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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.
| 301. |
Electrons with de- Broglie wavelength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays isA. `lambda_(0)=(2mclambda^(2))/h`B. `lambda_(0)=(2h)/(mc)`C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/h^(2)`D. `lambda_(0)=lambda` |
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Answer» Correct Answer - A `(hc)/lambda_(c)=KE_(e)=p_(e)^(2)/(2m_(e))=(h/lambda)^(2)/(2m)=h^(2)/(2mlambda^(2))Rightarrow lambda_(c)=(2hcmlambda^(2))/h^(2)Rightarrow lambda_(c)=(2cmlambda^(2))/h` |
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| 302. |
Electrons with de- Broglie wavelength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays isA. `lambda_(0) = (2 mc lambda^(2))/(h)`B. `lambda_(0) = (2 h)/(mc)`C. `lambda_(0) = (2 m^(2)c^(2) lambda^(2))/(h^(2))`D. `lambda_(0) = lambda` |
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Answer» Correct Answer - C The cut-offwavelength is given by `lambda_(0) = (hc)/(eV)` According to de Broglie equation. `lambda = (h)/(P) = (h)/(sqrt(2 meV)) implies lambda^(2) = (h^(2))/(2 meV)` `implies V = (h^(2))/(2 me lambda^(2))` From (i) and (ii) `lambda_(0) = (hc xx 2 me lambda^(2))/(e h^(2)) = (2 me lambda^(2))/(h)` |
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| 303. |
Statement 1: In process of photoelectric emission, all emitted electrons do not have same kinetic energy. Statement 2: If radiation falling on photosensitive surface of a metal consists of different wavelengths, then energy acquired by electrons absorbing photons of different wavelengths shall be different.A. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - B Both statement 1 and 2 are true, but even it radiation of single wavelength is incident on photosensitive surface, electrons of different KE will be emitted. |
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| 304. |
Statement -1: In process of photoelectric emission , all emitted electrons donot have sam Kinetic energy Statement-2: If radiation falling on photosensitive surface of a metal consits of different wavelength, then energy acquired by electrons absorbing photons of different wavelength shall be different. |
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Answer» Correct Answer - B Both statement `I` and `II` are true , but even it radiation of single wavelength is incident on photosensitive surface, electron of different `KE` will be emitted |
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| 305. |
X-ray falling on a meterialA. exterts a force itB. transfers energy to itC. transfers momentum to itD. transfers impulse to it |
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Answer» Correct Answer - A::B::C::D Photon exerts force due to change in momentum photon transfers its energy to the material. Photon transfers its energy to the material. Since, it exerts force, hecnce imaprts impulse also. |
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| 306. |
The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation,A. the intensity increasesB. the minimum wavelength increasesC. the intensity remain unchangedD. transfers impulse to it |
| Answer» Correct Answer - A::D | |
| 307. |
The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation,A. the intensity increasesB. the minimum wavelength increasesC. the intensity remains unchangedD. the minimum wavelength decreases |
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Answer» Correct Answer - A::D |
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| 308. |
The ratio of the energies of the hydrogen atom in its first to second excited state isA. `3//1`B. `1//4`C. `4//9`D. `9//4` |
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Answer» Correct Answer - D `E_(n)=-13.6Z^(2)//n^(2)` |
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| 309. |
The ratio of the energies of the hydrogen atom in its first to second excited state isA. `1//4`B. `4//9`C. `9//4`D. `4` |
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Answer» Correct Answer - C First excited state i.e., second orbit `(n = 2)` Second excited state i.e., third orbit `(n = 3)` `:. E = -(13.6)/(n^(2)) rArr (E_(2))/(E_(3)) = ((3)/(2))^(2) = (9)/(4)` |
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| 310. |
The ionization potential for second `H`e electron isA. `13.6 eV`B. `27.2 eV`C. `54.4 eV`D. `100 eV` |
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Answer» Correct Answer - C For the ionization of second He eelctron. `He^(+)` will act as hydrogen like atom. Hence ionization potetnial `Z^(2) xx 13.6 "volt" = (2)^(2) xx 13.6 = 54.4 V` |
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| 311. |
The energy required to remove an electron in a hydrogen atom from `n = 10` state isA. `1.36eV`B. `0.0136eV`C. `13.6eV`D. `0.136eV` |
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Answer» Correct Answer - D Energy requried=`E_(a)-E_(10)`=`(13.6)/10^(2)`=`0.136eV` |
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| 312. |
The kinetic energy of the electron in an orbit of radius `r` in hydrogen atom is (`e =` electronic charge)A. `(e^(2))/(r^(2))`B. `(e^(2))/(2r)`C. `(e^(2))/(r )`D. `(e^(2))/(2r^(2))` |
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Answer» Correct Answer - B Potential energy of electron in nth orbit of radius `r` in `H`-atom `U = - (e^(2))/(r )` (in `CGS`) `:. K.E. = (1)/(2) |P.E.| rArr K = (e^(2))/(2r)` |
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| 313. |
The energy required to remove an electron in a hydrogen atom from `n = 10` state isA. `13.6 eV`B. `1.36 eV`C. `0.136 eV`D. `0.0136 eV` |
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Answer» Correct Answer - C Energy required `= (13.6)/(n^(2)) = (13.6)/(10^(2)) = 0.136 eV` |
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| 314. |
A light beam of wavelength `400 nm` is incident on a metal of work- function `2.2 eV`. A particular electron absorbs a photon and makes 2 collisions before coming out of the metal (a) Assuming that 10% of existing energy is lost to the metal in each collision find the final kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions find the maximum number of collisions, the electron should suffer before it becomes unable to come out of the metal. |
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Answer» Correct Answer - `4` `E_(lambda)="Energy of the photon" =(hc)/(lambda)=(1.24xx10^(4))/(4000)eV` `KE " " "of" " "e=3.1 eV` For `e^(-)` not to come out, its energy should be less then `2.2 eV`. i.e., `(3.1)xx(0.9)^(n) lt 2.2 n gt 4` |
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| 315. |
The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to (where `e, h` and `c` have their usual meanings)A. `2pihc//e^(2)`B. `e^(2)h//2pic`C. `e^(2)c//2pih`D. `2pie^(2)//hc` |
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Answer» Correct Answer - D Speed of electron in nth orbit `("in" CGS) v_(n) = (2piZe^(2))/(nh) (k = 1)` For first orbit `H_(2) , n = 1` and `Z = 1` So `v = (2pie^(2))/(h) rArr (v)/(c ) = (2pie^(2))/(hc)` |
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| 316. |
Consider atoms `H, He^(+), Li^(++)` in their ground states. Suppose `E_(1), E_(2)` and `E_(3)` are minimum energies required so that the atoms `H He^(+), Li^(++)` can achieve their first excited states respectively, thenA. `E_(1) = E_(2) = E_(3)`B. `E_(1) gt E_(2) gt E_(3)`C. `E_(1) lt E_(2) lt E_(3)`D. `E_(1) = E_(2) = E_(3)` |
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Answer» Correct Answer - C `DeltaE prop z^(2)` i.e, greatest fro `Li^(++)` and minimum fo `H`. |
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| 317. |
In X-ray tube , when the accelerating voltage `V` is halved, the difference between the wavelength of `K_(alpha)` line and minimum wavelength of continuous X-ray spectrumA. romains contantB. becomes more than two timesC. becomes halfD. becomes less than two times |
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Answer» Correct Answer - D |
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| 318. |
In an `x` ray, if the accelerating potential different is changed , then:A. The frequency of characterictic `x`-rays of a material will get changedB. number of electrons emitted will changeC. the difference between` lambda_(0)` (minimum wavelength ) and `lambda _(kalpha)` (wavelength of `K_(alpha) `x` ray) will get changedD. difference between `lambda_(k alpha)` and `lambda_(k beta)` will get changed . |
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Answer» Correct Answer - C As the accelerating proportional difference is changed only the minimum wavelength changes. It has no effect on wavelengths of characteristic x-ray(whether they are produced or not ). |
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| 319. |
The colour of the positive column ina gas discharge tube depends onA. the type of glass used to construct the tubeB. the gas in the tubeC. the applied voltageD. the meterial of the cathode |
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Answer» Correct Answer - B The colour of the positive coulum in a discharge tube depends on the type of gas e.g., for air, colour is purple red, for `H_(2)`, colour is blue etc. |
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| 320. |
In an experiment for positive ray analysis with Thomson method, two identical parabola are obtianed when applied electric fields are `3000` and `2000 V//m`. The particles are singly ionised particles assuming same magnetic field :A. `1 : 3`B. `2 : 4`C. `3 : 1`D. `4 : 2` |
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Answer» Correct Answer - A For same magnetic field `(y^(2))/(x) prop (1)/(E) ((e)/(m))` For singly ionised particle, `(y^(2))/(x) prop (1)/(E_(1)) (e)/(m_(1))` Fro doubly ionised particle, `(y^(2))/(x) prop (1)/(E_(2)) (2e)/(m_(2))` Since the parabolas for both the particles are identical, `(1)/(E_(1)) (e)/(m_(1)) = (1)/(E_(2)) (2e)/(m_(2))` So `(m_(1))/(m_(2)) = (E_(2))/(2E_(1)) = (2000)/(2 xx 3000) = (1)/(3)` |
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| 321. |
A beam of electrons striking a cooper target produces X-rays. Its spectrum is as shown. Keeping the voltage same if the copper target is replaced with a different metal, the cut-off wavelength and characteristic lines of the new spectrum will change in comparison with old as:A. cut-off wavelength will remain unchanged while characteristic lines will be different.B. Both cut-off wavelength and characteristic lines will remain unchanged.C. Both Cut-off wavelength and characteristic lines will be differentD. Cut-off wavelength will be different while characteristic lines will remain unchanged |
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Answer» Correct Answer - A Cut-off wave length depend on accelerating voltage and characteristic wavelength depends on nature of the target material. |
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| 322. |
In Thomson experiment of finding `e//m` for electrons, been of electron is replaced by that of muons (particle with same charges as of electrons but mass `208` times that of electrons). No deflection condition in this case satisfied ifA. `B` is increased `208` timesB. `E` is increased `208` timesC. `B` is increased `14.4` timesD. None of these |
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Answer» Correct Answer - C In the condition of no deflection `(e)/(m) = (E^(2))/(2VB^(2)) rArr` If `m` is increased by `208` times then `B` should be increased `sqrt(208) = 14.4` times. |
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| 323. |
The energy required to excition an electron from `n=2` to `n=3` energy state is `47.2eV`. The charge number of the nucleus, around which the electrons revolving will beA. `5`B. `10`C. `15`D. `20` |
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Answer» Correct Answer - A `E=E_(3)-E_(2)=(13.6Z^(2))/4-(13.6Z^(2))/9=47.2` ` Rightarrow Z=?` |
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| 324. |
A closest distance of approach of an `a` particle travelling with a velocity `V` towards `Al_(13)` nucleus . `d` . The closest distacne of approach of an alpha particle travelling with velocity `4V` toward `Fr_(26)` nucleus isA. `d//2`B. `d//4`C. `d//8`D. `d//16` |
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Answer» Correct Answer - C `1/2mv^(2)=1/4piepsilon_(0)xx(q_(1)q_(2))/r: raz^(2)/v^(2)` |
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| 325. |
Cathode ray tube is operating at`5KV.` . Then the `K.E.` acquire by the electrons isA. `5eV`B. `5MeV`C. `5KeV`D. `5V` |
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Answer» Correct Answer - C `K.E.=Ve "` where `V` is `P.D` |
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| 326. |
The electron in a hydrogen atom makes a transition `n_(1) rarr n_(2)`, where `n_(1)` and `n_(2)` are the principle quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. the possible values of `n_(1)` and `n_(2)` areA. `n_(1)=4, n_(2)`B. `n_(1)=8,n_(2)=2`C. `n_(1)=8,n_(2)=1`D. `n_(1)=6,n_(2)=3` |
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Answer» Correct Answer - A::D `T_(n)propr_(n)/V_(n)RightarrowT_(n)propn^(3)` given `t_(n1)=8T_(n2)Rigtharrown_(1)=2n_(2)` |
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| 327. |
In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first excited state. Subsequently `H` atoms transfer their total excitation energy to `He+` ions (by collisions) Assume that the bohr model of atom is exactly valid. The wavelength of light emitted in the visible region by `He+` lons after collisions with `H` atoms is -A. `6.5 xx 10^(-7) m`B. `5.6 xx 10^(-7) m`C. `4.8 xx 10^(-7) m`D. `4.0 xx 10^(-7) m` |
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Answer» Visible light lies in the range, `lambda_(1) = 4000 Å to lambda_(2) = 7000 Å`. Energy of photons corresponding to these wavelength (in eV) would be : `E_(1) = (12375)/(4000) = 3.09 eV`, and `E_(2) = (900)/(11 R) = 1.77eV` From energy level diagram of `He^(+)` atom, we can see that in transition from `n = 4 to n = 3`, energy of photon released will lie between `E_(1)` and `E_(2)`. `Delta E_(43) = - 3.4 (- 6.04)` `= 2.64 eV` wavelength to photon corresponding to this energy. `lambda = (12375)/(264) Å = 4687.5 Å` `= 4.68 xx 10^(-7) m` |
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| 328. |
In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first excited state. Subsequently `H` atoms transfer their total excitation energy to `He+` ions (by collisions) Assume that the bohr model of atom is exactly valid. The ratio of the kinetic energy of the `n = 2` electron for the `H` atom to the of `He^(+)` lon is -A. `(1)/(4)`B. `(1)/(2)`C. 1D. 2 |
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Answer» kinetic energy, `K prop Z^(2)` `(K_(H))/(K_(He^(+))) = ((1)/(2)^(2) = (1)/(4)` |
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| 329. |
In a mixture of `H- He^(+)` gas (`He+ `is singly ionized He atom), `H` atom and `He+` ions are excited to their respective first excited state. Subsequently `H` atoms transfer their total excitation energy to `He+` ions (by collisions) Assume that the bohr model of atom is exactly valid. The quantum number `n` of the state finally populated in `He^(+)` inos is -A. `2`B. `3`C. `4`D. `5` |
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Answer» Correct Answer - C `DeltaE_(He)=13.6xx4(1/4-1/n^(2))eV,DeltaE_(He)=DeltaE(H)` |
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| 330. |
A hydrogen-like gas is kept in a chamber having a slit of width `d = 0.01 mm`. The atom of the gas are continuously excited to a certain energy state . The excite electron make transition to lower levels , From the initial excite state to the second excited state and then from the second excited state ground state. In the process of deexcitation, photons are emitted and come out of the container through a slit. The intensity of the photons is observed on a screen placed parallel to the plane of the slit . The ratio of the angular width of the central maximum corresponding to the two transition is `25//2`. The angular width of the central maximum due to first transition is `6.4 xx 10^(-2)` radian. Find the atomic number of the gas and the principal quantum number of the inital excited state. |
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Answer» Correct Answer - 2 `(1)/(lambda) = R Z ((1)/(n_(f))^(2) - (1)/(n_(1) ^(2)))` `(lambda_(2))/(lambda_(1)) = ((1)/(9) - (1)/(n^(2)))/(8//9)` where `n` is the principal quantum number of the initial excited state . Angular width `theta = (2 lambda)/(d)` `theta _(2))/(theta_(1)) = (lambda_(2))/(lambda_(1)) = (2)/(25) = (n^(2) - 9)/(8 n^(2)) implies n = 5` `(2 lambda_(1))/(d) = theta_(1). (1)/(lambda_(1)) = (2)/(d xx theta_(1) = (2)/(6.4 xx 10^(7) m^(-1)` `(1)/(lambda_(1)) = R Z^(2) ((1)/(9) - (1)/(n^(2)))` `(2)/(6.4 xx 10^(7) = 1.097 xx 10^(7) z^(2) ((1)/(9) - (1)/(25))` `z^(3) = (2)/(6.4) xx (225)/(16 xx 1.097) implies z = 2` |
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| 331. |
A gas containining hydrogen-like ions, with atomic number `Z`, emits photons in transition `n + 2 to n, where n = Z` These photons fall on a metallic plate and eject electrons having minimum de Broglie wavelength `lambda of 5 Å`. Find the value of `Z` , if the work function of metal is `4.2 eV` |
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Answer» Correct Answer - 2 `E = - (13.6 e V) Z^(2) ((1)/((Z + 2)^(2)) - (1)/(Z^(2)))` `= - 13.6 xx ((Z^(2) - (Z + 2)^(2))/((Z + 2)^(2))) = (4( Z + 1) xx 13.6)/((Z + 2)^(2)) eV` (i) [Now energy of electron is `K = (h^(2))/(2 lambda^(2) m)` Solving `K = 6 eV` ltrbgt ` (4( Z + 1) xx 13.6)/((Z + 2)^(2)) = 6 + 4.2 = 10.2 eV` `( Z + 1)/(((Z + 2)^(2)) = (3)/(16) implies (Z - 2) (3 Z + 2) = 0` So , the value of `Z = 2` (neglecting the negative//fractional value) |
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| 332. |
A hydrogen atom `(mass = 1.66 xx 10^(-27)` kg "ionzation potential" = 13.6eV)`, moving width a velocity `6.24 xx 10^(4) m s^(-1)` makes a completely inelastic head-on collision with another stationary hydrogen atom. Both atoms are in the ground state before collision . Up to what state either one atom may be excited? |
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Answer» Correct Answer - 2 Using `COM: mu = 2 mv` `v = (u)/(2) , Delta E = (1)/(2) mu^(2) - 2 xx (1)/(2) m ((u)/(2))^(2)` `implies Delta E = (1)/(4) mu^(2) = (1)/(4) xx 1.66 xx 10^(-27) xx (6.24 xx 10^(4)) ^(2)` `= 13.6 ((1)/(l^(2)) - (1)/(n^(2)))` `= (1.66 xx 10^(-27) xx 6.24 xx 10^(4)) ^(2)/(4 xx 1.6 xx 10^(-19)) implies n = 2` |
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| 333. |
The energy required to separate a hydrogen atom into a proton and an electron is `1.6eV` then the velocity of electron in a hydrogen atom isA. `2.2xx10^(4)m//s`B. `2.2xx10^(2)m//s`C. `2.2xx10^(6)m//s`D. `2.2xx10^(10)m//s` |
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Answer» Correct Answer - C Velocity of revolving electron electron is given as `(mv^2)/r=1/(4piepsilon_(0)) e^(2)/r^(2),thereforeV=(e)/sqrt(4piepsilon _(0) mr)` |
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| 334. |
For the structural analysis of crystals, X-rays are used becauseA. X-rays have wavelength of the order of the inter-atomic spacingB. X-rays are highly penetrating radiationsC. X-rays are coherent radiationsD. |
| Answer» Correct Answer - A | |
| 335. |
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)` are |
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Answer» Correct Answer - 2 `(T_(1))/(T_(2)) = (n_(1)^(3))/(n_(1)^(3)) implies 8_(1)/(8_(2)) = (n_(1)^(3))/(n_(1)^(3)) implies (n_(1))/(n_(2)) = (2)/(1)` |
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| 336. |
A hydrogen-like atom (aatomic number `Z`) is in a hight excited state of quantum number `n` . This excited atom can make a transition to the first excited state by successively emitting two proton of energy `10.20 e V and 17.00 e V`, respectively. Alternatively , the atom from the same excited state can make a transition to the second excited state by successively emitting two proton of energy `4.25 e V and 5.954 e V`, respectively. Determine the value of `n and Z` (ionization energy of hydrogen atom `= 13.6 eV)` |
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Answer» For the transition from `a` higher state `n` to the first excited state `n_(1) = 2`, the total energy released is `10.2 + 17.0 e V` or `27.2 e V` Thus , for `Delta E = 27.2 e V, n_(1) = 2 and n_(2) = n`, we have `27.2 = 13.6 Z^(2) [(1)/(4) - (1)/(n^(2))]` ...(i) For the eventual transition to the second excited state `n_(1) = 3`, the total energy released is `(4.25 + 5.95) e V or 10.2 e V` Thus, `10.2 = 13.6 Z^(2) [(1)/(9) - (1)/(n^(2))]` ...(ii) Driving Eqs. (i) and (ii), we get `(27.2)/(10.2) = (9 n^(2) - 36)/(4 n^(2) - 36)` Solving we get `n^(2) = 36` or `n = 6` Substituting `n = 6` in any of the above equations, we obtain `Z^(2) = 9` or `Z = 3` Thus, `n = 6` and `Z = 3` |
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| 337. |
What is the ionisation potential of hydrogen atom?A. `12.97V`B. `10.2V`C. `13.6V`D. `27.2V` |
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Answer» Correct Answer - C `E_("ionisation")=E_(alpha)-E_(1)` |
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| 338. |
`alpha`-particles of enegry `400 KeV` are boumbardel on nucleus of `._(82)Pb` . In scattering of `alpha`-particles, it minimum distance from nucleus will beA. `0.59 nm`B. `0.59 Å`C. `0.59 p m`D. `0.59 p m` |
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Answer» Correct Answer - D Suppose closet distance is `r`, according to conservation of energy. `400 xx 10^(3) xx 1.6 xx 10^(-19) = 9 xx 10^(9) ((ze)(2e))/(r )` `rArr 6.4 xx 10^(-14)` `= (9 xx 10^(9) xx (82 xx 1.6 xx 10^(-19))xx (2 xx 1.6 xx 10^(-19)))/(r )` `rArr r = 5.9 xx 10^(-13) m = 0.59p m` |
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| 339. |
The transition from the state `n=4` to `n=3` in a hydrogen is ultraviolet radiation. Infrared radiation will be obtained in the transition from:A. `3 rarr 2`B. `4 rarr 2`C. `5 rarr 4`D. `2 rarr 1` |
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Answer» Correct Answer - C `5rarr4` Transition energy from `5` to `4` will be lessthan from `4 rarr3`. All other transition energy are higher than that for `4 rarr 3`. |
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| 340. |
Suppose an electron is attracted toward the origin by a force`(k)/(r )` where `k` is a constant and `r` is the distance of the electron from the origin .By applying Bohr model to this system the radius of the `n^(th)` orbital of the electron is found to be `r_(n)` and the kinetic energy of the electron to be `T_(n)` , Then which of the following is true ?A. `T_(n)` independent of `n, r_(n)prop n`B. `T_(n) prop(1)/(n),r_(n) prop n`C. `T_(n) prop(1)/(n) n_(1),r_(n) prop n^(2)`D. `T_(n) prop (1)/(n^(2)),r_(n)prop n^(2)` |
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Answer» Correct Answer - A `(mv^(2))/(r )=(K)/(r )`.....(1) `Mvr=(nh)/(2pi)`....(2) Solve the equation |
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| 341. |
A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of `10fm` to the nucleus. The de - Broglie wavelength (in units of fm) of the proton at its start is take the proton mass, `m_p = 5//3xx10^(-27) kg, h//e = 4.2xx10^(-15) J-s//C`, `(1)/(4piepsilon_0) = 9xx10^9m //F, 1 fm = 10^(-15)`. |
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Answer» `hv=W+13.6eV` and in second case `KE_(max)=(12420)/(1215)=10.2eV` ` therefore5/6hv=W+10.2eVRightarrowv=5xx10^(15)H_(z)` |
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| 342. |
An image of the sun is formed by a lens, of the focal length of 30 cm, on the metal surface of a photoelectric cell and a photoelectric current I is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photoelectric current in this case isA. `I//2`B. `2I`C. `I`D. `4 I` |
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Answer» Correct Answer - C The current `I` is proportional to light energy falling on the lens per second with is same in the two cases. Hence same `I`. |
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| 343. |
The work function of a certain metal is `(hC)/lambda_(0)`. When a monochromatic light of wavelength `lambda lt lambda_(0)` is incident such that the plate gains a total power `P`. If the efficiency of photoelectric emission is `eta%` and all the emitted photoelectrons are captured by a hollow conducting sphere of radius `R` already charged to potential `V`, then neglecting any interaction of potential of the sphere at time `t` is:A. `V+(100eta lambda Pet)/(4piepsilon_(0)RhC)`B. `V=(etalambdaPet)/(400pi epsilon_(0)RhC)`C. `V`D. `(lambdaPet)/(4 pi epsilon_(0)RhC)` |
| Answer» Correct Answer - B | |
| 344. |
The ratio of the largest to shortest wavelength in Balmer series of hydrogen spectra is,A. `25/9`B. `17/6`C. `9/5`D. `5/4` |
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Answer» Correct Answer - C `1/lambda_(max)=RZ^(2)(1/n_(1)^(2)-1/n^(2_2))` where `n_(1)=2,n-(2)=3` ` 1/lambda_(min)=RZ^(2)(1/n_(1)^(2)-1/n^(2_(2)) Where n_(1)=2,n_(2)=a` ` 1/lambda_(min)=RZ^(2)(1/n_(1^(2))-1/2^(2_(2)))` where `n_(1)=2,n_(2)=a` |
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| 345. |
Suppose in certine condition only those transition are allowed to hydrogen atoms in which the principal quantum number a changes by`2` (a) Find the smaller wavelength emitted by hydrogen (b) list the wavelength emitted by hydrogen in the visible range `(380 nm to 780 nm)` |
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Answer» Correct Answer - (a) `(9)/(8R)=103nm` (b) `(16)/(3R)=487 nm` Allowed energy of hyrogen atom are `{:(E_(1)=-13.6 eV",",E_(2)= -3.4 eV",",E_(3)=-1.5eV",",E_(4)= -0.85 eV",",),(E_(5)=-0.54eV",",E_(6)=37 eV",",E_(7)=-27eV",",E_(8)=0.2eV",",):}` For emission of the smallest wavelength we must wave `DeltaE_(max)` `DeltaE_(max)=E_(3)-E_(1)` `13.6-1.5=12.1eV` corresponding `lambda=(hc)/(DeltaE)=(1242eV-nm)/(12.1)=103 nm`. Any transition from `n=4` to `n=2` will be falling in the visible range `Delta E-E_(4)-E_(2)=2.55eV` corresponding `lambda=(hc)/(DeltaE)=487 nm`. |
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| 346. |
An electron in a hydrogen atoms makes a transition form `n_1` to `n_2` where `n_1 and n_2` are two principal quantum numbers of two states. If time period of electron in state `n_1` is 8 times the time period in state `n_2`, find the ratio `(n_2//n_1)`, assuming Bohr model to be true.A. `8:1`B. `4:1`C. `2:1`D. `1:2` |
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Answer» Correct Answer - C Time period of revolution of electron in the nth orbit is `T` a `jn^(3) ` |
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| 347. |
A uniform magnetic field and a uniform electric field are produced, pointing in the same direction. An electron is projected with its velocity pointed in the same direction. What will be the effect on electron?A. the electron will turn to its rightB. the electron will turn to its leftC. the electron velocity will increase in magnitudeD. the electron velocity will decrease in magnitude |
| Answer» Correct Answer - D | |
| 348. |
A metal is illumimated by light of two different wavelength `248nm` and `310 nm` . The maximum speeds of the photoelectrons corresponding in these wavelength are `u_(1) and u_(2)` respectively . If the ratio `u_(1) : u_(2) = 2 : 1` and `hc = 1240 eVnm `, the work function of the meal is nearlyA. `3:7 eV`B. `3.2 eV`C. `2.8 eV`D. `2.5 eV` |
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Answer» Correct Answer - A `248 nm=1240//248 ev-= 5ev` `310nm -=1240//310ev=4 ev` `(K.E_(1))/(K.E_(2))=(4)/(1)=(5ev-W)/(4ev-W)` `rArr 16-4W=S-W` `rArr11-3W` `rArrW=(11)/(3)=3.67 ev ~=3.7 ev` |
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| 349. |
(a) Find the maximum wavelength `lambda_(0)` of light which can ionize a hydrogen atom in its ground (b) light of wavelength `lambda_(0)` is inclined on a hydrogen atom which is in its first excited state find the kinetic energy of the electron coming out |
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Answer» Correct Answer - `[(a) 913 Å, (b) 10.2eV]` |
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| 350. |
Consider a hydrogen atom with its electron in the `n^(th)` orbital An electomagnetic radiation of wavelength `90 nm` is used to ionize the atom . If the kinetic energy of the ejected electron is `10.4 eV` , then the value of `n` is ( hc = 1242 eVnm) |
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Answer» Correct Answer - `2` `(hc)/(lambda)-{13.6eV.(1)/(2)}=10.4` `rArr(124eV)/(90)-(13.6)/(n^(2))=10.4` `rArr(41.4)/(3)-(13.6)/(n^(2))=10.4` `rArr13.8-10.4=(13.6)/(n^(2))` `rArr3.4=(13.6)/(n^(2))` `rArrn^(2)=4` `rArr n=2` |
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