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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.
| 101. |
An electron in an atom jumps in such a way that its kinetic energy changes from x to `x/4`. The change in potential energy will be:A. `+(3)/(2)x`B. `-(3)/(8)x`C. `+(3)/(4)x`D. `-(3)/(4)x` |
| Answer» Correct Answer - 1 | |
| 102. |
The number of spectral lines produced when an electron jumps from `5^(th)` orbit to `2^(nd)` orbit in the hydrogen atom is. |
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Answer» Correct Answer - 7 no. of spectral lines `=(Deltan(Deltan+1))/(2)` `Delta n = n_(2) - n_(1)` |
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| 103. |
Which hydrogen -like species will have the same r adius as that of Bohr orbit of hydrogen atom ?A. `n = 2, Li^(2+)`B. `n = 2, Be^(3+)`C. `n = 2, He^(+)`D. `n = 3, Li^(2+)` |
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Answer» Correct Answer - B |
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| 104. |
When the electron of a hydrogen atom jumps from the n=4 to the n=1 state , the number of all pos - sible spectral lines emitted is :-A. 15B. 6C. 3D. 4 |
| Answer» Correct Answer - 2 | |
| 105. |
An electron an a hydrogen like species makes a transition from the oth Bohr orbit to the next outer Bohr `(-=n + 1)` .Find an appocimate relation between the dependence of the frequency of the photon abserbed as a function of n .Assume n its to have a large value `(ngt gt1)` |
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Answer» `overset(Delta E_("energy difference"))underset((n rarr n + 1))=hv` `= 2.18 xx 10^(-18) xx Z^(2)((1)/(n^(2)) - (1)/((n+1)^(2)))J` `rArr hv = 2.18 xx 10^(-18) xx Z^(2)((2n + 1)/(n^(2)(n+1)^(2))) J` Since `n gt gt1` (given) `n + 1~ n, 2n +1 ~~ 2n` `rArr hv= 2.18 xx 10^(-18)Z^(2) xx (2n)/(n^(4)) J` `rArr v prop n^(-3)` |
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| 106. |
Calculate the frequency of light amitted in an electron transition from the sixth to the second orbit of a hydrogen atom .In what region of the specturm does this frequency accur? |
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Answer» We know that `v = 3.289 xx 10^(15)((1)/(2^(2)) - (1)/(6^(2)))` `= 3.289 xx 10^(15) cm s^(-1) (0.2222 cm^(-1))` `= 7.31 xx 10^(4) s^(-1)` The line is in the visible specturm of hydrogen it is a part of the balmer series because the electron has a transition from a higher orbit to the second orbit |
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| 107. |
When an excited hydrogen atom returned to its ground state, some visible quanta were observed along with other quanta. Which of the following transitions must have occurred ?A. `2 to 1`B. `3 to 1`C. `3 to 2`D. `4 to 2` |
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Answer» Correct Answer - A |
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| 108. |
When an excited hydrogen atom returned to its ground state, some visible quanta were observed along with other quanta. Which of the following transitions must have occurred ?A. `2 rarr 1`B. `3 rarr 1`C. `3 rarr 2`D. `4 rarr 2` |
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Answer» Correct Answer - A (a) Since some visible quanta were observed along with other quanta, electrons must have made transition from some higher state to `n = 2` and then from `n = 2 "to" n= 1`. `:.` Transition from `2 rarr 1` is compulsory, because electron from `n = 2` will finally fall into `n = 1`. |
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| 109. |
The electron present in 5th orbit in excited hydrogen atoms returned back to ground state. The no. of lines which appear in Lyman series of hydrogen spectrumA. 5B. 10C. 4D. 6 |
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Answer» Correct Answer - C Lines in a series `= n_(2) - n_(1)` |
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| 110. |
An electron in the third energy level of an excited `He^(o+)` ion return back to the ground sate .The photon emitted in the process is absorbed by a stationary hydrogen atom in the process is absermine by a stationary hydrogen atom in the ground state .Determine the velocity of the photoelectron ejected from the hydrogen atom in metre per second |
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Answer» Transition occurs from `n = 2` to `n = 1` energy released during transition `Delta E = 13.6(Z)^(2)[(1)/(1^(2)) - (1)/(3^(2))]" "He^(o+)(Z = 2)` `= 13.6 xx 2^(2)((8)/(9)) = 13.6 xx 4 xx (8)/(9) = 48.35 J` The energy is abserbed by a stationary hydrogen atom `Delta E = H underset ((n = 1))"atom"]` = e("electron ejected") `:. (Delta E - phi) = KE = (1)/(2) mv^(2)` or `(48.35 - 13.6) = (1)/(2) xx 9.1 xx 10^(-31) xx v^(3)` `or v = ((34.73)/(4.05 xx 10^(-31)))^(1//2) = 3.49 xx 10^(6)ms^(-1)` |
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| 111. |
Outermost electronic comfiguration of the highest electronegative atom is ………..A. `ns^(2)np^(3)`B. `ns^(2)np^(4)`C. `ns^(2)np^(5)`D. `ns^(2)np^(6)` |
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Answer» Correct Answer - C |
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| 112. |
Electronic configuration calcium atom can be written asA. [Ne] `4p^(2)`B. [Ar] `4s^(2)`C. [Ne]`4s^(2)`D. [Kr]`4p^(2)` |
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Answer» Correct Answer - B (b) To write the electronic configuration of an atom, it is better if we remember the atomic number of noble gases and the orbitals follow the noble gas. The atomic number of Ca is 20 and its nearest noble gas is argon (Ar = 18). |
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| 113. |
The outermost electronic configuration of the most electronegative element isA. `ns^2 np^3`B. `ns^2 np^4`C. `ns^2 np^5`D. `ns^2 np^6` |
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Answer» Correct Answer - C The most electronegative element is F, whose outermost electronic configuration is `2s^2 2p^5` |
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| 114. |
Write the electronic configuration of the following and report the number of unpaired electron in eacha. `Mn^(2+)` b.`Cr^(2+)` c`Fe^(2+)` d.`Ni^(2+)` e.`CI^(2+)` f.`Zn^(2+)` g.`Fe^(2+)` h`Na` i.`Mg` .j`Cr(3+)` |
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Answer» Electronic configuration Number of ampaired electron a. `_(23) Mn^(4+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5) 3` b.`_(24) Cr^(2+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6) 4` c.`_(26) Fe^(3+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5) 5` d.`_(28) Ni^(2+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5) 2` e.`_(17) Ci^(0 ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6) 0` f.`_(30) Zn^(2+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10) 0` g.`_(26) Fe^(2+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6) 4` h.`_(17) Na^(4+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(2) 1` i.`_(12) Mg^(4+ ) 1s^(2)2s^(2)2p^(6)3s^(2) 0` j.`_(24) Cr^(4+ ) 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5) 3` |
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| 115. |
The electronic configuration of calcium ion `(Ca^(2+))` isA. `1s^2, 2s^2 2p^6, 3s^2 3p^6 , 4s^2`B. `1s^2, 2s^2 sp^6, 3s^2 3p^6 , 4s^1`C. `1s^2, 2s^2 2p^6, 3s^2 3p^6 , 3d^2`D. `1s^2, 2s^2 2p^6, 3s^2 3p^6 , 4s^0` |
| Answer» Correct Answer - D | |
| 116. |
An ion `(Mn^(a+))` has the magnetic moment equal to `4.9` B.M` What is the value of (a) :A. `3`B. `4`C. `2`D. `5` |
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Answer» Correct Answer - A Magnetic moment ` = sqrt (n(n+2)) = 4.9`, `:. N=4` Thus Mn ion bas four unpaied eelctron `(n) Mn^(3+) : 1s^2 . 2s^2 , 2s^2 2p^y , 3s^2 3p^2 3p^6 3d^4`. |
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| 117. |
The electronic configuration (outermost ) of `Mn^(+2)` ion (atomic number of Mn=25) in its ground state isA. `3d^5 , 4s^0`B. `3d^4, 4s^1`C. `3d^3 , 4s^2`D. `3d^2, 4s^2 4p^2` |
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Answer» Correct Answer - A `._25Mn=[Ar]3d^5 4s^2 underset"+2 electrons"overset"-2 electrons"hArr Mn^(2+) =[Ar]3d^5 4s^0` |
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| 118. |
The electronic configuration of the `Mn^(4+)` ion is -A. `3d^(4)4s^(0)`B. `3d^(2)4s^(1)`C. `3d^(1)4s^(2)`D. `3d^(3)4s^(0)` |
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Answer» Correct Answer - D `._(25)Mn=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(5),4s^(2)` `Mn^(+4)=1s^(2),2s^(2),2p^(6),3s^(2)3p^(6)3d^(3),4s^(0)` |
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| 119. |
A hydrogen atom in its ground state is irradiated by light of wavelength `970Å` Taking `hc//e = 1.237 xx 10^(-6)`eV m and the ground state energy of hydrogen atom as ` - 13.6 eV` the number of lines present in the emmission spectrum is |
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Answer» Correct Answer - 6 Absorbed quantum energy in `(eV)=(hc)/(elamda)` `=(1.237xx10^(-6))/(970xx10^(-10))` `=12.75eV` Energy of nth shell `=-13.6+12.75` `=0.85`eV We know `E_(n)=(E_(1))/(n_(2))` `thereforen^(2)=(E_(1))/(E_(n))=(-13.6)/(-0.85)=16` i.e., `n=4` Number of spectral emission lines `=(n(n-1))/(2)=(4xx3)/(2)=6` |
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| 120. |
A light source of wavelength `lambda` illuminates a metal and ejects photoelectron with `(KE)^(max) = 1 eV.` Another light source of wave length `(lambda)/(3),` ejects photoelectrons from same metal with `(KE)^(max)=5 eV.` Find the value of work function (eV) of metal. |
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Answer» Correct Answer - 1 |
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| 121. |
A light source of wavelength `lambda` illuminates a metal and ejects photo-electrons with `(K.E.)_(max) =1.eV` Another light source of wavelength `(lambda)/(3)`, ejects photo-electrons from same metal with `(K.E.)_(max) = 4eV` Find the value of work function?A. `1eV`B. `2eV`C. `0.5eV`D. `1.5eV` |
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Answer» Correct Answer - C `(hc)/(lambda) = phi +KE` |
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| 122. |
Visible spectrum contains light of following colours 'Violet - Indigo - Blue - Green - Yellow - Orange - Red' (VIBGYOR). Its frequency ranges from violet `(7.5xx10^(14) Hz)` to re `(4xx10^(14) Hz.)` Find out the maximum wavelength `("in" "Å")` in this range. |
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Answer» Correct Answer - 7500 |
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| 123. |
Which state of triply ionised Beryllium `(Be^(+++))` the same orbital radius as that of the ground state hydrogen ? |
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Answer» Correct Answer - 2 |
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| 124. |
An ion `(Mn^(a+))` has the magnetic moment equal to `4.9` B.M` What is the value of (a) : |
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Answer» Correct Answer - 3 `:.` Magnetic moment ` sqrt(n( n + 2)) = 4.9` ` :. n =4` (where (n) no. ofunpared elctron 0 thus `Mn^(a+) ` ion has hour unpaired electron . ` ._(25) Mn^(3+) : 1s^2, 2^2 2p^6 . 3s^ 3p^2 3p^6 3d4` ` :. A=3` . |
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| 125. |
An oil drop has ` 8.01 xx 10^(-19) C` charg .Calculate the number of electrons in this drop. |
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Answer» Correct Answer - 5 Charge on lil drop ` = 8. 01 xx 10^(-19) C` ` :. 1 . 602 xx 10^(-19 ) C` is charge on one electron ` :. 8. 01 xx 10^(-19) C` charge on ` = ( 8. 01 xx 10^(-19))/(1. 602 xx 10^(-19)) = 5` electrons . |
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| 126. |
The ratio of `(E_2 - E_1) "to" (E_4 - E_3)` for the hydrogen atom is approximately equal to.A. 10B. 15C. 17D. 12 |
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Answer» Correct Answer - A (a) `(E_4 - E_3)/(E_2 - E_1) = ((-(1)/(16)) -(-(1)/(9)))/((-(1)/(4))-(-1)) = ((1)/(9) - (1)/(16))/((3)/(4))` =`(7)/(144) xx (4)/(3) = (7)/(78) = (1)/(10) J`. |
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| 127. |
Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon ?A. 3sB. `2p`C. `1s`D. `3d` |
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Answer» Correct Answer - C `E-1` for 1s-orbital is minimum. |
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| 128. |
Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon ?A. `3 s`B. `2 p`C. `2 s`D. `1 s` |
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Answer» Correct Answer - D (d) 1 s-orbital is of lowest energy. Absorption of photon can raise the electron in higher energy state but emission is not possible. |
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| 129. |
Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon ?A. 3 sB. `2 p`C. 2 sD. 1 s |
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Answer» Correct Answer - D (d) Energy is emitted when electron falls from higher energy level to lower energy is absorbed when electron moves from lower level to higher level. `1s` is the lowest energy level of electron in an atom. `:.` An electron in `1 s` level of hydrogen can absorb energy but cannot emit energy. |
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| 130. |
What is the wavelength of the radiation with photon energy which is the mean value of photon energies of radiations with wavelength `lamda_1 = 4000 Å` and `lamda_2 = 6000 Å` ?A. `5000 Å`B. `5200 Å`C. `5600 Å`D. `4800 Å` |
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Answer» Correct Answer - D (d) `(hc)/(lamda_3) = ((hc)/(lamda_1) + (hc)/(lamda_2))/(2)` `rArr (1)/(lamda_3) = (1)/(2) ((1)/(lamda_1) + (1)/(lamda_2))` =`(1)/(2) ((1)/(4000) +(1)/(6000))` `rArr lamda_3 = 4800 Å`. |
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| 131. |
The energies `E_1` and `E_2` of two radiations are `25 eV` and `50 e V` respectively. The relation between their wavelengths, i.e., `lamda_1` and `lamda_2` will be.A. `lambda_1=1/2 lambda_2`B. `lambda_1 =lambda_2`C. `lambda_1 = 2lambda_2`D. `lambda_1=4lambda_2` |
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Answer» Correct Answer - C `lamda_1=2lambda_2` |
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| 132. |
The energies `E_1` and `E_2` of two radiations are `25 eV` and `50 e V` respectively. The relation between their wavelengths, i.e., `lamda_1` and `lamda_2` will be.A. `lamda = lamda_2`B. `lamda_1 = 2 lamda_2`C. `lamda_1 = (1)/(2) lamda_2`D. `lamda_1 = 4 lamda_2` |
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Answer» Correct Answer - B (b) `E_1 = (hc)/(lamda_1)` and `E_2 = (hc)/(lamda_2)` `:. (E_1)/(E_2) = (lamda_2)/(lamda_1)` `(25)/(50) = (lamda_2)/(lamda_1)` `:. lamda_1 = 2 lamda_2`. |
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| 133. |
The wave function of `2s` electron is given by `W_(2s) = (1)/(4sqrt(2pi))((1)/(a_(0)))^(3//2)(2 - (r )/(a_(0)))e^(-1 a0)` It has a node at `r = r_(p)` .Find the radiation between `r_(p)` and a |
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Answer» The wave function of `2s` electron is `W_(2s) = (1)/(4sqrt(2pi)) ((1)/(a_(0)))^(3//2)(2 =- (r )/(qa_(0))) = e^(-e//aq)` Thus the probability of finding of `2s` electron at a point is `W_(2s)^(2)[(1)/(4sqrt(2pi))((1)/(a_(0)))^(3//2)(2 - (r )/(a_(0)))e^(-r//aq)]^(2)` `= (1)/(2pi)((1)/(a_(0)))^(1) (2 - (r )/(a_(0)))^(2)e^(-2raq)` Node is the point ar which the probability of finding an electron is zero means the value of `W_(2s)^(2)` is zero when `r = r_(p)`so `(1)/(32pi)((1)/(a_(0)))^(2)(2 - (r_(0))/(a_(0))) e^(-2q//4b) = 0` `rArr 2 - (r_(0))/(a_(0)) = 0` `:. r_(0) = 2a_(0)` |
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| 134. |
A metal surface is exposed to solar radiations. Which of the following is true?A. The emitted electrons have energy less than a maximum value of energy depending upon frequency of incident radiationsB. The emitted electrons have energy less than maximum value of energy depending upon intensity of incident radiationsC. The emitted electrons have zero energyD. The emitted electrons have energy equal to energy of photons of incident light. |
| Answer» Correct Answer - B | |
| 135. |
The ionisation potential of H-atom is `13.6eV`. It is exposed to electromagnetic radiation of wavelength `1028A^(@)` and gives out induced radiations, thenA. Longest wavelength of induced is `6568A^(@)`B. Lowest wavelength of induced radiation in `102A^(@)`C. Longest wavelength of induced radiation is `3252A^(@)`D. Longest wavelength of induced is `1216A^(@)` |
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Answer» Correct Answer - A `(1)/(lambda_(1)) = (1)/(lambda_(2)) +(1)/(lambda_(3))` |
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| 136. |
The rato of the energies of two different radiations whose frequencies are `3 xx 10^(14)Hz` and `5 xx 10^(14)Hz` isA. `3:5`B. `5:3`C. `3:1`D. `5:1` |
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Answer» Correct Answer - A `(E_(1))/(E_(2)) = (v_(1))/(v_(2))` |
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| 137. |
The wave function orbital of H-like atoms is given as onder `psi_(2s) = (1)/(4sqrt(2pi)) Z^(3//2) (2 - Zr)^(Zr//2)` Given that the radius is in `Å` then which of the following is the radius for nodal surface for `He^(Theta)` ion ?A. `1au`B. `2 au`C. `2.5 au`D. `4 au` |
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Answer» `(2- Zr) = 0 (therefore" Z = 20 "for" He^(o+))` `:. Zr = 2` `r = (2)/(Z) = (2)/(2) = 1 au` |
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| 138. |
2.4 mole of `H_2` sample was taken . In one experiment ` 60%` of the sample was exposed to conitnuous radiations of frequency ` 4.47 xx 10 Hz`, of which all he electrons are removed from the atom . In another experiment remaining sample was irradiated with light of wavelength ` 600Å` , when all the electrons are removed from the surface , Calculate the ratio of maximum velocity of the ejected electron in the two cases . Also report the velocity of ejected electron in each case . Assume that ejected electrons does not interact wih any photon . (Ionization potential of ` H= 13. 6 eV)`. |
| Answer» Correct Answer - ` 0. 83 or 1.22 ,` | |
| 139. |
The expression for Bohr radius of nth orbit of an atoms isA. `r=(n^(2)h^(2))/(4pi^(2)mZe^(2))`B. `r=(nh)/(4pi^(2)mZe^(2))`C. `r=(n^(2)h^(2))/(4pi^(2)mZ)`D. `r=(n^(2)h^(2))/(4pi^(2)me^(2))` |
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Answer» Correct Answer - A `r=(n^(2)h^(2))/(4pi^(2)mZe^(2))=R_(0)n^(2)/Z=0.529n^(2)/Z Å` |
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| 140. |
The differnce between nth and `(n+1)` the Bohr radius of B atom is equal to be its `(n-1)` th Bohr radius .The value of n isA. 1B. 2C. 3D. 4 |
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Answer» `r_(2) prop n^(2)` `But r_(n + 1) - r_(n) = r_(n-1)` `(n+1)^(2) - n^(2) = (n - 1)^(2)` `n = 4` |
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| 141. |
The differnce between nth and `(n+1)` the Bohr radius of B atom is equal to be its `(n-1)` th Bohr radius .The value of n isA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - D `r_n =r_1 xxn^2 , r__(n+1) = r_1 xx (n+1)^2` ` r_(n-1) = r_n (n-1)^2` `:. R_(n+1) =r_n = r_1 [n^2 + 2 n+ 1 -n^2]` ` =r_1 (2 n+1)` ` r_( n-1) = r _(n+1)^(-r_n)` ltbRgt ` :. (n-1)^2 = 2n +1` ` :. N=4`. |
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| 142. |
The particles present in the nucleus of an atom are-A. the proton and the electronB. the electron and the neutronC. the proton and the neutronD. none of these |
| Answer» Correct Answer - C | |
| 143. |
The approximate size of the nucleus of `._28^64 Ni` is :A. `3fm`B. `4fm`C. `5fm`D. `2fm` |
| Answer» Correct Answer - C | |
| 144. |
The approximate size of the nucleus of `._28^64 Ni` is :A. 3 fmB. 4 fmC. 5 fmD. 2 fm |
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Answer» Correct Answer - C |
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| 145. |
A n `alpha`-particle of velocity `1.6xx10^(7) m s^(-1)` approaches a gold nucleii `(Z=79)`. Calculate the distance of closest approach. Mas of an `alpha`-particle `=6.6xx10^(-27) kg`. |
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Answer» Correct Answer - `4.3xx10^(-14) m` |
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| 146. |
Radial probability distribution curve is shown for s-orbital. The curve is:A. 1sB. 2sC. 3sD. 4s |
| Answer» Correct Answer - A | |
| 147. |
Match the column: |
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Answer» Correct Answer - (A)S:(B)P,Q,R;(C)R;)D)Q,R,S (A) `to S` Number of radial nodes =2 curve starts from origin `therefore` s orbital `n-0-1=2` `n=3" ,"3s` (B) ` to P,Q,R` xy plame acts as nodal plane for `p_(y),d_(xy),d_(y,z)` (c) ` to R As 33,[Ar^(18)] 3d^(10) 4s^(2) 4p^(3)` `32^(nd)` electron present in `4p_(y)` (D)` to Q,R,S` number of radial noles =2 `-l-1=2` if `l=0,n=3` `l=n,1=n-4` `l=2,n=5` |
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| 148. |
Match the factors( in column-II) with properticlues ( in column -I) which undergo change with charge in ltbgt factors in column -II. `{:(,"Column I", , "Column II"),((A),"Work function",(P),"Wavelength of photon used" ("less than threshold wavelength")),((B),"Threshold frequency",(Q),"Photo intensity of light"),((C),"Stopping potential",(R),"Distance of point source from metal plate"),((D),"Photo current" (v gt v_(0)),(S),"Metal"):}` |
| Answer» Correct Answer - (A)S,(B)S,(C)P,S(D)Q,R | |
| 149. |
In hydrogen spectum the limiting line the value of n ………… |
| Answer» Correct Answer - Infinity | |
| 150. |
Which of the following series of transitions in the spectrum of hydrogen atom falls in visible region?A. Lyman seriesB. Balmer seriesC. Paschen seriesD. Brackett series |
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Answer» Correct Answer - B |
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