InterviewSolution
Saved Bookmarks
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.
| 451. |
As the electron in the Bohr orbit is hydrogen atom passes from state `n = 2` to `n = 1` , the `KE (K) and PE (U)` change asA. `K` two-fold, `U`four-foldB. `K` four-fold, `U`two-foldC. `K` four-fold, `U` also four-foldD. `K` two-fold, `U`also four-fold |
|
Answer» Correct Answer - C `U = 2E, K=- E` and `E=- (13.6)/(n^(2)) = eV` |
|
| 452. |
The wavelength of `k_(alpha) ` X- rays produced by an X - rays tube is `0.76 Å` . The atomic number of the anode material of the tube is …….A. 30B. 40C. 50D. 60 |
|
Answer» Correct Answer - B `(1)/(lambda) = Z^(2) R_(oo) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` for `K_(alpha) line , n_(1) = 1 and n_(2) = 2` `(1)/(lambda) = Z^(2) R_(oo) (3)/(4)` `Z = sqrt((4)/(3lambda R_(oo))) = 39.9 = 40` |
|
| 453. |
The wavelength of `k_(alpha) ` X- rays produced by an X - rays tube is `0.76 Å` . The atomic number of the anode material of the tube is ……. |
|
Answer» `K_(alpha)` X-ray are produced when an electron makes a transition from `n = 2 to n = 1` to fill a vacancy in `K` shell . The wavelength of X-ray lines is given by `(1)/(lambda_(K_alpha)) = R(Z - 1)^(2) ((1)/(1^(2)) - (1)/(2^(2))) = (3)/(4) R(Z - 1)^(2)` or `(Z - 1)^(2) = (4)/(3 R lambda_(K_alpha))` `= (4)/(3 xx (1.097 xx 10^(7)) xx (0.76 xx 10^(-10)))` or `(Z - 1)^(2) = 1600` or `Z - 1 = 40` or `Z = 41` |
|
| 454. |
The element which has a `K_(alpha)` X-rays line of wavelength `1.8 Å` is `(R = 1.1 xx 10^(7) m^(-1), b = 1 and sqrt(5//33) = 0.39)`A. `C o, Z = 27`B. `Iron, Z = 26`C. `M n, Z = 25`D. `N i, Z = 28` |
|
Answer» Correct Answer - A `(1)/(lambda_(alpha)) = (3R)/(4) (Z - 1)^(2)` `(Z - 1) = sqrt((4)/(3 R lambda_(alpha))) = sqrt((4)/(3 xx 1.1 xx 10^(7) xx 1.8 xx 10^(-10)))` `= (200)/(3) sqrt((5)/(33)) = (78)/(3) = 26 implies Z = 27` |
|
| 455. |
the photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen like atom X in the second excited state. Then, the hydrogen-like atom X makes a transition of nth orbit.A. `X=He^(+),n=4`B. `X=Li^(++),n=6`C. `X=He^(+),n=6`D. `X=Li^(++),n=9` |
|
Answer» Correct Answer - D Energy of` n_(th)` state in hydrogen is same as energy of `3n^(th)` state in `Li^(++)`.therefore3to1 transaction I H would give same energy as the `9to3` transaction in `li^(++)`. |
|
| 456. |
the photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen like atom X in the second excited state. Then, the hydrogen-like atom X makes a transition of nth orbit.A. `X=He(+),n=4`B. `X=Li^(++),n=6`C. `X=He^(+),n=6`D. `X=Li^(++),n=9` |
|
Answer» Correct Answer - D `13.6xx(1/1-1/9)=13.6xxZ^(2)(1/3^(2)-1/n^(2))` Rightarrow is satisfied for `Z=3` and `n=9`n of the given option |
|
| 457. |
The photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen like atom X in the second excited state. Then, the hydrogen-like atom X makes a transition of nth orbit.A. `X = He^(+) , n = 4`B. `X = Li^(+ +) , n = 6`C. `X = He^(+) , n = 6`D. `X = Li^(+ +) , n = 9` |
|
Answer» Correct Answer - D Energy of `n ^(th)` state in hydrogen is same as to `3 n ^(th)` state in `L i^(+ +)`. 3 to 1 transition in H would give same energy as the ` 3 xx 3` to `1 xx 3` i.e. , 9 to 3 transition in `L i^(+ +)`. |
|
| 458. |
If element with particle quantum number `ngt4` were not allowed in nature , the number of possible element would beA. 32B. 60C. 64D. 4 |
|
Answer» Correct Answer - B `N = Sigma 2 n^(2)` `N = 2 (1^(2) + 2^(2) +3^(2) +4^(2)) = 60` |
|
| 459. |
The figure represents the observed intensity of `X` - rays emitted by an `X` - ray tube as a function of wavelength . The sharp peaks `A` and `B` denote A. band spectrumB. continuous spectrumC. characteristic radiationsD. white radiations |
|
Answer» Correct Answer - C |
|
| 460. |
Let `E=(-1me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))` be the energy of the `n^(th)` level of H-atom state and radiation of frequency `(E_(2)-E_(1))//h` falls on it ,A. it will not be absorbed at allB. some of atoms will moves to the first excited state .C. all atom will be excited to the `n=2` stateD. no atom will make a transition to the `n=3` state. |
|
Answer» Correct Answer - B::D Given `E_(n)=(-1me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))` Energy of incident radiation `E_(2)-E_(1)=hv=-3.40-(-13.6)=10.2eV` the energy is sufficient to move some of atom in ground state `(E_(1)` to the first excited state `(E_(2)`, not atom shall make transition to `n=3` state be -cause energy required for transition to `n=3` is `12.09eV` |
|
| 461. |
The electron in a hydrogen atom make a transtion `n_(1) rarr n_(2)` where `n_(1) and n_(2)` are the priocipal quantum number of the two states . Assume the Bohr model to be valid . The time period of the electron in the initial state is eight time that in the final state . The possible values of `n_(1) and n_(2)` areA. `n_(1) = 4, n_(2) = 2`B. `n_(1) = 8, n_(2) = 2`C. `n_(1) = 8, n_(2) = 1`D. `n_(1) = 6, n_(2) = 2` |
|
Answer» Correct Answer - A `T prop n^(3)`. Given `T_(n1) = 8T_(n2)`, hence `n_(1) = 2n_(2)` Therefore, option (a) and (d) both are correct. |
|
| 462. |
The electron in a hydrogen atom make a transtion `n_(1) rarr n_(2)` where `n_(1) and n_(2)` are the priocipal quantum number of the two states . Assume the Bohr model to be valid . The time period of the electron in the initial state is eight time that in the final state . The possible values of `n_(1) and n_(2)` areA. `n _(1) = 4, n _(2) = 2`B. `n _(1) = 8, n _(2) = 2`C. `n _(1) = 8, n _(2) = 3`D. `n _(1) = 6, n _(2) = 2` |
|
Answer» Correct Answer - A `T^(2) prop R^(3) and r prop n^(2) implies T^(2) prop n^(6) implies T prop n^(3)` `(T_(1))/(T_(2)) = ((n_(1))/(n_(2)))^(3) implies 8 = ((n_(1))/(n_(2)))^(3) or (n_(1))/(n_(2)) = 2` Only a satisfies the above, hence this is right choice. |
|
| 463. |
An electron revolving in an orbit of radius `0.5 Å` in a hydrogen atom executes per secon. The magnetic momentum of electron due to its orbital motion will beA. `1.256 xx 10^(-23) A m^(2)`B. `653 xx 10^(-26) A m^(2)`C. `10^(-3) A m^(2)`D. `256 xx 10^(-26) A m^(2)` |
|
Answer» Correct Answer - A `M = 1 A = e f pi r^(2)` `= 1.6 xx 10^(-19) xx 10^(16) xx 3.14 xx (0.5 xx 10^(-10)) ^(2) A m^(2)` `= 1.256 xx 10^(-23) A m^(2)` |
|
| 464. |
The total energy of an electron in the ground state of hydrogen atom is `- 13.6 eV`. The potiential energy of an electron in the ground state of `Li ^(2+)` ion will beA. `122.4 eV`B. `- 122.4 eV`C. `244.8 eV`D. `-244.8 eV` |
|
Answer» Correct Answer - D `E_(P) = - k e^(2)/( r) , E = - k e^(2)/(2 r)` So, `E_(P) = 2 E = 2 (- 13.6) eV = - 27.2 eV` Potential energy of electron in the ground state of `L i^(2 +)` ion is `= - 3 xx 27.2 eV or - 244.8 eV` |
|
| 465. |
A physicist wishes to eject electrons by shining light on a metal surface. The light source emits light of wavelength of `450 nm`. The table litsts the only available and their work fucntions. `{:("Metal",,,,W_(0)(eV)),("Barium",-,,,2.5),("Lithium",-,,,2.3),("Tantalum",-,,,4.2),("Tungsten",-,,,4.5):}` Which option correctly identifies the metal that will produce the most energetic electrons and their energies?A. Lithium, `0.45 eV`B. Tungston, `1.75 eV`C. Lithium, `2.30 eV`D. Tungston,`2.75 eV` |
| Answer» Correct Answer - A | |
| 466. |
A physicist wishes to eject electrons by shining light on a metal surfac. The light source emits light of wavelenght of 450 nm. The table lists the only available metals and their work functions. `{:("Metal",W_(0)(eV)),("Barium",2.5),("Lithium",2.3),("Tantalum",4.2),("Tungsten",4.5):}` Suppose photoelectric experiment is done separately with these metals with light of wavelenght 450 nm. The maximum magnitude of stopping potential amongst all the metals. is-A. `2.75 "volt"`B. `4.5 "volt"`C. `0.45 "volt"`D. `0.25 "volt"` |
|
Answer» Correct Answer - C The maximum magnitude of stopping potential will be for metal of least work function. `:.` required stopping potential is `V_(s)=(hv-phi_(0))/(e )=0.45 "volt"` |
|
| 467. |
The figure shows an energy level diagram for the hydrogen atom. Several transitions are marked as I, II, III,_____. The diagram is only indicative and not be scale. In which transitions is a Balmer photn absorbed?A. IIB. IIIC. IVD. VI |
| Answer» Correct Answer - D | |
| 468. |
A protons , a deuteron and an alpha-particle are accelerated through the same p.d of `V` volt. The velocities acquired by them are in the ratioA. `1:1:sqrt2`B. `1:sqrt2:1`C. `1:1:1`D. `sqrt2:1:1` |
|
Answer» Correct Answer - D `v=sqrt((2eV)/m) Rightarrow vpropsqrt(e/m)` |
|
| 469. |
A proton of mass `m` moving with a speed `v_(0)` apporoches a stationary proton that is free to move. Assuming impact parameter to be zero., i.e., head-on collision. How close will be incident proton go to other proton ?A. `e^(2)/(4piepsilonmv_(0)^(2)`B. `e^(2)/(piepsilon_(0)mv_(0)^(2)`C. `e^(2)/(mv_(0)^(2)`D. zero |
|
Answer» Correct Answer - B use : LCLM and conservation of energy . `Ie:mv_(0)=2mu-(1)` ` (mv_(0)^(2))/2=2xx1/2m(u_(0)/2)=e^(2)/(4piepsilonr)` |
|
| 470. |
When two electrons enter into a magnetic field with different velocities , they defect in different circular parts , in such a way that radius of one path is double that of the other. `1xx10^(7)ms^(-1)` is the velocity of electron in smaller circle of radius `2 xx 10^(-3)m`. The velcoity of electron in the other circular path is:A. `4xx10^(7)ms^(-1)`B. `2xx10^(6)ms^(-1)`C. `2xx10^(7)ms^(-1)`D. `2 xx 10^(6) ms^(-1)` |
|
Answer» Correct Answer - C `v=(mv)/(Bq)` `6pinrV_(0)=mg:Eq=mg` |
|
| 471. |
An electron is accelerated in an electric field of `40V cm^(-1)`. If `e//m` of electron is `1.76 xx10^(11) Ckg^(-1)` , then its acceleration isA. `14.0xx10^(14)ms^(2)`B. `14.0xx10^(10)ms^(-2)`C. `7.0xx10^(10)ms^(-2)`D. `7.04xx10^(14)ms^(-2)` |
|
Answer» Correct Answer - D `a=(Ee)/m` |
|
| 472. |
The wave length of `Kbeta` X-ray of certain metal is `12.42`pm . It takes `10Kev` to remove the electron from M shell of an atom of that metal the minimum accelerating voltage that should be applied across the X-ray tube , so that a `K_(a)` X-ray would be produced is (`hc=1242eVnm`)A. `10KV`B. `100KV`C. `110KV`D. `90KV` |
|
Answer» Correct Answer - C `E_(Kbeta)=(1242eV-nm)/(12.42xx10^(-3nm))=100KeV` and energy of `M` shell is `E_(M) =-10KeV` for K_(alpha) to be produced , electron from k shell is to be knocked out therefore kinetic energy of the striking electron should be `ge` energy of K shell `i.e K_(Kbeta) = E_(M)-E_(K) Rightarrow E_(K)=E_(M)-E(Kbeta)` `i.e e_(K) =-10-100=-110eV` therefore kinetic energy of striking electron =110eV `eV=110eV RightarrowV=110V` |
|
| 473. |
Any radiation in the ultra violet region of Hydrogen from a metal . Then the metal is, nearlyA. `3.3xx10^(15) Hz`B. `2.5xx10^(15)Hz`C. `4.6xx10^(14)Hz`D. `8.2xx10^(14)Hz` |
|
Answer» Correct Answer - A `thereforev=c/lambda` |
|
| 474. |
In a cathods ray tube ,a `p.d` of `3000V` is maintained betweens the deflector plates whose seperation is `2 cm`. A magnetic filed of `2.5x10^(3)` at right angle to the electric field gives no deflection of electern beam , which received an initial acceleration by a `p.d` of `10,000 V` , the `e//m` of electron is |
|
Answer» `V_(1)` is `p.d` between the deflector plates and `V_(2)` be the potential through which electrons are accelerated `1/2m upsilon^(2) = V_(2)q 1/2m. V_(1)^(2)/(d^2 B^(2))=v^(2)q,q/m=V_(1)^(2)/(2d^2B^(2)V_(2))` `(q)/(m)= (9xx10^(6))/(2xx4xx10^(-4)xx6.25xx10^(-6)xx10^(4))=1.8xx10^(11)c//kg` |
|
| 475. |
In the photoelectric experiment, if we use a monochromatic light, the I-V curve is as shown. If work function of the metal is 2eV, estimate the power of light used. (Assume efficiency of photo emission `=10^(-3)%`, i.e. number of photoelectrones emitted are `10^(-3)%` of number of photons incident on metal.) A. 2WB. 5WC. 7WD. 10 W |
|
Answer» Correct Answer - C The energy of incident photons is given by `hv=eV_(s)+phi_(0)=2+5=7 eV` (`V_(s)` is stopping potential and `phi_(0)` is work function) Saturation current `=10^(-5) A =(etaP)/(hv)e=(10^(-5)P)/(7xxe) e` (`eta` is photo emission efficiency) `:.P=7 W` |
|
| 476. |
Yellow light of 557 nm wavelength is incident on a cesium surface. It is found that no photo electrons flow in the circuit when the cathode-anode voltage drops below 0.25V. Then the threshold wavelength for photo electric effect from cesium isA. 577 nmB. 653 nmC. 734 nmD. 191 nm |
|
Answer» Correct Answer - B `(hc)/(lambda) =phi+ev_(s) , phi=2.23 =1.98 eV rArr lambda_(th)=(hc)/(phi)=627 nm` |
|
| 477. |
A point source causes photoelectric effect from a small metal plate. Which of the curves in Fig. may represent the saturation photo-current as a function of the distance between the source and the metal?A. AB. BC. CD. D |
|
Answer» Correct Answer - D `Iprop I prop 1/("distance"^(2)` |
|
| 478. |
Photo electrons are liberated by ultraviolet light of wavelength `3000 Å` from a metalic surface for which the photoelectric threshold wavelength is `4000 Å`. Calculate the de Broglie wavelength of electrons emitted with maximum kinetic energy. |
| Answer» Correct Answer - `lambda_(d)=sqrt((hlambda.lambda_(th))/(2m.c(lambda_(th)-lambda)))=sqrt((6xx10^(-7)h)/(m_(e )c))m=12.08Å` | |
| 479. |
Suppose the wavelength of the incident light is increased from `3000 Å` to `3040Å`. Find the corresponding change in the stopping potential. [Take the product `hc=12.4xx10^(-7)eVm)`] |
|
Answer» Correct Answer - `dV_(s)=(hc)/(e ).(dlambda)/(lambda^(2))= -(hc)/(228e)xx10^(7)= -5.5xx10^(-2)"volt"` `V_(s)=(hc)/(elambda)-(W)/(e )` On differentating `dV_(s)= -(hc)/(elambda^(2))(dlambda)= -5.5xx10^(-2) "volt"` `dV_(s)=(hc)/(e).(Deltalambda)/(lambda^(2))=(hc)/(228e)xx10^(7)=5.5xx10^(-2)"Volt"` |
|
| 480. |
When photon of wavelength `lambda_(1)` are incident on an isolated shere supended by an insulated , the corresponding stopping potential is found to be `V`. When photon of wavelength `lambda_(2)` are used , the orresponding stopping potential was thrice the above value. If light of wavelength `lambda_(3)` is used , carculate the stopping potential for this case.A. `(hc)/(e) [(1)/(lambda_(3)) + (1)/(2 lambda_(2)) - (1)/(lambda_(1))]`B. `(hc)/(e) [(1)/(lambda_(3)) + (1)/(2 lambda_(2)) - (3)/(lambda_(1))]`C. `(hc)/(e) [(1)/(lambda_(3)) + (1)/( lambda_(2)) - (1)/(lambda_(1))]`D. `(hc)/(e) [(1)/(lambda_(3)) - (1)/( lambda_(2)) - (1)/(lambda_(1))]` |
|
Answer» Correct Answer - B `K E_(lambda_1) = (hc)/(lambda_(1)) - psi = e Delta V` `K E_(lambda_2) = (hc)/(lambda_(2)) - psi = 2 e Delta V` `implies 3 ((hc)/(lambda_(1)) - psi) = (hc)/(lambda_(2)) -psi` `psi = hc ((3)/(2 lambda_(1)) - (1)/(2 lambda_(2)))` `implies K E_(lambda_3) = (hc)/(lambda_(3)) - hc [(3)/(2 lambda_(1)) - (1)/(2 lambda_(2))]` ` = hc [(1)/(lambda_(3)) + (1)/(2 lambda_(2)) - (3)/(2 lambda_(1))]` `e Delta V = hc [(1)/(lambda_(3)) + (1)/(2 lambda_(2)) - (3)/(2 lambda_(1))]` ` Delta V = (hc)/(e) [(1)/(lambda_(3)) + (1)/(2 lambda_(2)) - (3)/(lambda_(1))]` |
|
| 481. |
a. X-ray are electromagnetic waves because these are producted by the deceleration of fast moving electron . b. X-ray are electromagnetic waves because these produce line spectrum. c. the cut-off wavelength depends on the target metarial . d. The intensity of `K_(alpha)` transition is more probable then of `K_(beta)` e The transition. e. Frequency of `K_(alpha)` radiation is more than that of `K_(beta)`. |
|
Answer» a. True, According to the theory of electromagnetion , electromagnetic waves are generated by the acceleration or relardation of charges. b. True, Line spectrum of electromagnetic waves are producedced due to the transition of electron from the higher level to the lower level. c. False, The cut-off wavelength depends only on the kinetic energy of the striking electrons. d. True, The `K_(alpha)` transition is more probable then the `K_(beta)` transition. e. False Frequency of `K_(beta)` is more than that of `K_(alpha)` becouse `lambda_(B) gt lambda_(A)` as shown: `(1)/(lambda_(beta)) = R_(oo) (Z-1)^(2) [(1)/(1^(2)) - (1)/(3^(2))] = (8)/(9) R_(oo) (Z-1)^(2)` and `(1)/(lambda_(alpha)) = R_(oo) (Z-1)^(2) [(1)/(1^(2)) - (1)/(2^(2))] = (3)/(4) R_(oo) (Z-1)^(2)` |
|
| 482. |
Find the cut off wavelength for the continuous X - rays coming from an X-ray tube operating at 40 kV. |
|
Answer» We know that `lambda_(min) - (hc)/(e V) = (12400)/(V) = (12400)/(40 xx 10^(3)) = 0.31 Å`. |
|
| 483. |
Determine the minimum wavelength that hydrogen in its ground state can obsorb. What would be the next smaller wavelength that would work?A. `133 nm`B. `13.3 nm`C. `10.3 nm`D. `103 nm` |
|
Answer» Correct Answer - D When wavelength is maximum, the energy is maximum. Hence, this is from the ground state to the first excited state, for which the energy is `13.6 eV - 3.4 eV = 10.2 eV` hence, the required wavelength is `122 nm` The next possibility is to jump from the ground state to the second excited state which requires `=m 13.6 xx 1.5 = 12.1 eV`. Hence, it correcponding to a wavelength `lambda = ( c)/(v) = (hc)/(E_(3) - E_(1)) = ((6.63 xx 10^(-34)) xx(3 xx 10^(8)))/((12.1) xx (1.6 xx 10^(-19))) = 103 nm` |
|
| 484. |
An X-ray tube is operating at 150 kV and 10 mA. If only 1% of the electric power supplied is converted into X-rays, the rate at which the target is heated in calories per second isA. 3.57B. 35.7C. 4.57D. 15 |
|
Answer» Correct Answer - A `P= VI = 150 xx 10^(3) xx 10 xx 10^(-3) = 1500 Js^(-1)` Heating rate of target `= (1500)/(4.2) xx (1)/(100) = 3.57 cal s^(-1)` |
|
| 485. |
The potential different across the Coolidge tube is `20 kV and 10 m A` current flows through the voltage supply. Only `0.5%` of the energy carried by the electrons striking the largest is converted into X-ray. The power carried by the X-ray beam is `p`. ThenA. `P = 0.1 W`B. `P = 1 W`C. `P = 2 W`D. `P = 10 W` |
|
Answer» Correct Answer - B `P = VI` Therefore, total power down by Coolidge tube `P_(T) = VI = 200 W`. As `0.5 %` of the energy is corried by X-ray is `0.5 % of P_(T) = (0.5)/(100) xx 200 = 1 W` |
|
| 486. |
If the potential difference applied across a Coolidge tube is increased , thenA. wavelength of `K_(alpha)` will increaseB. `lambda_(min)` will increaseC. different between wavelength of `K_(alpha)` and `lambda_(min)` increaseD. none o these |
|
Answer» Correct Answer - C The characteristic X-ray depends on the material used. |
|
| 487. |
The potential different across the Coolidge tube is `20 kV and 10 m A` current flows through the voltage supply. Only `0.5%` of the energy carried by the electrons striking the largest is converted into X-ray. The power carried by the X-ray beam is `p`. ThenA. `0.1`B. `1W`C. `2W`D. `10W` |
|
Answer» Correct Answer - B power drawn by coolidge tube =`20xx10^(3)xx10xx10^(-3)=200W` Power of x-rays =`(0.5)/(100)xx200=1W` |
|
| 488. |
The energy levels of a hypotherical one electron atom are shown in figure If an electron with initial kinetic energy `6 eV` is to interact with this hypothetical atom , what minimum energy will this electron carry after interaction?A. `2 eV`B. `3 eV`C. `6 eV`D. `0 eV` |
|
Answer» Correct Answer - C Because excitation energy for the second shell is more than `6 eV`, hence electron having initial kinetic energty of `6 eV`will not interact with the atom .Because it cannot transfer its energy to the electron in the atom. |
|
| 489. |
The energy levels of a hypotherical one electron atom are shown in figure The initial kinetic energy of an electron is `11 eV` and it interact with the above said hypothetical one electron atom , the minimum energy carried by the electron after interaction isA. `0.7 eV`B. `0.3 eV`C. `0.9 eV`D. `1 eV` |
|
Answer» Correct Answer - A As `E_(2) - E_(1) = 15.6 - 5.3 = 10.3 eV` hence energy of the electron after interaction is `11 - 10.3 eV = 0.7 eV`. |
|
| 490. |
The energy levels of a hypotherical one electron atom are shown in figure Find the ionization potential of the atom.A. `11.2 eV`B. `13.5 eV`C. `15.6 eV`D. `12.6 eV` |
|
Answer» Correct Answer - C Given that`E_(1) = - 15.6 eV, E_(oo) = 0 eV` Ionzation energy of the aton: `E_(oo) - E_(1) = 0 - (- 15.6 eV) = 15.6 eV` So, ionization potential `= 15.6 eV` |
|
| 491. |
The energy levels of a hypotherical one electron atom are shown in figure Find the excitation potential for the state `n = 3`.A. `14.64 eV`B. `9.93 eV`C. `12.52 eV`D. `10.04 eV` |
|
Answer» Correct Answer - C The excitation energy for the `n = 3` state is `Delta E = E_(3) - E_(1) = 15.6 - 3.08 = 12.52 eV` Excitation potential `= 12.52 V` |
|
| 492. |
Two hydrogen atom in ground state are moving on opposite direction with the same speed and collide head on . The minimum kinetic energy of each hydrogen atom for the collision to be inelastic so that both the atoms are eccited isA. `13.6eV`B. `10.2eV`C. `20.4eV`D. `27.2eV` |
|
Answer» Correct Answer - B Let K by the initial energy of each atoms . Maximum loss in Ke takes place in perfectly inelastic collision. From conservation of linear momentum `"mu"+(-"mu")=2mvRightarrowv=0` Where v=common velocity after collision. therefore maximum possible loss in KE is ` DeltaK=1/2"mu"^(2)xx2-1/22mxx0^(0)=21/2"mu"^(2)=2K` Minimum energy required to excite both the electrons is `DeltaE=10.2+10.2=20.4eV`. For inelastic collision with two electron excited `DeltageDeltaERightarrow2Kge20.4RightarrowK_(min)=10.2eV` |
|
| 493. |
Light form a dicharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find (a) the energy of the photons causing the photoelectrons emission. (b) the quantum numbers of the two levels involved in the emission of these photons. (c ) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and (d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (lonization potential of hydrogen is 13.6 eV.) |
|
Answer» `E_(K) = 0.73 e V, W = 1.82 e V` Lonization energy of `H` atom `= 13.6 e V` a. `hv = W + E_(K) = 1.82 e V = + 0.73 e V = 2.55 e V` b. The electronic energy levels of `H` -atom are given by `E_(n) = - (Rhc)/(n^(2)) = - (13.6)/(n^(2)) e V` `For n = 1, E_(1) = - 13.6 e V` `For n = 2, E_(2) = - 3.4 e V` `For n = 3, E_(3) = - 1.51 e V` `For n = 4, E_(4) = - 0.85 e V` Clearly `E_(4) - E_(2) = - 0.85 e V - (-3.4 e V) = 2.55 e V` i.e. Quantum of electron in `H` atom `J = n (h)/(2 pi)` For `n = 4 J_(1) = 4 (h)/(2 pi) = (2 h)/(pi)` For `n = 2 J_(2) = 2 (h)/(2 pi) = (2 h)/(pi)` Therefore change angular momentum, `Delta J = J_(1) - J_(2) = (2h)/( pi) - (h)/(pi) = (h)/(pi)` d. According to conservation of momentum, `(hc)/(c) + mv = 0` `:. v = - (hc)/(cm) = - (2.55)/((3 xx 10^(8)) xx (1.67 xx 10^(-27))` `= - (2.55 xx 1.6 xx 10^(-19))/(3 xx 10^(8) xx 1.67 xx 10^(-27))` `= - 0.814 m s^(-1)` Therefore , recoil speed of `H` atom `= 0.814 m s^(-1)` |
|
| 494. |
Light form a dicharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find (a) the energy of the photons causing the photoelectrons emission. (b) the quantum numbers of the two levels involved in the emission of these photons. (c ) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and (d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (lonization potential of hydrogen is 13.6 eV.)A. `4 to 2`B. `3 to 1`C. `3 to 2`D. `4 to 3` |
|
Answer» Correct Answer - A `E_(n) = (-13.6 z^(2))/(n^(2)) eV` `E_(1) = - 13.6 eV` `E_(2) = - 3.4 eV` `E_(3) = - 1.51 eV` `E_(4) = - 0.85 eV` `E_(4) - E_(2) = (0.85) - (-3.4) = 2.55 eV` |
|
| 495. |
Assertion: According to classical theory, the proposed path of an electron in Rutherford atom model will be parabolic. Reason: According to electromagnetic theory an accelerated particel continuosly emits radiation.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. |
|
Answer» Correct Answer - D According to classical electromagnetic theory, an accelerated change continuously emits radiation. As electron revolving in circular paths are constantly experiencing centripetal acceleration, hence they will be losing their energy continuously and the orbital radius will go on decreasing and from spiral and finally the electron will fall on the nucleus. |
|
| 496. |
Ionization energy of a hydrogen like A is greater than that of another hydrogen like ion Let r, u, E and L represent the radius of the orbit, speed of the electron energy of the atom and orbital angular momentum of the electron respectively, in ground stateA. `r_(A) gt r_(B)`B. `u_(a) gt u_(B)`C. `E_(A).E_(B)`D. `L_(A) gt L_(B)` |
| Answer» Correct Answer - B | |
| 497. |
Calculate (a) the wavelength and (b) the frequencey of the `Hbeta` line of the Balmer series for hydrogen. |
|
Answer» a. `H_(beta)` line of Balmer series corrresponds to the transition from `n = 4` to `n = 2`level . The corrresponds wavelength for `H_(beta)` line is ` (1)/(lambda) = (1.097 xx 10^(7)) [(1)/(2^(2)) - (1)/(4^(2))]` `= 0.2056 xx 10^(7) m` `lambda = 4.9 xx 10^(-7)) m` ltrbgt b. `f = (c )/(lambda) = (3.0 xx 10^(8))/(4.9 xx 10^(-7))` `= 6.12 xx 10^(14) Hz` |
|
| 498. |
The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true?A. Its Kinetic energy increase and its potential and total energies decreaseB. Its Kinetic energy decrease , potential energy increases and its total remian the sameC. Its Kinetic and total energies decrease and its potential energy increasesD. Its Kinetic , potential and total energies decrease |
|
Answer» Correct Answer - A Kitnetic energy `(K.E)=(13.6z^(2))/(n^(2)eV)` Ptential `(P.E)=(-2(13.6)z^(2))/n^(2)eV` Total energy `(T.E) =(-13.6z^(2))/n^(2)eV` When an electron in H-atom makes a transition from an excited state to the ground state, value of n decrease, hence `K.E` & `T.E`. decrease |
|
| 499. |
The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true?A. Its kinetic energy increases and its potential and total energies decrease.B. Its kinetic energy decreases, potential increases and its total energy remain tha sameC. Its kinetic energy and total energies decrease, and its potential, energy increases.D. Its kinetic potential and total energies decrease. |
|
Answer» Correct Answer - A We know that as the electron comes nearer to the nucleus, the potential energy decrease ((-KZe^(2))/( r) = PE and - r decreases))` The `KE will increase [:. KE = (1)/(2) |PE| = (1)/(2) KZe^(2))/( r)]` The Total energy decrease [TE = (1)/(2) KZe^(2))/( r)]` |
|
| 500. |
The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true?A. Its kinetic energy increases and its potential and total energies decreaseB. Its kinetic energy decreases, potential energy increases and its total energy remains the sameC. Its kinetic and total energies decreases and its potential energy increasesD. Its kinetic, potential and total energies decrease |
|
Answer» Correct Answer - A `E_(k) = (ke^(2))/(2r)` `E_(p) = (ke^(2))/(2r)` As the hydrogen atom makes transition to the ground state, `r` dereases. `E_(k)` increases, `E_(p)` and `E` decrease. |
|