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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. |
The binding energy of a H-atom considering an electron moving around a fixed nuclei (proton), is `B = - (me^(4))/(8n^(2)epsi_(0)^(2)h^(2))` (m= electron mass) If one decides to work in a frame of refrence where the electron is at rest, the proton would be movig around it. By similar arguments, the binding energy would be : `B = - (me^(4))/(8n^(2)epsi_(0)^(2)h^(2))` (M = proton mass) This last expression is not correct, because |
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Answer» Correct Answer - `C` when one decide to work in a frames of reference where the electron is at rest, the given expression is n to true as it forms the non-inertial frame of refrence. |
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| 352. |
`O_(2)`molecules consists of two oxygen atoms. In the molecules , nuclear force between the nuclei of the two atomsA. is not important because nuclear force are short-rangeB. is as important as electrostatic force for binding the two atomsC. Cancels the repulsive electrost atic force between the nucleiD. is not important because oxygen nucleus have equal number of neutrons and protons |
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Answer» Correct Answer - A In the molecules, nuclear force between the nuclei of the two atom is not important because nuclear forces are short--ranged . |
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| 353. |
Two particles of masses `m_(1),m_(2)` move with initial velocities `u_(1)` and `u_(2)`. On collision, one of the particles get excited to higher level, after absording enegry. If final velocities of particles be `v_(1)` and `v_(2)` then we must haveA. `m_(1)u_(1)^(2) + m_(2)u_(2)^(2)-epsilon = m_(1)upsilon_(1)^(2) + m_(2)upsilon_(2)^(2)`B. `(1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2) = (1)/(2)m_(1)upsilon_(1)^(2) + (1)/(2)m_(2)upsilon_(2)^(2) - epsilon`sC. `(1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2) -epsilon = (1)/(2)m_(1)upsilon_(1)^(2) + (1)/(2)m_(2)upsilon_(2)^(2)`D. `(1)/(2)m_(1)^(2)u_(1)^(2) + (1)/(2)m_(2)^(2)u_(2)^(2) + epsilon = (1)/(2)m_(1)^(2)upsilon_(1)^(2) + (1)/(2)m_(2)^(2)upsilon_(2)^(2)` |
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Answer» Correct Answer - C `KE_("Before collision") = (1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2)` `KE_("After collision") = (1)/(2)m_(1) upsilon_(1)^(2) + (1)/(2)m_(2)upsilon_(2)^(2)` ` = (1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2) - epsilon` `KE_("Before collision") = KE_("After collision")` `(1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2) - epsilon = (1)/(2)m_(1)upsilon_(1)^(2) + (1)/(2)m_(2)upsilon_(2)^(2)` |
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| 354. |
Electrons in hydrogen like atom `(Z= 3)` make transition from the fifth to the fourth orbit and from the third orbit. The resulting radiation are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Calculate the work function of the metal and the stopping potential for the photoelectron ejected by longer wavelength (Rydberg constant` = 1.094 xx10^(7) m^(-1)` |
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Answer» Correct Answer - `2` The stopping potential for shorter wavelength is `3.95` volt i.e., maximum kinetic energy of photoelectrons corresponding to shorter wavelength will be `3.95 eV`. Further energy of incident photons corresponding to shorter wavelength will be in transition from `n=4` to `n=3` `E_(4-3)=E_(4)-E_(3)=(-(13.6)(3)^(2))/((4)^(2))-[(-(13.6)(3)^(2))/((3)^(2))]=5.95 eV` Now from the equation, `K_(max)=E-W` we have, `W=E-K_(max)=E_(4-3)-K_(max)` `=(5.95-3.95)eV` `=2eV` |
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| 355. |
An electron having energy `20 e V` collides with a hydrogen atom in the ground state. As a result of the colllision , the atom is excite to a higher energy state and the electron is scattered with reduced velocity. The atom subsequentily returns to its ground state with emission of rediation of wavelength `1.216 xx 10^(-7) m`. Find the velocity of the scattered electron. |
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Answer» The energy lost by the electron in exciting the hydrogen atom equals the energy corresponding to `lambda = 1.216 xx 10^(7) m` `hv = (hc)/(lambda) = (6.63 xx 10^(-34) xx 3.0 xx 10^(8))/(1.216 xx 10^(-7))` Now , the initial energy of electron . `E = 32 xx 10^(-19) J` Hence, the kinetic energy of the scattered electron, `E = 32 xx 10^(-19) J - 16.36 xx 10^(-19) J = 15.65 xx 10^(-19) J` The velocity `v` of the scattered electron is given by `(1)/(2) mv^(2) = E` or `v = ((2 E)/(m))^(1//2) = ((2 xx 15.64 xx 10^(-19))/(9.11 xx 10^(-31))) ^(1//2)` `= 1.86 xx 10^(6) ms^(-1)` |
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| 356. |
The ratio of wavelength of the lest line of Balmer series and the last line Lyman series is:A. 2B. 1C. 4D. `0.5` |
| Answer» Correct Answer - C | |
| 357. |
The minimum enegry required to excite a hydrogen atom from its ground state isA. `13.6 eV`B. `-13.6 eV`C. `3.4 eV`D. `10.2 eV` |
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Answer» Correct Answer - D Minimum energy required to excite from ground state `= 13.6 [(1)/(1^(2)) - (1)/(2^(2))] = 10.2 eV` |
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| 358. |
Ratio of the wavelength of first line of Lyaman series and first line of Balmer series isA. `1 : 3`B. `27 : 5`C. `5 : 27`D. ` 4 : 9` |
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Answer» Correct Answer - C `(1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` For first line of Lymen series `n_(1) = 1` and `n_(2) = 2` For first line of Balmer series `n_(2) = 2` and `n_(2) = 3` So, `(lambda_("Lymen"))/(lambda_("Balmen")) = (5)/(27)` |
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| 359. |
Electrons with energy `80 keV` are incident on the tungsten target of an X - rays tube , k- shell electrons of tungsten have `72.5 keV` energy X- rays emitted by the tube contain onlyA. a continuos X-ray spectrum (Bremsstrablung) with a minimum wavelength of `-0.155 Å`B. a continuos X-ray spectrum (Bremsstrablung) all wavelengthsSC. a continuos X-ray spectrum of tungstenD. a continuos X-ray spectrum (Bremsstrablung) with a minimum wavelength of `-0.155 Å` and the charateristic X-ray spectrum of tungsten |
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Answer» Correct Answer - D As `lambda_(0) = (hc)/(E) = 1.55 xx 10^(-31) m` `:. lambda_(0) =0.155 Å` Which is the minimum wavelength of continuous X-rays which carry every equivalent electrons. Now, as the energy of incident radiation is more than of K-shell electrons, the characteristic X-ray appear as peaks on the continuous spectrum. |
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| 360. |
In the diagram a graph between the intensity of `X`-rays emitted by a molybdenum target and the wavelength is shown , when electrons of `30 keV` are incident on the target. In the graph one peak is of `K_(alpha)` line and the other peak is of `K_(beta)` line A. first peak is of `K_(alpha)` line at `0.6 Å`B. highest peak is of `K_(alpha)` line at `0.6 Å`C. if the energy of incident particle is increased, then the peak will shift toward leftD. If the energy of incident particle is increased then the paak will shift toward right |
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Answer» Correct Answer - B |
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| 361. |
If the energy of a hydrogen atom in `nth` orbit is `E_(n)`, then energy in the nth orbit of a singly ionised helium atom will beA. `4E_(n)`B. `E_(n)//4`C. `2E_(n)`D. `E_(n)//2` |
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Answer» Correct Answer - A We have `E_(n) = (-2pi^(2)mK^(2)Z^(2)e^(4))/(n^(2)h^(2))`. For helium `Z = 2` hence requisite answer is `4E_(n)` |
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| 362. |
The transition from the state `n = 4` to `n = 3` in a hydrogen-like atom results in ultraviolet radiation. Infared radiation will be obtained in the transitionA. `2 rarr 1`B. `3 rarr 2`C. `4 rarr 2`D. `5 rarr 4` |
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Answer» Correct Answer - D As the transition `n = 4` and `n = 3`, results in `UV` radiation and infared radiation involves smaller amounts of energy `UV`. So we require a transition involving initial value of `n` greater than `4` e.g. `5 rarr 4`. |
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| 363. |
A hydrogen atom and a `Li^(2+)` ion are both in the second excited state. If `l_H` and `l_(Li)` are their respective electronic angular momenta, and `E_H and E_(Li)` their respective energies, then(a) `l_H gt l_(Li) and |E_H| gt |E_(Li)|`(b) `l_H = l_(Li) and |E_H| lt |E_(Li)|` (C ) `l_H = l_(Li) and |E_H| gt |E_(Li)|`(d) `l_H lt l_(Li) and |E_H| lt|E_(Li)|`A. `l_(H) gt l_(Li)` and `|E_(H)|gt|E_(Li)|`B. `l_(H) = l_(Li)` and `|E_(H)|lt|E_(Li)|`C. `l_(H) = l_(Li)` and `|E_(H)|gt|E_(Li)|`D. `l_(H) lt l_(Li)` and `|E_(H)|lt|E_(Li)|` |
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Answer» Correct Answer - B In second excited state `n = 3` So `l_(H) = l_(Li) = 3((h)/(2pi))` While `E prop Z^(2)` and `Z_(H) = 1, Z_(Li) = 3` So `|E_(Li)| = 9|E_(H)|`or `|E_(H)|lt|E_(Li)|` |
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| 364. |
Consider a hydrogen-like atom whose energy in nth excited state is given by `E_(n) = (13.6 Z^(2))/(n^(2))` When this excited makes a transition from excited state to ground state , most energetic photons have energy `E_(max) = 52.224 eV`. and least energetic photons have energy `E_(max) = 1.224 eV` Find the atomic number of atom and the intial state or excitation.A. `Z=2,n=5`B. `Z=2,n=4`C. `Z=3,n=6`D. `Z=4,n=6` |
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Answer» Correct Answer - A maximum energy is librated for transaction `E_(n-1)to _(1)`, and minimum energy `E_(n)toE_(n-1)` hence` DeltaE=(Rhc)Z^(2)[1/n_(1)^(2)-1/n_(2)^(2)]` |
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| 365. |
Consider a hydrogen-like atom whose energy in nth excited state is given by `E_(n) = (13.6 Z^(2))/(n^(2))` When this excited makes a transition from excited state to ground state , most energetic photons have energy `E_(max) = 52.224 eV`. and least energetic photons have energy `E_(max) = 1.224 eV` Find the atomic number of atom and the intial state or excitation. |
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Answer» Maximum energy is liberated for transition `E_(n)` to `E_(n-1)` and Maximum energy for `E_(n) to E_(n-1)` Hence , `E^(1)/n^(2)-E=52.224eV`…(1) and `E_(1)/n^(2) - E_(1)/(n-1)^(1) = 1.224eV`.....(2) Solving above equation simultaneously, we `E_(1)=- 54.4eV` and `n=5` Now `E_(1)=-(13.6Z^(2))/1^(2)=-54.4eV n=5` energy state . |
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| 366. |
The time of revolution of an electron around a nucleus of charge `Ze` in `n`th Bohr orbit is directly proportional toA. `n`B. `(n^(3))/(Z^(2))`C. `(n^(2))/(Z)`D. `(Z)/(n)` |
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Answer» Correct Answer - B `T = (2pi r)/(v) , r =` radius of `nth` orbit `= (n^(2)h^(2))/(pi mZe^(2))` `v =` speed of `bar(e)` in nth orbit `= (ze^(2))/(2 epsilon_(0)nh)` `:. T = (4epsilon_(0)^(2)n^(3)h^(3))/(mZ^(2)e^(4)) rArr T prop (n^(2))/(Z^(2))` |
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| 367. |
The frequency of revolution of an electron in nth orbit is `f_(n)`. If the electron makes a transition from nth orbit to `(n = 1)` th orbit , then the relation between the frequency `(v)` of emitted photon and `f_(n)` will beA. `v = f_(n)^(2)`B. `v = sqrt f_(n)`C. `v = (1)/(f_(n)`D. `v = f_(n)` |
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Answer» Correct Answer - D `mr omega^(2) = (ke^(2))/(r ^(2)) or omega^(2) = (ke^(2))/(mr ^(3))` or `4 pi ^(2) f_(n)^(2) = (ke^(2))/(mr ^(3)) or f_(n)^(2) = (ke^(2))/(4 pi^(2) mr ^(3))` But, `r = (1)/(k) xx (n^(2) h^(2))/(4 pi^(2) me ^(2))` `:. f_(n)^(2) = (ke^(2)(k xx 4 pi^(2) me ^(2))^(2))/(4 pi^(2) m (n ^(2) h^(2))^(3))` or `f_(n)^(2) = (k^(4) e^(8) (4 pi^(2))^(2) m^(2))/(n^(2) h^(2))^(3)` or `f_(n) = ( 4 pi ^(2)k^(2) me^(4))/(n^(3) h ^(3))` Again , `hv = h^(2) = k^(2) ( 2 pi^(2) me^(4))/(h^(2)) [(1)/((n - 1)^(2)) - (1)/(n^(2))]` or `v = k^(2) ( 2 pi^(2) me^(4))/(h^(2)) [(n^(2) - (n - 1)^(2))/( n^(2) (n - 1)^(2))]` `= k^(2) ( 2 pi^(2) me^(4))/(h^(3)) [((2 n - 1))/( n^(2) (n - 1)^(2))]` If `n` "is very large , then" `v = k^(2) (2 pi^(2) ke^(4))/(h^(3)) xx (2 n)/(n^(4))` `= (4 pi ^(2) k^(2) me^(4))/(n^(3) h^(3)) = f_(n)` |
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| 368. |
In hydrogen atom, the radius of `n^(th)` Bohr orbit is `V_(n)`. The graph Beetween `log .(r_(n)/r_(1))` will beA. B. C. D. |
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Answer» Correct Answer - A `r_(n)propn^(2)` |
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| 369. |
If, in a hydrogen atom, radius of nth Bohr orbit is `r_(n)` frequency of revolution of electron in nth orbit is `f_(n)` and area enclosed by the nth orbit is `A_(n)` , then which of the pollowing graphs are correct?A. |
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Answer» Correct Answer - A::B::C `r_(n)propn^(2) "therefore"` graph between `r_(n)`,n will be parabola `I_(n)propn^(2) Rightarrow r_(n)/r_(1)=2`log n hence the graph is a straigth line passing through origin `r_(n)propn^(2)RightarrowA_(n)propn^(4)RightarrowA_(n)/A_(1)=n^(n)` ` log(A_(n)/A_(1))=4log` n: i.e the graph is a straight line passing through origin. `fprop1/n^(3) Rightarrow f_(n)/f_(1)Rightarrow1^(3)/n^(3)Rightarrow(f_(n)/f_(1))=-3logn`brgt i.e the graph is a straight line passing origin with negative slope. |
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| 370. |
If `E_n` and `L_n` denote the total energy and the angular momentum of an electron in the nth orbit of Bohr atom, thenA. `E_(n)propj_(n)^(2)`B. `E_(n)prop1/j_(n)^(2)`C. `E_(n)propj_(n)`D. `E_(n)prop1/J_(n)` |
| Answer» Correct Answer - B | |
| 371. |
An electron in a Bohr orbit of hydrogen atom with the quantum number `N_(2)` has an angular momentum `4.2176 xx 10^(-34) kg m^(2) s^(-1)`. If the electron drops from this level to the next lower level , the wavelength of this lines isA. `18 nm`B. `187.6 pm`C. `1876 Å`D. `1.876 xx 10^(4) Å` |
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Answer» Correct Answer - D Angular momentum, `L = 4.2176 xx 10^(-34) = (n_(2) h)/(2 pi)` `implies n_(2) = 4` For the transtion from `n_(2) = 4 to n_(1) = 3`, the wavelength spectral line`= lambda` `(1)/(lambda) = (13.6)/(hc) ((1)/(3^(2)) - (1)/(4^(2)))` `= (13.6 eV)/(1240 eV nm) ((7)/(9 xx 16))` `lambda = (1240 xx 144)/(13.6 xx7) = 1876 nm = 18760 Å` `= 1.876 xx 10^(4) Å` |
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| 372. |
The force felt by an electron on entering into a magnetic field is indepent of itsA. chargeB. strength of the fieldC. masD. direction ot its velocity |
| Answer» Correct Answer - C | |
| 373. |
Assertion: In the duration electron jumps from fist excited state to ground state in a stationary isolated hydrogen atom, angular momentum of the electron about the nucleus is conserved. Reason: As the electron jumps from first excited state to ground state, in a hydrogen atom, the electrostatic force on electron is always directed twoards the nucleus. |
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Answer» Correct Answer - D As electron jumps from `n-2` to`n=1`, angular momentum `((nh)/(2pi)) ` does not remain conserved. |
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| 374. |
A nagatively charge electroscope with zinc disc discharge when irradiated by an ultraviolet lamp. What caused this ?A. `a` - particle from the source combine with electron of the discB. electron escape from the disc when ultraviolet radiation fall on it,C. ultraviolet rays ionize the air surrounding the electroscopD. the disc becomes hot and thermionic emission take palce |
| Answer» Correct Answer - B | |
| 375. |
Electron in hydrogen atom first jumps from third excited state to second excited state and then form second excited state to first excited state. The ratio of wavelength `lambda_(1): lambda_(2)` emitted in two cases isA. `7//5`B. `27//20`C. `27//5`D. `20//7` |
| Answer» Correct Answer - D | |
| 376. |
Suppose frequency of emitted photon is `f_(0)` when the electron of a stationary hydrogen atom jumps from a higher state `m` to a lower state `n`. If the atom is moving with a velocity `v (lt lt c)` and emits a photon of frequency `f` during the same transition, then which of the following statement are possible?A. f may be equal to `f_(0)`B. f may be greater than `f_(0)`C. f may be less than `f_(0)`D. f cannot be equal to `f_(0)` |
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Answer» Correct Answer - A::B::C When a stationary hydrogen atom emits a photon, then energy of the emitted photon will be equal to the difference of the energy of the two levels involved in the transition. Hence, energy of emitted photon will be equal to `(E_(m) - E_(n))` If a hydrogen atom is moving and a photon is emitted by it along the same direction in which it is moving , due to momentum of the emitted photon , the momentum of hydrogen atom will get decreased. Therefore energy of the emitted photon will be equal to `(E_(m) - E_(n)` + "loss of KE of the hydrogen atom"). But if the photon is emitted in a direction normal to the motion, of the hydrogen atom, then the frequency of the emitted photon will be equal to `f_(0)`. Hence, option (a) is correct, Obviously, option (d) is wrong. If the photon is emitted by the hydrogen atom in the direction opposite to its motion, then frequency of the emitted photon will be less than `f_(0)`. Hence, option (c) is correct. |
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| 377. |
If `a_(0)` is the Bohr radius, the radius of then `n=2` electronic orbit in triply ionized beryllium isA. `4a_(0)`B. `a_(0)`C. `a_(0)//4`D. `a_(0)//16` |
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Answer» Correct Answer - B `r=a_(0)=(n^(2))/(Z)=a_(0).(2^(2))/(4)=a_(0)` |
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| 378. |
Ground state energy of H-atom is -13.6 eV. The energy needed to ionise H-atom from its second excited state isA. `1.51` eVB. `3.4` eVC. `13.6` eVD. `12.1` eV |
| Answer» Correct Answer - A | |
| 379. |
The spectrum obtained from a sodium vapour lamp is an example ofA. band spectrumB. continuous spectrumC. emission spectrumD. absorption spectrum |
| Answer» Correct Answer - C | |
| 380. |
In Rutherford scattering experiment, what will b ethe correct angle for `alpha` scattering for an impact parameter `b = 0` ?A. `90^(@)`B. `270^(@)`C. `0^(@)`D. `180^(@)` |
| Answer» Correct Answer - D | |
| 381. |
The electron in a hydrogen atom jumps from ground state to the higher energy state where its velcoity is reduced to one-third its initial value. If the radius of the orbit in the ground state is `r` the radius of new orbit will beA. `3 r`B. `9r`C. `(r )/(3)`D. `(r )/(9)` |
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Answer» Correct Answer - B `v = (1)/(137)(c )/(n)` or `prop(1)/(n)` Since `v` is reduced to one-third therefore `n = 3` Now, `r prop n^(2)`. |
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| 382. |
In terms of Bohr radius `a_(0)`, the radius of the second Bohr orbit of a hydrogen atom is given byA. `4a_(0)`B. `8a_(0)`C. `sqrt2a_(0)`D. 2 `a_(0)` |
| Answer» Correct Answer - A | |
| 383. |
when a hydrogen atom is raised from the ground state to an excited stateA. PE decreases and KE increasesB. PE increases and KE decreasesC. both KE and PE decreaseD. absorption spectrum |
| Answer» Correct Answer - C | |
| 384. |
The radius of hydrogen atom in its ground state is `5.3 xx 10^(-11)m`. After collision with an electron it is found to have a radius of `21.2 xx 10^(-11)m`. What is the principle quantum number of `n` of the final state of the atom ?A. n = 4B. n = 2C. n = 16D. n = 3 |
| Answer» Correct Answer - B | |
| 385. |
When hydrogen atom is in first excited level, its radius is….its ground state radiusA. four times, its gound state radiusB. twice, its gound state radiusC. same as its ground sate radiusD. half of its ground state radius |
| Answer» Correct Answer - A | |
| 386. |
Which of the following statements is true of hydrogen atom ?A. Angular momentum `prop (1)/(n)`B. Linear momentum `prop (1)/(n)`C. Radius `prop (1)/(n)`D. Energy `prop (1)/(n)` |
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Answer» Correct Answer - B We know that angular momentum `L = mvr = (nh)/(2pir)r = mv = (nh)/(2pir)` Now, `r prop n^(2)` `mv = p prop (nh)/(2pi xx n^(2))` `p prop (h)/(2pin)` or `p prop (1)/(n)` |
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| 387. |
In the Bohr model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electrons. If `a_(0)` is the radius of the ground state orbit, `m` is the mass and `e` is the charge on the electron and `e_(0)` is the vacuum permittivity, the speed of the electron isA. `0`B. `(e)/(sqrt(epsilon_(0)a_(0)m))`C. `(e)/(sqrt(4pi epsilon_(0)a_(0) m))`D. `sqrt(4pi epsilon_(0)a_(0)m)/(e)` |
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Answer» Correct Answer - C `(mv^(2))/(a_(0)) = (1)/(4pi epsilon_(0)) (e^(2))/(a_(0)^(2)) rArr v = (e)/(sqrt(4pi epsilon_(0)a_(0)m))` |
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| 388. |
In the Bohr model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electrons. If `a_(0)` is the radius of the ground state orbit, `m` is the mass and `e` I sthe charge on the electron and is the permittivity of vacuum, the speed of the elctron is :A. `0`B. `(e)/(sqrt(epsilon_(0)a_(0)m)`C. `(e)/(sqrt(4pi epsilon_(0)a_(0) m)`D. `sqrt((4pi epsilon_(0)a_(0)m)/(e))` |
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Answer» Correct Answer - C `(mv^(2))/(a_(0)) = (1)/(4pi epsilon_(0))(e^(2))/(a_(0)^(2))` `:. v = (e)/(sqrt(4pi epsilon_(0)a_(0)m)` |
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| 389. |
In the Bohr model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electrons. If `a_(0)` is the radius of the ground state orbit, `m` is the mass and `e` is the charge on the electron and `e_(0)` is the vacuum permittivity, the speed of the electron isA. zeroB. `(e)/(sqrt(epsilon_(0)a_(0)m))`C. `(e)/(sqrt(4pi epsilon_(0)a_(0)m))`D. `(sqrt(4piepsilon_(0)a_(0)m))/(e)` |
| Answer» Correct Answer - C | |
| 390. |
Imagine an atom made up of a proton and a hypotnerical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypotnetical photon that will be emitted has wavelength `lambda` (given in terms of the Rydberg constant `R` for the hydrogen atom) equal toA. `9//(5R)`B. `36//(5R)`C. `18//(5R)`D. `4//R` |
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Answer» Correct Answer - C In hydrogen atom `E_(n) =- (Rhc)/(n^(2))` Also `E_(n) prop m`, where `m` is the mass of th electron. Here the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in `nth` orbit will be given by `E_(n) =- (2Rhc)/(n^(2))` This given `lambda_(max) = (18)/(5R)`. |
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| 391. |
In the Bohr model of a `pi-mesic` atom , a `pi-mesic` of mass `m_(pi)` and of the same charge as the electron is in a circular orbit of ratio of radius `r` about the nucleus with an orbital angular momentum `h//2 pi`. If the radius of a nucleus of atomic number `Z` is given by `R = 1.6 xx 10^(-15) Z^((1)/(3)) m`, then the limit on `Z` for which `(epsilon_(0) h^(2)//pi me^(2) = 0.53 Å and m_(pi)//m_(e) = 264) pi-mesic` atoms might exist isA. `lt 105`B. `gt 105`C. `lt 37`D. `gt 37` |
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Answer» Correct Answer - C The angular momentum is `m v r = (n h)/(2 pi) implies n = 1` centripetal force, `(m v^(2))/( r) = (Ze^(2))/(4 pi epsilon_(0) r^(2))` `r = (epsilon_(0) n^(2) h^(2))/(pi m_(pi) e^(2) Z) = ((psilon_(0) h^(2))/(pi m_(e) e^(2))) ((m_(e))/(m_(pi))) (1)/(Z)` `= (0.53 xx 10^(-10))/(264 Z) = (200 xx 10^(-15))/(Z)` `[:. m_(pi)/(m_(e)) = 264]` Since `r` cannot be less than nuclear radius, `r gt 1.6 z^((1)/(3)) xx 10^(-15) m` or `(200 xx 10^(-15))/(Z) gt 1.6 xx 10^(-15) Z^((1)/(3))` `implies Z lt ((200)/(1.6))^((3)/(4)) lt 37` |
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| 392. |
As the electron in the Bohr orbit is hydrogen atom passes from state `n = 2` to `n = 1` , the `KE (K) and PE (U)` charge asA. K two-fold. U also two-foldB. K four-fold. U also four-foldC. K four-fold. U also two-foldD. K two-fold. U also four-fold |
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Answer» Correct Answer - B `KE prop (1)/(n^(2)) and PE prop (1)/(n^(2))` |
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| 393. |
If the atom`(_100)Fm^(257)` follows the Bohr model the radius of `_(100)Fm^(257)` is `n` time the Bohr radius , then find `n` .A. `100`B. `200`C. `4`D. `1/4` |
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Answer» Correct Answer - D `100=2+8+18+32+40 (n=1)(n=2) (n=3)(n=4)(n=5)` `r_(n) =0.53xxn=0.53xxn^(2)/zÅ=0.53xx(25)/(100)` `Rightarrow 0.53xxn=0.53xx1/4Rightarrown=1/4(therefore radius =0.53Å)` |
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| 394. |
Hydrogen aton in a sample are excited to `n = 5` state and it is found that photons of all possible wavelength are present in the emission spectra. The min imum number of hydrogen atom in the sample would be |
| Answer» `Iomega=(nh)/(2pi)Rightarrowmr^(2)2pif=(nh)/(2pi)Rightarrowf=(nh)/(4pi^(2)mr^(2))r=r_(0)n^(2)and N=ft=8xx10^(6)` | |
| 395. |
A beam of `13.0 eV` electrons is used to bombard gaseous hydrogen. The series obtained in emission spectra is // areA. Lyman seriesB. Balmer seriesC. Brackett seriesD. All of these |
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Answer» Correct Answer - D As the electron beam is having energy of `13 eV` , it can excite the atom to the states whose energy is less than or equal to `0.6 eV (13.6 - 13, E_(5) = 0.544 eV and E_(4) = 0.85 eV` . So , the electron beam can excite the hydrogen gas maximum to `4^(th)` energy state , hence the transit electron can come back to ground state from either of three excited states, thus emitting Lyman, Balmer and Paschen series. |
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| 396. |
A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)A. 6 and 6B. 3 and 3C. 6 and 3D. 3 and 6 |
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Answer» Correct Answer - C In first case , transition is from `n^(th)` state to `2^(nd) (1 ^(st excite)` state. `:. (10.2 + 17.0) eV = 13.6 xx Z^(2) [(1)/(2^(2))/(1)/n^(2)]` In `2^(nd)` case , transition is from `n^(th)` state to `3^(rd)`state. `:. (4.25 + 5.95) eV = 13.6 xx Z^(2) [(1)/(3^(2))/(1)/(n^(2))]` Solving above equation , we get `n = 6 and Z = 3`. |
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| 397. |
The third line of the Balmer series spectrum of a hydrogen-like ion of atomic number `Z` equals to `108.5 nm`. The binding energy of the electron in the ground state of these ions is `E_(n)`. ThenA. `Z = 2`B. `E_(B) = 54.4 eV`C. `Z = 3`D. `E_(B) = 122.4 eV` |
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Answer» Correct Answer - A::B For the third line of Balmer series, `n_(1) = 2, n_(2) = 5` `:. (1)/(lambda) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = RZ^(2) ((1)/(2^(2)) - (1)/(5^(2))) = (21 RZ^(2))/(100)` `E = - 13.6 eV` `Z^(2) xx (21)/(100) = (hc)/(lambda) = (1242 eV nm)/(108.5 nm)` `Z^(2) = (1242 xx 100)/(108.5 xx 21 xx 13.6) = 4 implies Z = 2` Binding energy of an electron in the ground state of hydrogen - like ion `= 13.6 Z^(2)//n^(2) = 54.4 eV (n = 1)`. |
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| 398. |
Figure shown the electron energy level , referred to the ground state (the lowest possible energy) as zero , for five different isoleted atoms. Which atom can produce radiation of the shortest wavelength when atoms in the ground state are bombarded with electrons energy `W`? A. AB. BC. CD. D |
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Answer» Correct Answer - B From the graphs given , atom in graph `B` will absorb must of the energy `W` from the electron and re-radiate, in all directions, rediation of shortest wavelength when the atom returns to its ground state. |
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| 399. |
Which of the following atoms has the lowest ionization potential ?A. `._(8)^(16)O`B. `._(7)^(14)N`C. `._(55)^(133)Cs`D. `._(18)^(40)Ar` |
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Answer» Correct Answer - C Since the `._(55)^(133) Cs` has larger size among the four atoms gives, thus the electrons present in the outermost orbit will be away from the nucleus and the electronstaitc force experienced by electrons due to nucleus will be minimum. Therefore the enegry required to librate electron from outher will be minimum in the case of `._(55)^(133) Cs`. |
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| 400. |
The ground state energy of hydrogen atom is `-13.6 e V`. a. What is the kinetic energy of electron in the second excited state? b. If the electron jumps to the ground state from the second excited state , calculate the wavelength of the spectral line emittted. |
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Answer» a. The second excited state corresponds to `n = 3`. Now, `E_(3) = - (13.6)/(n^(2)) = (13.6)/(3^(2)) = - (13.6)/(9) = 1.51 e V` The kinetic energy of an electron in the second excited state`= - (-1.51 - 1.51 e V` b. Energy emitted , when the electron jumps from the second excited state to the ground state. `E = - 1.51 - (-13.6) = 12.09 e V = 12.09 xx 1.6 xx 10^(-19)J` The wavelength of the spectral line emitted. lambda `= (hc)/(E) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(12.09 xx 1.6 xx 10^(-19))` `= 1.027 xx 10^(-7) m = 1.027 Å`. |
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