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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.
| 251. |
The speed of electron in the second orbit of `Be^(3+)` ion will beA. `( C)/(137)`B. `( 2 C)/(137)`C. `(3 C)/(137)`D. `(4 C)/(137)` |
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Answer» Correct Answer - B `v = Z [(1)/(137) (c)/(n)]` `= 4 xx (1)/(137) xx (c)/(n)` `(2 c)/(137)` |
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| 252. |
If the radius of an orbit is `r` and the velocity of electron in it is `v` , then the frequency of electron in the orbit will beA. `2 pi r v`B. `(2 pi)/(v r)`C. `(v r)/(2 pi)`D. `(v)/(2 pi r)` |
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Answer» Correct Answer - D `v = 2 pi r f` `implies f = (v)/(2 pi r)` |
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| 253. |
The velocity of an electron in the second orbit of sodium atom (atomic number `= 11`) is `v`. The veocity of an electron in its fifth orboit wil beA. `v`prop`V`B. `(22v)/5`C. `5/2v`D. `2/5v` |
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Answer» Correct Answer - D `v_(n)alpha1/n` |
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| 254. |
The potential energy of an electron in the fifth orbit of hydrogen atom isA. `0.54 eV`B. `- 0.54 eV`C. `1.08 eV`D. `- 1.08 eV` |
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Answer» Correct Answer - D `E = - (13.6)/(5^(2)) eV = - 0.544eV` `E_(p) = - 2 xx 0.544 eV = - 1.088 eV` |
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| 255. |
The velocity of an electron in the second orbit of sodium atom (atomic number `= 11`) is `v`. The veocity of an electron in its fifth orboit wil beA. `v`B. `(22)/(5)v`C. `(5)/(2)v`D. `(2)/(5)v` |
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Answer» Correct Answer - D `v_(n) prop (1)/(h) rArr (v_(5))/(v_(2)) = (2)/(5) rArr v_(5) = (2)/(5) v_(2) = (2)/(5) v` |
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| 256. |
The threshold frequency for a metallic surface corresponds to an energy of `6.2eV` and the stopping potential for a radiation incident on this surface is `5 V` . The incident radiation lies inA. X-ray regionB. ultra-violet regionC. infra-red regionD. visible region |
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Answer» Correct Answer - B `lambda=(1242eVnm)/(11.2)~~1100 B232Å` Ultraviolet region |
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| 257. |
The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion x. Calculate energies of the first four levels of x. |
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Answer» Correct Answer - `- 54.4 e V, - 13.6 e V, - 6.04 e , - 3.52 e V ,54.4 e V` We know that `(1)/(lambda) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` For first line of Lyman series in hydrogen atom, `(1)/(lambda_(1)) = R ((1)/(1^(2)) - (1)/(2^(2))) = (3 R)/(4)` (i) For second line of Balmer series of hydrogen-like ion `X` `(1)/(lambda) = Z^(2) R ((1)/(2^(2)) - (1)/(4^(2))) = (3 Z^(2) R)/(16)` (ii) given that `lambda_(1) = lambda_(2)` `:. (3 R)/(4) = (3 Z^(2) R)/(16) or Z = 2` Thus, the ion `X` is singly ionized helium atom . Energy of `n^(th)` state of ion `X` is given by `E_(S) = - (13.4)/(n^(2)) xx Z^(2)` `:. E_(X)_(1) = - (13.4 xx 4)/(1) = - 54.4 e V` `E_(X)_(2) = - (13.4 xx 4)/(4) = - 13.6 e V` `E_(X)_(3) = - (13.4 xx 4)/(9) = - 6.04 e V` `E_(X)_(4) = - (13.4 xx 4)/(16) = - 3.52 e V` |
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| 258. |
X-ray are emitted by a tube containing the electron gt Noibum `(Z = 41)` as anticathode and the `K_(alpha)` X-ray an allowed to be incident on an unknown gas containing hydrogen- like atom ions. It is found that the X-ray cause the emission of photoelectrons with an energy of `2.7 keV` from these ions. Find a. the minimum voltage at which the X-ray tube should be opreated so that the momentum of the emitted photoelectrons is doubled. b. the approximate value of `Z` for the target of the anticathode after the momentum has been doubled. c. the approximate value of `Z` the atomic number of the atom of the gas. d. the number of the ion produced in the gas caused by X-ray , if the inetensity of X-ray is `100 mWm^(-2)` and `1%` of the X-rays cause ionization. |
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Answer» Correct Answer - `24.5 keV, = 850 per m^(3)` b. 50 c. 32 The wavelength of the `K_(alpha)` radiation from `Nb` is `lambda = (4 hc)/(3 R) (1)/((Z - 1)^(2)) = (4 hc)/(3 R) (1)/(40^(2))` `E_(K_alpha) = 1.63 keV` `Z_(gas)^(2) = 1000 implies Z_(gas) = 32` If the momentum of the electron is doubled, then the `KE` is increased by `4` times Therefore , the minimum energy of the X-rays is `(16.3 + 3 xx 2.7) keV = 24.5 keV`. Since `E_(K_alpha) prop (Z - 1)^(2)` `((Z - 1)/(40))^(2) = ((24.5)/(16.3)) implies Z = 50` The voltage at which the tube should be operated should be greater than `24.5 kV`. The number of X-ray photons incident`= 100 mWm^(-2)` `implies l = (100 xx 10^(-3) xx 10^(19))/(1.6 xx 24.5 xx 10^(3)) implies m^(-2)` Number of ions produced in `1 s (n) = 1% of(l)/(C )` `(100 xx 10^(-3) xx 10^(-2) xx 10^(19))/(1.6 xx 24.5 xx 10^(3) xx 3 xx 10^(8)) m^(-3) = 850 per m^(-3)` |
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| 259. |
when the voltage applied to an X-ray tube increases from `V_(1) = 10 kV` to `V_(2) = 20 kV`, the wavelenght interval between `K_(alpha)` line and cut-off wavelenght of continuous spectrum increases by a factor of `3`. Atomic number of the metallic target is |
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Answer» Correct Answer - 30 `lambda_(C_2) = (h)/(eV) = (12375)/(10 xx 10^(3)) Å = 1.2375 Å` (i) `lambda_(C_1) = 0.61875 Å` (ii) `(1)/(lambda_(k_alpha)) = (Z - 1)^(2) {(1)/(1) - (1)/(4)} xx 10^(7)` (iii) It is give `3 xx (lambda_(k_alpha) - 1.2375) = lambda_(k_alpha) - 0.61875` `implies lambda_(k_alpha) = 1.54338 Å` (iv) Putting this value in (iii). `Z - 1 = 29` `implies Z = 30` |
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| 260. |
one of the following radiation are not emitted by electron transition in atom , choose the optionA. ultra violet raysB. infrared radiationsC. visible raysD. `alpha-rays` |
| Answer» Correct Answer - D | |
| 261. |
The energy emitted by a source is in the form ofA. photonsB. electronsC. protonsD. neutrons |
| Answer» Correct Answer - A | |
| 262. |
The wavelength involved in the spectrum of deuterium `_(1)^(2)D` are slightly different from that of hydrogen spectrum becauseA. the size of the nuclei are differentB. the nuclear forces are different in two casesC. the mases of the two nuclei are differentD. the attraction and the nucelus is different in the two cases |
| Answer» Correct Answer - C | |
| 263. |
The binding energy of the electron in the ground state of `He` atom is equal to `E_(0)=24.6 eV`. Find the energy required to remove both the electrons from the atom.A. `24.6eV`B. `79.0eV`C. `54.4eV`D. `none of these |
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Answer» Correct Answer - B The total energy required to make the electron free from nucleus is the sum of the energy required to separate the electron from the influence of each other and the energy required to separate the electron from the influence of nucleus , i.e total required energy energy=`BE` of electron in `He` atom + ionization energy of He atom =`(24.6+13.6xx2^(2))eV` ` (24.6+54.4)eV=79eV` |
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| 264. |
Hydrogen atom will be in its ground state , if its electrons is inA. any energy levelB. the lowest energy stateC. the highest energy stateD. the intermediate state |
| Answer» Correct Answer - B | |
| 265. |
The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum isA. `1//2`B. `2//137`C. `1//137`D. `1//237` |
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Answer» Correct Answer - C Speed of electron in nth orbit of hydrogen atom `v = (e^(2))/(2 epsilon_(0)nh)` In ground state `n = 1 rArr v = (e^(2))/(2 epsilon_(0) nh)` `rArr (v)/(c ) = (e^(2))/(2 epsilon_(0) nh)` `= ((1,6 xx 10^(-19))^(2))/(2 xx 8.85 xx 10^(-12) xx 3xx 10^(3) xx 6.6 xx 10^(-34))` `= (1)/(137)` |
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| 266. |
The wavelength of Lyman series isA. `(4)/(3 xx 10967) cm`B. `(3)/(4 xx 10967) cm`1C. `(4 xx 10967)/(3) cm`D. `(3)/(4) xx 109767 cm` |
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Answer» Correct Answer - A `(1)/(lambda) = R_(H) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`. For Lyman series `n_(1) = 1` and `n_(2) = 2`, ltbegt `3.4.` When `n_(2) = 2`, we get `lambda = (4)/(3R_(H)) = (4)/(3 xx 10967) cm` |
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| 267. |
The first line of Balmer series has wvaelength `6563 Å`. What will be the wavelength of the ifrst member of Lyman series?A. `1215.4 Å`B. `2500 Å`C. `7500 Å`D. `600 Å` |
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Answer» Correct Answer - A `(1)/(lambda_("Balmer")) = R [(1)/(2^(2)) - (1)/(3^(2))] = (5R)/(36), (1)/(lambda_("Lyman")) = R [(1)/(1^(2)) - (1)/(2^(2))] = (3R)/(4)` `:. Lambda_("Lyman") = lambda_("Balmer") xx (5)/(27) = 1215.4 Å` |
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| 268. |
If `lambda_(max)` is `6563 Å`, then wave length of second line of Balmer series will beA. `lambda = (16)/(3R)`B. `lambda = (36)/(5R)`C. `lambda = (4)/(3R)`D. None of the above |
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Answer» Correct Answer - A For Balmer series `(1)/(lambda) = R ((1)/(2^(2)) - (1)/(n^(2)))` where `n = 3, 4, 5` For second line `n = 4` So `(1)/(lambda) = R ((1)/(2^(2)) - (1)/(4^(2))) = (3)/(16) R rArr lambda = (16)/(3R)` |
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| 269. |
The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen isA. `(4)/(3)`B. `(525)/(376)`C. `25`D. `(900)/(11)` |
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Answer» Correct Answer - D Shortest wavelength comes from `n_(1) = oo` to `n_(2) = 1` and longest wavelength comes from `n_(1) = 6` to `n_(1) = 5` in the given. Case. Hence `(1)/(lambda_(min)) = R ((1)/(1^(2)) - (1)/(oo^(2)))= R` `(1)/(lambda_(max)) = R ((1)/(5^(2)) - (1)/(6^(2)))= R ((36 - 25)/(25 xx 36)) = (11)/(900)R` `:. (lambda_(max))/(lambda_(min)) = (900)/(11)` |
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| 270. |
`29` electron are remove from `Zn` atom `(Z=30)` by certain means . The minimum energy needed to remove the `30th` electron , will be :A. `12.24keV`B. `408keV`C. `0.45keV`D. `765keV` |
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Answer» Correct Answer - A After removing 29 electrons, Zn atom will become `Zn^(+29) therefore E=13.6Z^(2)eV=12,24KeV` |
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| 271. |
Consider a gas consisting `Li^(+2)` (which is hydrogen like ion). (a) Find the wavelength of radiation required to excite the electron in `Li^(++)` from `n=1` and `n=3`. (Ionisation energy of the hydrogen atom equals `13.6 eV`). (b) How many spectral lines are observed in the emission spectrum of the above excited system? |
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Answer» Correct Answer - (a) `(hc)/(13.6xx8e)=113.7Å` (b) `3` (a) `DeltaE= 13.6xx3^(2)xx((1)/(1)(1)/(3^(2)))eV` `13.6xx8eV` `lambda=(hc)/(DeltaE)=(12400)/(13.6xx8)Å=113.7Å` (b) no. of spectral lines `= .^(n)C_(2)` `= .^(3)C_(2)=3` |
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| 272. |
The number of different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to third excited state isA. `3`B. `4`C. `5`D. `6` |
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Answer» Correct Answer - A `N=(n(n-1))/2: "where"n=3` |
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| 273. |
What is the minimum energy that must be given to a H atom in ground state so that it can emit an `H gamma` line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such `H gamma` photon ?A. `10.2eV`B. `12.1eV`C. `13.06eV`D. `13.6eV` |
| Answer» Correct Answer - C | |
| 274. |
The ratio of minimum to maximum wavelength in Balmer series isA. `5 : 9`B. `5 : 36`C. `1 : 4`D. `3 : 4` |
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Answer» Correct Answer - A `(1)/(lambda) = R[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] rArr (lambda_(min))/(lambda_(max)) = [[(1)/(2^(2)) - (1)/(3^(2))])/[[(1)/(2^(2)) - (1)/(oo)]) = (5)/(9)` |
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| 275. |
The first line in the Lyman series has wavelength `lambda`. The wavelegnth of the first line in Balmer series isA. `(2)/(9)lambda`B. `(9)/(2)lambda`C. `(5)/(27)lambda`D. `(27)/(5)lambda` |
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Answer» Correct Answer - D For first line in Lyman series `lambda_(L_(1)) = (4)/(3R) ………(i)` For first line in Balmer series `lambda_(B_(1)) = (36)/(5R)……..(ii)` From equations (i) and (ii) `lambda_(B_(1))/(lambda_(L_(1))) = (27)/(5) rArr lambda_(B_(1)) = (27)/(5) lambda` |
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| 276. |
The Lyman series of hydrogen spectrum lies in the region :A. InfraredB. VisibleC. UltravioletD. Of-X-rays |
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Answer» Correct Answer - C Lyman series lies in the `UV` region. |
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| 277. |
Which of the following is true?A. Lyman series is a continuous spectrumB. Paschen series is a line spectrum in the infraredC. Balmer series is a line spectrum in the ultravioletD. The spectral series formula can be drived from the Rutherford modelof the hydrogen atom |
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Answer» Correct Answer - B Paschen series lies in the infrared region. |
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| 278. |
The time taken by a photoelectron to come out after the photon strikes is approximatelyA. `10^(-1)s`B. `10^(-4)s`C. `10^(-10)s`D. `10^(-16)s` |
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Answer» Correct Answer - C `10^(-10) sec`. |
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| 279. |
If the kinetic energy of a free electron doubles , its de - Broglie wavelength changes by the factorA. `(1)/(2)`B. `2`C. `(1)/(sqrt(2))`D. `sqrt(2)` |
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Answer» Correct Answer - C We know `lambda=(h)/(m v)` and `K=(1)/(2) mv^(2)=((3v)^(2))/(2m)` `rArrmv=sqrt(2mK)` Thus `lambda=(h)/(sqrt(2mK))` `rArr lambda prop (1)/(sqrt(K))` `:. (lambda_(2))/(lambda_(1))=sqrt(K_(1))/(sqrt(K_(2)))= sqrt(2K_(1))/(sqrt(2K_(1))) " "( :. K_(2)=2K_(1))` `rArr (lambda_(2))/(lambda)=(1)/(sqrt(2))` `:. lambda_(2)=(1)/(sqrt(2))` |
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| 280. |
The maximum wavelength of Brackett series of hydrogen atom will be _____`A^(@)`A. `35,890`B. `14,440`C. `62,160`D. `40,477` |
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Answer» Correct Answer - D `RZ^(2)(1/n_(1)^(2)-1/n^(2_(2)))=1/lambda "where" n_(1)=4:n_(2)=5` |
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| 281. |
The angular momentum of the electron in third orbit of hydrogen atom, if the angular momentum in the second orbit of hydrogen is `L` isA. `L`B. `3L`C. `(3//2)L`D. `2//3L` |
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Answer» Correct Answer - C `L=(nh)/(2pi)` |
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| 282. |
The `De-Broglie wave length of electron in second exited state of hydrogen atom isA. `3.33A^(@)`B. `6.66A^(@)`C. `9.99A^(@)`D. `1.06A^(@)` |
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Answer» Correct Answer - B `2pir=nlambdaRightarrowlambda=(2pir)/n=6pix` |
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| 283. |
Hydrogen atom in this ground state is excited by means of monochromatic radiation of wavelength `975Å`. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom is `13.6eV`A. Total number of lines in emission spectrum would be `6`.B. Energy difference between `3^(rd)` and `4^(th)` orbit is `0.66 eV`.C. longest wavelength is emission spectrum would be `1.875 mu m`.D. smallest wavelength in emission spectrum would be `975 Å`. |
| Answer» Correct Answer - A::B::C::D | |
| 284. |
Consider an electron obrbiting the nucleus with speed `v` in an orbit of radius `r`. The ratio of the magetic moment to the orbtial angular momentum of the electron is independent of:A. radius `r`B. speed `v`C. charge of electron `e`D. mass of electron `m_(e )` |
| Answer» Correct Answer - A::B | |
| 285. |
In a hydrogen like atom electron make transition from an energy level with quantum number `n` to another with quantum number `(n - 1)` if `n gtgt1` , the frequency of radiation emitted is proportional to :A. `(1)/(n^(2)`B. `(1)/(n^(3))`C. `n^(2)`D. `(1)/(n^(4))` |
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Answer» Correct Answer - B `E_(n)-E_(n-1)` `=(-13.6 Z^(2))/(n^(2))+(-13.6Z^(2))/((n-1)^(2))=[(1)/((n-1)^(2))-(1)/(n^(2))]` `hv=13.6 Z^(2)[(n^(2)-(n-1)^(2))/(n^(2)(n-1)^(2))]` `=(13.6Z^(2))/(n^(2))[(1)/((1-1//n)^(2))-1]` `=(13.6Z^(2))/(n^(2))[a+(2)/(n)-1]=(2xx13.6Z^(2))/(n^(3))v prop(1)/(n^(3))` (b) |
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| 286. |
In a hydrogen like atom electron make transition from an energy level with quantum number `n` to another with quantum number `(n - 1)` if `n gtgt1` , the frequency of radiation emitted is proportional to :A. `(1)/(n)`B. `(1)/(n^(2))`C. `(1)/(n^(3)2)`D. `(1)/(n^(3)2)` |
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Answer» Correct Answer - B::D `DeltaE=hv` `v=(DeltaE)/(h)=k[(1)/((n-1)^(2))-(1)/(n^(2))]=(k2n)/(n^(2)(n-1)^(2))` `~~(2K)/(n^(3))prop(1)/(n^(3))` `r=(mv)/(eB)rArr(r^(2)e^(2)B^(2))/(2)=(m^(2)v^(2))/(2)` `(r^(2)e^(2)B^(2))/(2m)=(mv^(2))/(2)` `1.89-phi=(r^(2)e^(2)B^(2)1)/(2m)(1)/(e)eV=(r^(2)eB^(2))/(2m)eV=(100xx10^(-6)xx1.6xx10^(-19)xx9xx10^(-8))/(2xx9.1xx10^(-31))` `phi=1.89-(1.6xx9)/(2xx9.1)` `=1.89=0.79~=1.1 eV` |
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| 287. |
Hydrogen `(._(1)H^(1))`, Deuterum `(._(1)H^(2))`, singly ionised Hellium `(._(2)He^(4))^(+)` and doubly ionised lithium `(._(3)Li^(6))^(++)` all have one electron around the nucleus. Consider an electron tranition from `n=2` to `n=1`. If the wave lengths of emitted radiation are `lambda_(1),lambda_(2),lambda_(3)` and `lambda_(4)` respectively then approximately which one of the follwing is correct ?A. `4lambda_(1)=2lambda_(2)=2lambda_(3)=lambda_(4)`B. `lambda_(1)=2lambda_(2)=2lambda_(3)=lambda_(2)`C. `lambda_(1)=lambda_(2)=4lambda_(3)=9lambda_(4)`D. `lambda_(1)=2lambda_(1)=3lambda_(3)=4lambda_(2)` |
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Answer» Correct Answer - B `(1)/(lambda)=Rz^(2)((1)/(1^(2))-(1)/(2^(2)))` `(1)/(lambda_(1))=R(1)^(2)(3//4) (1)/(lambda_(2))=R(1)^(2)(3//4)` `(1)/(lambda_(3))R2^(2)(3//4)(1)/(lambda_(4))=R3^(2)(3//4)` `(1)/(lambda_(1))=(1)/(4lambda_(3))=(1)/(9lambda_(4))=(1)/(lambda_(2))` So option `(3)` is correct |
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| 288. |
The graph between the stopping potential `V_(0)` and frequency `v` for two different metal plates `P` and `Q` are shown in the figure. Which of the metals has greater threshold wavelength and work function. |
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Answer» `eV_(s)= hv-W` `W=(hc)/(lambda_(th))` From graph, `W_(Q) gt W_(p)` So work function of `Q` is greater than `P` `lambda_(th p) gt lambda_(th Q)` |
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| 289. |
An oil drop of mass m fall through air with a terminal velocity in the presence of upward electric field of intensity `E`. the drop carries a charge `+q`. `R` is the viscous drag force and `f` is the buoyancy force . Them for the motion of the drop.A. there is a net force directed upwardB. there is a net force directed downwardC. `mg=Eq+F+R`D. `mg+Eq=F+R` |
| Answer» Correct Answer - C | |
| 290. |
An oil drop of radius r carrying a charge q remain stationary in the presence of electric field of intensity E. If the density of oil `rho` , thenA. `E=4/3pi r^(3)rhogq`B. `E=4/3pir^(3)rho g`C. `E=4/3pir^(3)rho g//q`D. `E=4/3pir^(3)rho//g^(3)` |
| Answer» Correct Answer - C | |
| 291. |
An oil drop of mass m fall through a viscous medium. The viscous drag force. `F` is proportional to the velocity of the drop . At the instant it begins to fall the force that acts on the oil drop is (neglect buoyancy)A. mgB. `mg-F`C. `F-mg`D. F |
| Answer» Correct Answer - A | |
| 292. |
An oil drop of mass m and charge +q is balanced in vaccum by a uniform electric field of intensity `E`. the direction of this field should beA. Vertically upwardsB. verticallu downC. horizontal perpendicular to the direction of motion of the elctron beamD. inclined at `45^(0)` to the horizontal |
| Answer» Correct Answer - A | |
| 293. |
An oil drop of mass m and falls through a medium that offers a viscous drag force. F. if the velocity of the drop is constant it means thatA. `F gt mg`B. `F lt mg`C. `F gt mg`D. `F=mg` |
| Answer» Correct Answer - B | |
| 294. |
Light described at a place by the equation `E= (100 V(m^-1)) [sin (5 xx (10^15)m(s^-1))t + sin (8 xx (10^15)(s^-1)t]` falls on a metal surface having work function `2.0 eV`. Calculate the maximum kinetic energy of the photoelectrons. |
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Answer» The light contains two different frequencies. The one with larger frequency will case photoelectrons with largest kinetic energy. This larger frequency is `v=(omega)/(2pi)=(8xx10^(15)s^(-1))/(2pi)` The maximum kinetic energy of the photoelectron is `K_(max)=hv-W` `=(4.14xx10^(15)eV-s)xx((8xx10^(15))/(2pi)s^(-1))-2.0 eV` `5.27 eV-2.0eV=3.27 eV`. |
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| 295. |
A plate of mass `10 gm` is in equillibrium in air due to the force exterted by light beam on plate. Calculate powder of beam. Assume plates is perfectly absorbing. |
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Answer» Since plate is in air, so gravitational force will act on this `F_("gravitaional")=mg` (dowmward) `=10xx10^(-3)xx 10` `10^(-1)N` for the equilibrium force exterted by light beam should be equal to `F_("gravitational")` `F_("Photon")=F_("gravitaional")` Let power of light beam be `P` `:. F_("photon")=(P)/(c )` `:. (P)/(c )=10^(-1) P=3.0xx10^(8)xx10^(-1)` `P=3xx10^(7)W` |
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| 296. |
The work function for sodium surface is 2.0eV and that for aluminium surface is 4.2eV. The two metals are illuminated with appropriate radiations so as to cause protoemission. ThenA. Both aluminium and sodium will have the same threshold frequencyB. The threshold frequency of aluminium will be more than that of sodiumC. The threshold frequency of aluminium will be less than that of sodiumD. The threshold wavelength of aluminium will be more than that of sodium |
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Answer» Correct Answer - B The threshold frequency for `Al` must be greater it has higher work function. |
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| 297. |
Positive rays consist ofA. electronsB. neutronsC. positive ionsD. electromagnetic waves |
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Answer» Correct Answer - C Positive rays consist of positive ions. |
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| 298. |
Find the kinetic energy, potential energy and total energy in first and second orbit of hydrogen atom if potential energy in first orbit is taken to be zero. |
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Answer» `E_(1)= -13.60eV K_(1)=-E_(1)=13.60 eV U_(1)2E_(1)=-27.20eV` `E_(2)=(E_(1))/((2)^(2))= -3.40 eV K_(2)=3.40eV` and `U_(2)= - 6.80eV` Now, `U_(1)=0,i.e.,` potential energy has been increased by `27.20 eV` while kinetic energy will remain unchanged. So values of kinetic energy, potential energy and total energy in first orbit are `13.60 eV, 0,13.60` respectively and for second orbit these valeus are `3.40 eV, 20.40 eV` and `23.80 eV`. |
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| 299. |
The energy levels of a hypothetical one electron atom are shown in the figure. (a)Find the ionization potential of this atom. (b) Find the short wavelength limit of the series terminating at n=2 (c ) Find the excitaion potential for the state n=3. (d) find wave number of the photon emitted for the transition n=3 to n=1n=`oo`__________________0eV n=5 __________________ -0.80 eV n=4 ______________ -1.45 eVn=3__________________ -3.08 eV n=2_______________ -5.30 eVn=1 ____________________ -15.6 eV. |
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Answer» (a) Ionization potential `=15.6 V` (b) `lambda_(min)=(12400)/(5.3)=2340Å` (c )`DeltaE_(31) = -3.08-(-15.6)=12.52eV` Therefore, excitation potential for state `n=3` is `12.52` volt. (d) `(1)/(lambda_(31))=(DeltaE_(31))/(12400)A^(-1)=(12.52)/(12400)A^(-1)` (e ) (i) `E_(2)-E_(1)=10.3eV lt 6eV`. (ii) `E_(2)-E_(1)=10.3 eV lt 11 eV`. Hence electron can excite the atoms. So `K_(min)=(11-190.3)=0.7 eV` |
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Four lowest energy levels of `H`-atom are shown in the figure. The number of possoble emission lines would be A. `3`B. `4`C. `5`D. `6` |
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Answer» Correct Answer - D Number of possible emission lines `= (n(n - 1))/(2)` where `n = 4`, Number `= (4(4 - 1))/(2) = 6` |
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