :A. ` 10^(-10) m`B. ` 10^(-20) m`C. ` 10^(-30) m`D. ` 10^(-40)m` |
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Answer» Correct Answer - C ` lambda = h/(mu) = ( 6.6 xx 10^(-34) xx 3600)/( 200 xx 10^(-3) xx5)` |
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| 394. |
In H-atom if `r1` is the radius fo first Bohr orbit de-Broglie wavelength of an elecrton in ` 3^(rd)` orbit is :A. `3 pi x `B. `6 pi x`C. `(9x)/(2)`D. `(x)/(2)` |
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Answer» Correct Answer - B |
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| 395. |
Choose the correct statement(s):A. For a particular orbital in hydrogen atom, the wave function may have negative valueB. Radial probability distribution function may have zero value but can never have negative valueC. `3d_(x^(2)-y^(2))` orbital has two angular nodes and one radial node.D. yz and xz planes are nodal planes for `d_(xy)` orbital |
| Answer» Correct Answer - A::B::D | |
| 396. |
A gas absorbs a photon fo ` 355nm` and emits at two wavelengths . If one fo the emission is at ` 6 80` nm , the other si at :A. ` n= 4 , l=1`B. ` n = 4, l=0`C. ` n = 3, 1=2`D. `n= 3, l=1` |
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Answer» Correct Answer - C ` E = hv = hc //lambda ` ` E =E_1 + E_2` or `( hc)/(lambda ) = (hc)/(lambda _1) + (hc)/(lambda _2)` ` rArr lambda = 1/lambda = 1/(lambda _1) +1/(lambda _2) rArr 1/( 355) = 1/(680) + 1/(lambda _2)` `:. lambda _2 = ( 355 xx 680)/( 680 -355) = 742.769 nm ~~ 743 nm`. |
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| 397. |
The outer electronic structure of `3s^2 3p^5` is possessed byA. ClB. OC. ArD. Br |
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Answer» Correct Answer - A The atomic number of chlorine is 17 its configuration is `1s^2 2s^2 2p^6 3s^2 3p^5` |
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| 398. |
Hydrogen `(_(1)H^(1))` Deuterium `(_(1)H^(2))` singly omised helium `(_(1)He^(1))` and doubly ionised lithium `(_(1)Li^(6))^(++)` all have one electron around the nucleus Consider an electron transition from `n = 2 to n = 1` if the wavelength of emitted radiartion are `lambda_(1),lambda_(2), lambda_(3),and lambda_(4)`, repectivelly then approximetely which one of the following is correct ?A. `lamda_(1)=lamda_(2)=4lamda_(3)=9lamda_(4)`B. `lamda_(1)=2lamda_(2)=3lamda_(3)=4lamda_(4)`C. `4lamda_(1)=2lamda_(2)=2lamda_(3)=lamda_(4)`D. `lamda_(1)=2lamda_(2)=2lamda_(3)=lamda_(4)` |
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Answer» Correct Answer - A We know `(1)/(lamda)=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]=RZ^(2)[(1)/(1^(2))-(1)/(2^(2))]` `lamda=(4)/(3RZ^(2))` `lamda_(1)=(4)/(3R),lamda=(4)/(3R)k,lamda_(3)=(4)/(12R),lamda_(4)=(4)/(27R)` `thereforelamda_(1)=lamda_(2)=4lamda_(3)=9lamda_(4)`. |
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| 399. |
What is the potential drop through which an electron , with a de Broglie wavelength of `1.5 "Å"` , should be accelerated , if its de Broglie wavelength should be reduced to `1 "Å"` ?A. 110 voltsB. 70 voltsC. 83 voltsD. 55 volts |
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Answer» Correct Answer - C |
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| 400. |
The follwing diagram indicates the energy levels of a certain atom when the system moves from `2E` level to `E`, a photon of wavelength `lambda` is emitted. The wavelength of photon produced during its transition from `(4E)/(3)` level to `E` is A. `(lambda)/(3)`B. `(3lambda)/(4)`C. `(4lambda)/(3)`D. `3 lambda` |
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Answer» Correct Answer - D `2E - E = E = (hv)/(lambda)` `(4)/(3)E = (E)/(3) =? :.` New wave length `= 3lambda` |
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