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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.
| 1. |
In hydrogen atom, when electron jupms from second to first orbit, then enrgy emitted isA. `-13.6 eV`B. `-27.2 eV`C. `-6.8 eV`D. None of these |
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Answer» Correct Answer - D `E_(n_(1)rarrn_(2)) =- 13.6 [(1)/(n_(2)^(2)) - (1)/(n_(1)^(2))], n_(1) = 2` and `n_(2) = 1` `rArr E_(II) rarr E_(I) =- 13.6 xx (3)/(4) =- 10.2 eV` |
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| 2. |
In hydrogen atom, when electron jupms from second to first orbit, then enrgy emitted isA. `-13.6 eV`B. `-27. 2eV`C. `-6.8 eV`D. None of these |
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Answer» Correct Answer - D `E_(n_(1) rarr n_(2)) = - 13.6 [(1)/(n_(2)^(2)) - (1)/(n_(1)^(2))] , n_(1) = 2` and `n_(2) = 1` `rArr E_(II) rarr E_(1) = - 13.6 xx (3)/(4) = - 10.2 eV` |
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| 3. |
The enrgy of an atom (or ion) in the ground state is `- 54.4 eV ` .If may beA. `He^(+)`B. `Li^(2+)`C. hydrogenD. deuterium |
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Answer» Correct Answer - A `E=(-13.6z^(2))/n^(2)` |
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| 4. |
Whenever a hydrogen atom emits a photon in the Balmer series .A. It may emit another photon in the Balmer seriesB. It must emit another photon in the lyman seriesC. the second photon, if emitted, will have a wavelength of about `122 nm`D. It may emit a second photon, the wavelength of this photon cannot be predicted |
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Answer» Correct Answer - B::C Any traansition causing a photon to be emitted in the Balmer series must end `n = 2`. The must be followed by the transition from `n = 2 to n = 1`, emitting a photon of energy `10.2 eV,which corresponds to a wavelength of about `122 nm`. This belongs to the Lyman series. |
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| 5. |
Electrons of energy `12.09 eV` can excite hydrogen atoms . To which orbit is the electron in the hydrogen atom raisd and what are the wavelengths of the radiations emitted as it dropes back to the ground state? |
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Answer» The energy of the electron in different states are: `E_(1) = - 13.6 e V` for `n = 1`. `E_(2) = - 3.4 e V` for `n = 2`. and `E_(3) = - 1.51 e V` for `n = 3` Evidently the energy needed by an electron to go to the `E_(3)` level `(n = 3` or M- level ) is `13.6 - 1.51 = 12.09 e V`. Thus , the electron ia reised to the third orbit of principal quantum number `n = 3` . Now , an electron in the `n = 3` level can return to the ground state making the following possible jumps: a. `a = 3` to `n = 2` and then from `n = 2 to n = 1` b. `n = 3 to n = 1` Thus, the corresponding wevelength emitting are a. For `n = 3` to `n = 2`: `(1)/(lambda_(1)) = R[(1)/(2^(2)) - (1)/(3^(2))] = (5 R)/(36)` or `lambda_(1) = (36)/(5 R) = (36)/(5 xx 1.097 xx 10^(7)) = 6563 Å` This wevelengths belongs to the Balmar series and lines in the visible region. b. For `n = 2 to n = 1:` `(1)/(lambda_(2)) = R[(1)/(1^(2)) - (1)/(2^(2))] = (3 R)/(4)` or `lambda_(1) = (4)/(4 R) = (4)/(3 xx 1.097 xx 10^(7)) = 1215 Å` `lambda_(2)` belongs to the Lyman series and lines in the ultraviolect region. c. For the direct jump `n = 3 to n = 1`: `(1)/(lambda_(3)) = R[(1)/(1^(2)) - (1)/(3^(2))] = (8 R)/(9)` or `lambda_(3) = (9)/(8 R) = (9)/(8 xx 1.097 xx 10^(7)) = 1026 Å` which also belongs to the Lyman series and lies in the ultraviolet. |
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| 6. |
Find the quantum number `n` corresponding to the excited state of He^(+) ion if on transition to the ground state that ion emits two photons in succession with wavelengths state that ion emits two photons in succession with wavelengths`1026.7 and 304Å. (R = 1.096 xx 10^(7)m^(_1)` |
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Answer» Given ` (1)/(lambda_(1)) + (1)/(lambda_(2)) = RZ^(2) (1 - (1)/(n^(2)))` `(1)/(n^(2)) = 1 - [(lambda_(1) +lambda_(2))/(lambda_(1) lambda_(2)) xx (1)/(RZ^(2))]` `= 1 - (1330.7 xx 10^(-10))/(1026.7 xx 304 xx 10^(-20) xx 4 xx 1.096 xx 10^(7))` Thus `(1)/(n^(2)) = 0.0275 implies 6.03` Hence the quantum number`= 6` |
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| 7. |
Find the quantum number `n` corresponding to the excited state of `He^(+)` ion, if on transition to the ground state that ion emits two photons in succession with wave lengths `108.5` and `30.4 nm`. |
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Answer» Correct Answer - `n=5` Total energy of radiation `E=(hc)/(lambda_(1))+(hc)/(lambda_(2))` `13.6xx4xx(1-(1)/n^(2))=((1240)/(108.5)+(1240)/(30.4))eV` `1-(1)/(n^(2))=(1)/(13.6xx4)xx52.22` |
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| 8. |
Electron of energies 10.20 eV and 12.09 eV` can cause radiation to be emitted from hydrogen atoms . Calculate in each case, the principal quantum number of the orbit to which electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state. |
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Answer» Correct Answer - 2 and 3 We know the orbital energy of an electron revolving in `n^(th)` orbit is given by `E_(n) = - (13.6)/(n^(2)) eV` where `n` is the principal quantum number When `n = 1, E_(1) = - 13.6 eV`, `n = 2, E_(2) = - 3.4 eV`, `n = 3, E_(3) = - 1.51 eV`. energy needed by an electron to go from `L toL` level is `(13.6 - 3.4) = 10.2 eV` and that required to go form `K to M level is (13.6 - 1.51) = 12.09 eV`. The corresponding quantum numbers are `n = 2 to n = 3`, respestively. Hence electron will be raised to principal quantum numbers `2 and 3` corresponding to energies `10.20 eV and 12.09 eV`, respestively. `n = 2 and 3` When electron are coming back from `n = 2 to n = 3` to the ground state , i.e., `n = 1`. That is the case of Lyman series. Hence . `(1)/(lambda_(1)) = R[(1)/(1^(2)) - (1)/(2^(2))] = (3 R)/(4)` ` implies lambda_(1) = (4)/(3 R) = (4)/(3 xx (10.97 xx 10^(6))) = 1216 Å` and `(1)/(lambda_(2)) = R[(1)/(1^(2)) - (1)/(3^(2))] = (8 R)/(9)` ` lambda_(2) = (9)/(8 R) = (9)/(8 xx (10.97 xx 10^(6))) = 1020 Å` `("Rounding off to nearest interger")` |
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| 9. |
A hydrogen atom ia in excited state of principal quantum number `n` . It emits a photon of wavelength `lambda` when it returnesto the ground state. The value of `n` isA. `sqrt(lambda R (lambda R - 1))`B. `sqrt((lambda (R - 1))/(lambda R))`C. `sqrt((lambda R)/(lambda R - 1))`D. `sqrt lambda (R - 1)` |
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Answer» Correct Answer - C `(hc)/lambda = R hc (1 - 1//n^(2))` or `n = sqrt((lambda R)/(lambda R - 1))` |
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| 10. |
The value of wavelength radiation emitted by an `H`-atom, if atom is excited to state with principal quantum number four is :A. `(5R)/(36)`B. `(16)/(15R)`C. `(36)/(5R`D. `(3R)/(16)` |
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Answer» Correct Answer - B `1/lambda=RZ^(2)(1/n^(2_(2))1/n_(2)^(2))` |
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| 11. |
Assertion: Hydrogen atom consists of anly one electron but its emission spectrum has may lines. Reason: Only Lyman series is found in the absorption spectrum of hydrogen atom whereas in the emission spectrum, all the series are found.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B When the atom gets appropriate energy from ouiside, then this electron rises to some higher energy level. Now, it can return either directly to the lower energy level or come to the lowest enegry level after passing through other lower energy ends, hence all possible transitions take place in the source and many lines are seen in the spectrum. |
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| 12. |
Assertion: The electron in the hydrogen atom passes from energy level `n = 4` to the `n = 1` level. The maximum and minimum number of photon that can be emitted are six and one respectively. Reason: The photons are emitted when electron make a transtition from the higher energy state to the lower energy state.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B Maximum number of photon is given by all the transitins possible `=^(4_(C_(2)) = 6`. Minimum number of transition `=1`, that is, directly jumps from `4` to `1`. |
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| 13. |
A sample of hydrogen gas in its ground state is irrdation with photon of `10.2eV` energies The radiation from the above the sample is used to irradiate two other the sample of excited ionized `He^(+)` and excited ionized `Li^(2+)`, respectively . Both the ionized sample absorb the incident radiation. A gas of monoatomic hydrogen is bombared with a stream of electron that have been accelerated from rest through a potential difference of `12.75` volt. in the emission spectrum, one cannot observed any line ofA. Lyman seriesB. Balmer seriesC. Paschen seriesD. Pfund series |
| Answer» Correct Answer - D | |
| 14. |
A hydrogen atom in ground state absorbs `10.2eV` of energy .The orbital angular momentum of the electron is increases byA. `1.05 xx 10^(-34) J s`B. `2.11 xx 10^(-34) J s`C. `3.16 xx 10^(-34) J s`D. `4.22 xx 10^(-34) J s` |
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Answer» Correct Answer - A `- 13.6 - (-10.2) = - 3.4 eV`. `(-13.6)/(n^(2)) = - 3.4 or n^(2) = (13.6)/(3.4) = 4` or `n = 2` Increase in angular momentum `= (2 h)/(2 pi) = (h)/(2 pi) = (h)/(2 pi)` `= (6.625 xx 10^(-34))/(2 xx 3.14) J s` `= 1.05 xx 10^(-34) J s` |
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| 15. |
A hydrogen atom in its ground state absorbs `10.2 eV` of energy. The orbital angular momentum is increased byA. `1.05xx10^(-34)` J-secB. `3.36xx10^(-34)` J-secC. `2.11xx10^(-34)` J-secD. `4.22xx10^(-34)` J-sec |
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Answer» Correct Answer - A |
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| 16. |
A hydrogen atom in its ground state absorbs `10.2 eV` of energy. The orbital angular momentum is increased byA. `1.05 xx 10^(-34)J-sec`B. `3.36 xx 10^(-34)J-sec`C. `2.11 xx 10^(-34)J-sec`D. `4.22 xx 10^(-34)J-sec` |
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Answer» Correct Answer - A Electron after absorbing `10.2 eV` energy goes to its first excited sate `(n = 2)` from ground state `(n = 1)`. `:.` Increase in momentum `= (h)/(2pi)` `= (6.6 xx 10^(-34))/(6.28) = 1.05 xx 10^(-34) j-sec` |
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| 17. |
An electron in the `n = 1` orbit of hydrogen atom is bound by `13.6 eV`. If a hydrogen atom I sin the `n = 3` state, how much energy is required to ionize itA. `13.6 eV`B. `4.53 eV`C. `3.4 eV`D. `1.51 eV` |
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Answer» Correct Answer - D Required eenrgy `E_(3) = (+13.6)/(3^(3)) = 1.51eV` |
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| 18. |
The relation between `lambda_(1)=` wavelength of series limit of Lyman series ,`lambda_(2)=`the wavelength of the series limit of Balmer sereis & `lambda_(3)=` the wavelength of first line of Lyman series:A. `lambda_(1)=lambda_(2)+lambda_(3)`B. `lambda_(3)=lambda_(1)+lambda_(2)`C. `lambda_(2)=lambda_(3)-lambda_(1)`D. `(1)/(lambda_(1))-(1)/(lambda_(2))=(1)/(lambda_(3))` |
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Answer» Correct Answer - D `(K)/(lambda_(1))=E_(oo)-E_(1) " "rArr(K)/(lambda_(2))= E_(oo)-E_(2)` `(K)/(lambda_(3))=E_(2)-E_(3) " "rArr(1)/(lambda_(1))-(1)/(lambda_(2))=(1)/(lambda_(3))` |
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| 19. |
If the series limit of Lyman series for Hydrogen atom is equal to the series limit Balmer series for a hydorgen like atom, then atomic number of this hydrogen-like atom will beA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - B By using `(1)/(lambda) = RZ^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` For Hydrogen atom `(1)/((lambda_(min))_(H)) = R [(1)/(1^(2))-(1)/(oo)] = R` `rArr(lambda_(min)_(H)) = (1)/(R ) …….(i)` For hydrogen-like atom `((1)/(lambda_(min)))_("atom") = RZ^(2)((1)/(2^(2))-(1)/(oo))` `rArr (lambda_(min))_("atom") = (4)/(RZ^(2)) .....(ii)` From equation (i) and (ii) `(1)/(R ) = (4)/(RZ^(2)) rArr Z = 2`. |
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| 20. |
Energy of an electron in an excited hydrogen atom is `-3.4eV`. Its angualr momentum will be: `h = 6.626 xx 10^(-34) J-s`.A. `2.11 xx 10^(-34)`B. `3 xx 10^(-34)`C. `1.055 xx 10^(-34)`D. `0.5 xx 10^(-34)` |
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Answer» Correct Answer - A `E_(n) = - 3.4 e V, E_(n) prop - (1)/(n^(2))` ` E_(1) = - 13.6 e V`. Clearly `n = 2` `= (n h)/(2 pi) = (2 h)/(2 pi) = (h)/(pi) = 2.11 xx 10^(-34) J s` |
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| 21. |
The wavelength of the first line of Lyman series in hydrogen atom is `1216`. The wavelength of the first line of Lyman series for `10` times ionized sodium atom will be addedA. `0.1 Å`B. `1000 Å`C. `100 Å`D. `10 Å` |
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Answer» Correct Answer - D `lambda prop (1)/(Z^(2))` Now, `lambda_(Na) = (1216)/(11 xx 11) = 10 Å` |
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| 22. |
The ionisation potential of hydrogen atom is `13.6` volt. The energy required to remove an electron in the `n = 2` state of the hydrogen atom isA. `27.2 eV`B. `13.6 eV`C. `6.8 eV`D. `3.4 eV` |
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Answer» Correct Answer - D Energy required to remove electron in the `n = 2` state `= + (13.6)/((2)^(2)) = + 3.4 eV` |
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| 23. |
Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximatelyA. `1.9 eV`B. `2.3 eV`C. `3.4 eV`D. `4.5 eV` |
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Answer» Correct Answer - A `13.6 ((1)/(2^(2)) - (1)/(3^(2))) eV = 1.9 eV` |
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| 24. |
The energy of hydrogen atom in its ground state is `-13.6 eV`. The energy of the level corresponding to the quantum number `n` is equal `5` isA. `-5.40 eV`B. `-2.72eV`C. `-0.85 eV`D. `-0.54 eV` |
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Answer» Correct Answer - D `E_(n) = - (13.6)/(n^(2)) eV rArr E_(5) = (-13.6)/(5^(2)) = (-13.6)/(25) = - 0.54 eV` |
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| 25. |
Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the wavelength of the resulting radiation?A. `lambda = (h)/(m_(0)c)`B. `lambda = (2 h)/(m_(0)c^(2))`C. `lambda = (h)/(2 m_(0)c^(2))`D. None of these |
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Answer» Correct Answer - A From conservation of momentum, two identical photons travel in opposite direction with equal magnitude of momentum and energy ` hc//lambda` from conservation of energy , we have `(hc)/(lambda) + (hc)/(lambda) = m_(0) C^(2) + m_(0) C^(2)` `lambda = (h)/(m_(0) C)` |
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| 26. |
The orbiting `v_(n) of e^(bar)` in the nth orbit in the case of positronium is x-field compared to that in the nth orbit in a hydrogen atom , where `x` has the valueA. 1B. `sqrt2`C. `1// sqrt2`D. 2 |
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Answer» Correct Answer - A `m v_(0)^(2) = K//r_(n)`. Since modified `m` is half and modified `r_(n)` is double, `v_(n)` remain the same as in `H`- atom. |
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| 27. |
Statement-1: The de-Broglie wavelength of a molecules (in a sample of ideal gas ) varies inversely as the square root of absolute temperature The arm velocity of a molecules (in a sample of ideal gas ) depend in temperature. |
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Answer» Correct Answer - B de-Broglie wavelength associated with gas molecules varies as `lambdaalpha 1/sqrtT. |
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| 28. |
If an electron in `n = 3` orbit of hydrogen atom jumps down to `n = 2` orbit, the amount of energy reeased and the and the wavelength of radiation emitted areA. `0.85 eV, 6566 Å`B. `0.89 eV, 1240 Å`C. `1.89 eV, 6566 Å`D. `1.5 eV, 6566 Å` |
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Answer» Correct Answer - C `(1)/(lambda) = R[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `(1)/(lambda) = R ((1)/(2^(2)) - (1)/(3^(2)))` implies `lambda = (36)/(5 R) = (36 xx 10^(-7))/(5 xx 1.097) = 6.566 xx 10^(-7)` `lambda = 6566 Å` `E = (h c)/(lambda) = (6.64 xx 10^(-34) xx 3 xx 10^(8))/(6.566 xx 10^(-7)) = 3.05 xx 10^(-19) J` `:. E = 1.89 e V` |
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| 29. |
(i) Calculate the first three energy levels for positronium. (ii) Fine the wavelength of the `H_(a)` line `(3 rarr2"transition")` of positronium. |
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Answer» In positromium electron and positron revolve around their centre of mass `(1)/(4 pepsilon_(0))(e^(2))/(r^(2))=(mv^(2))/(r//2)`.......(1) `(nh)/(2pi)=2xxmv_(k//2)` From (1) & (2) `V=(1)/(2).(1)/(4 piepsilon_(0)).(e^(2))/(nh)xx2p=(e^(2))/(4 epsilon_(0)nh)` `TE= -(1)/(2) mv^(2)xx2= -m (e^(4))/(16 epsilon_(0)^(2)n^(2)h^(2))` `= -6.8(1)/(n^(2))eV` (i) `E_(1)= -6.8 ev` `E_(2)= -6.8xx(1)/(2^(2))eV= -1.70 eV` `E_(3)= -6.8xx(1)/(3^(2))eV= -0.76-(-1.70)eV` The corresponding wave length `lambda(1.24xx10^(4))/(0.94)Å=1313Å` (i) `-6.8eV, -1.7 eV, -0.76 eV`, (ii) `1313Å` |
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| 30. |
In which of the following transition will the wavelength be minimum ?A. `n _(1) = 5, n _(2) = 4`B. `n _(1) = 4, n _(2) = 3`C. `n _(1) = 3, n _(2) = 2`D. `n _(1) = 2, n _(2) = 1` |
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Answer» Correct Answer - D In transition of electron from higher energy to lower energy level , the wavelength is given by `lambda = hc//Delta E, where Delta E` is the energy different between two levels, for minimum `lambda,Delta E` should be maximum , so (d) is the correct option. |
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| 31. |
Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in `.^(1)H` and `.^(2)H`. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass `mu`, revolving around the nucleus at a distance equal to the electron -nucleus separation. Here `mu = m_(e) M//(m_(e)+M)`, where M is the nuclear mass and `m_(e)` is the electronic mass. Estimate the percentage difference in wavelength for the `1st` line of the Lyman series in `.^(1)H` and `.^(2)H`. (mass of `.^(1)H` nucleus is `1.6725 xx 10^(-27)` kg, mass of `.^(2)H` nucleus is `3.3374 xx 10^(-27)` kg, Mass of electron `= 9.109 xx 10^(-31) kg`.)A. `8.1xx10^(-2)%`B. `2.7xx10^(-2)%`C. `0.81%`D. `0.27%` |
| Answer» Correct Answer - B | |
| 32. |
The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number of `Z` ) is the same as the shortest wavelength of the Balmer series of hydrogen atom. The value of `z` isA. 4B. 2C. 3D. 6 |
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Answer» Correct Answer - B `R [(1)/(1^(2)) - (1)/(oo^(2))] = R Z^(2) [(1)/(2^(2)) - (1)/(oo^(2))] or Z = 2` |
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| 33. |
In which of the following transition will the wavelength be minimum ?A. `n = 5 to n = 4`B. `n = 4 to n = 3`C. `n = 3 to n = 2`D. `n = 2 to n = 1` |
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Answer» Correct Answer - D `lambda _(min)` is found for `n = 2 to 1` since energy gap is maximum. |
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| 34. |
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15eV. What wil be observed by the detector? (a) 2 photons of energy 10.2 eV (b) 2 photons of energy 1.4 eV (c ) One photon of energy 10.2 eV and an electron of energy 1.4 eV (d) One photon of energy 10.2 eV and another photon of energy 1.4 eVA. Two proton of energy `10.2 eV`B. Two proton of energy `1.4 eV`C. One photon of energy `10.2 eV` and one electron of energy `1.4 eV`D. One electron having kinetic energy nearly `11.6 eV`. |
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Answer» Correct Answer - D Total energy received by the atom will be `25.2 eV, 13.6 eV` energy is needed to remove the electron from the atteraction of the nucleus. Rest of the energy will be almost available in the from of `KE` of electron |
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| 35. |
A monochromatic light soure of frequency f illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the whole experiment is repeated with an incident radiation of frequency `(5)/(6) f`, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength `1215 Å`. (a) What is the frequency of radiation? (b) Find the work- function of the metal. |
| Answer» Correct Answer - `upsilon=(6)/(h)[13.6-(hcxx10^(10))/(1215)]=5xx10^(15)Hz`, | |
| 36. |
A beam of ultraviolet radius hacking wavelength between `100nm and 200nm` is inclined on a sample of atomic hydrogen gas. Assuming that the atoms are in ground state which wavelength will have low intensity in the transmitted been? If the energy of a photon is equal to the ground state it hasl arge probability of being observed by on atom in the ground state |
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Answer» Correct Answer - `[lambda_(1) = 122 nm; lambda_(2) = 103 nm]` |
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| 37. |
Order of `q//m` ratio of proton, `alpha`-particle and electron isA. `e gt p gt alpha`B. `p gt alpha gt e`C. `e gt alpha gt p`D. None of these |
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Answer» Correct Answer - A `:. m_(e) lt m_(p) lt m_(alpha) rArr ((q)/(m))_(e) gt ((q)/(m))_(p) gt ((q)/(m))_(alpha)` |
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| 38. |
The energy of a photon is `E = hv` and the momentum of photon `p = (h)/(lambda)` , then the velocity of photon will beA. E/pB. EpC. `[E//p]^(2)`D. `[E//p]^(1//2)` |
| Answer» Correct Answer - A | |
| 39. |
The ratio between total acceleration of the electron in singly ionized helium atom and hydrogen atom (both in ground state) isA. 1B. 8C. 4D. 16 |
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Answer» Correct Answer - B `a = v^(2)/( r)` `:. A prop ((z)^(2))/((l//z))` `(for n = 1)` or `a prop z^(3)` `:. (a_(1))/(a_(2)) xx ((2)/(1)) ^(2) = 8` |
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| 40. |
The longest wavelength that the singly ionized helium atom in its ground state will obsorb isA. `912 Å`B. `304 Å`C. `606 Å`D. `1216 Å` |
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Answer» Correct Answer - B `lambda = (912 Å)/(Z^(2) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` for singly ionized helium atom `Z = 2` for the wavelength to be longest, `n_(1) = 1, n_(2) = 2` `:. lambda = (912 Å)/((2^(2))[1 - (1)/(4)]) = (912)/(3) = 304 Å` |
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| 41. |
The ratio between total acceleration of the electron in singly ionized helium atom and hydrogen atom (both in ground state) isA. `1`B. `8`C. `4`D. `16` |
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Answer» Correct Answer - B `a=v^(2)/rthereforeaprop(z)^(2)/((1//z))orthereforeapropz^(3)thereforea_(1)/a_(2)=8` |
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| 42. |
The circumference of the second Bohr orbit of electron in hydrogen atom is `600nm` . The potential difference that must be applied between the plates so that the electron have the de Broglie wavelength corresponding in this circumference isA. `10^(-5) V`B. `(5)/(3)10^(-5) V`C. `5 xx 10^(-5) V`D. `3 xx10^(-5) V` |
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Answer» Correct Answer - D de Brogile wavelength of electron in hydrogen atom`= (h)/(mv) = (2 pi r_(n))/(n)` For second Bohr orbit, `lambda = (600 xx 10^(-9))/(2) = 3000 xx 10^(-9) m` `lambda = sqrt((150)/(V)) Å = 300 Å` `:. V = (150)/((3000)^(2)) = (5)/(3) xx 10^(-5) V` |
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| 43. |
In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential eenrgy of the electron isA. `1:2`B. `-1:2`C. `2:1`D. `-2:1` |
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Answer» Correct Answer - B `PE=-2K.E` |
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| 44. |
The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of `3`:A. `6.57xx10^(15)`B. `6.57xx10^(13)`C. `1000`D. `6.57xx10^(14)` |
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Answer» Correct Answer - A `v_(1)(t)=n.2pir_(1)Rightarrown=?` |
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| 45. |
Statement I: In an X-ray tube , if the energy with which an electron strickes the metal target increases , then the wavelength of the characteristic X-rays also changes. Statement II : Wavelength of the characteristic X-rays depends only on the initial and final energy levels.A. Statement I is True , Statement II is True , Statement II is a correct explanation for Statement I.B. Statement I is True , Statement II is True , Statement II is NOT a correct explanation for Statement I.C. Statement I is True , Statement II is False.D. Statement I is False , Statement II is True. |
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Answer» Correct Answer - D As the energy of striking electron is increased, the wavelength of the characteristic X-ray does not change as characteristic X-ray are emitted when electrons are making transition from a higher energy level to a lower of characteristic X-ray is given by `E = (hc)/(lambda) = |E_(f)| - |E_(i)| = E_(i) - E_(f)` which does not depend at all on the energy of the striking electron. The only dependence is that the striking electron from the inner shell. |
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| 46. |
After absorbing a slowly moving neutrons of mass `m_(N)` (momentum ~0) a nucleus of mass M breaks into two nucleii of mass `m_(1) and 5m_(1)(6m_(1)=M+m_(N))`, respectively . If the de-Broglie wavelength of the nucleus with mass `m_(1) ` is ` lambda`, then de Broglie wavelength of the other nucleus will beA. `5 lambda`B. `lambda//5`C. `lambda`D. `25 lambda` |
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Answer» Correct Answer - C `P_(i)=0` `P_(f)=P_(1)+P_(2)` `P_(i)=P_(f)` `0=P_(1)+P_(2)` `(P_(1)=-P_(2))` `lambda_(1)=(h)/(P_(1))` `lambda_(2)=(h)/(P_(2))` `|lambda_(1)|=|lambda_(2)|` `lambda_(1)=lambda_(2)=lambda` |
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| 47. |
Assertion : An electron in hydrogen in hydrogen atom passes from `n=4` to `n=1` level . The mximum number of photons that can be emitted is 6. Reasoning : maximum nymber of photons omitted can only be `4`.A. If both assertion and reason are true and reason is the correct explaination of assertion.B. If Both assertion and reason are true but reason is not the Correct explaination of assertionC. If assertion is true and reason is falseD. If assertion is false but reason is true |
| Answer» Correct Answer - C | |
| 48. |
Monochromatic radiation of wavelength `lambda` is incident on a hydrogen sample in ground state. hydrogen atoms absorb a fraction of light and subsequently and radiation of six different wavelength .Find the value of `lambda`A. `203 nm`B. `95 nm`C. `80 nm`D. `73 nm` |
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Answer» Correct Answer - B In the emittion spectrum `10` lines are observed, so the energy level `n` to which the sample has been excited after absorbing the radiation is given by `(n (n - 1))/(2) = 10` `"which given"` `n = 5` so, `(h c)/(lambda) = 13.6 (1 - (1)/(5^(2))) eV` `(1242)/(lambda) eV- nm = 13.6 xx (24)/(25) eV` `:. lambda = 95 nm` |
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| 49. |
The average life time of an excited state of an electron in Hydrogen is of order of `10^(-8)s. It makes (xxx10^(6))` revolutions when it is in the state `n=2` and before it suffers a transition `n=1` state. Find the value of `xx` . Given `h/(4pi^(2)ma_(-0)^(2))=64xx10^(14)`, where m is mass of electron and `a_(0)` is Bohrs radius . |
| Answer» `r=(mv)/(Bq)=p/(Bq)RightarrowP=Bqr and lambda=h/p` i.e `lambda_(1)/lambda_(2)=P_(2)/p_(1)=(Bq_(2)r)/(Bq_(1)r)=q_(2)/q_(1)=2/1=2` | |
| 50. |
The de-Broglie wavelength of an electron in the first Bohr orbit isA. Equal to one fourth the circumference of the first orbitB. Equal to half the circyumference of the first orbitC. Equal to twice the circumference of the first orbitD. Equal to the circulference of the first orbit |
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Answer» Correct Answer - D `mvr_(n) = (nh)/(2pi) rArr pr_(n) = (nh)/(2pi) rArr (h)/(lambda) xx r_(n) = (nh)/(2pi)` `rArr lambda = (2pir_(n))/(n)`, for first orbit `n = 1` so `lambda = 2pi r_(1)` `=` circumference of first orbit |
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