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. |
An alpha particle after passing through a potential difference of `2xx10^(6)` volt falls ona silver foil. The atomic number of silver is 47. Calculate (i) the K.E. of the alpha-particle at the time of falling on the foil. (ii) K.E. of the `alpha`-particle at a distance of `5xx10^(-14)m` from the nucleus, (iii) the shortest distance from the nucleus of silver to which the `alpha`-particle reaches. |
|
Answer» Correct Answer - `6.4xx10^(-13)J,2.1xx10^(-13),3.4xx10^(-14)m` (i) K.E. of a-particlue for Al for K.E.=`q.V=6.4xx10^(-13)J" " (q=2xx1.6xx10^(-19))` (ii) `P.E=(kq_(1)q_(2))/(r)=(9xx10^(9)xx2xx1.6xx10^(19)xx47xx1.6xx10^(-19))/(5xx10^(-14))` `=4.33xx10^(-13)J` Net K.E.=K.l-P.E.`=(6.4-4.3)XX10^(-13)J` `2.07xx10^(-13)J` (iii) K.E=P.E. `6.4xx10^(-13)=(2.16xx10^(-26))/(r)` `r=3.4xx10^(-14)m` |
|
| 452. |
`Be^(7)` captures a K electron into its nucless .What is the mass number and atomic number of the nuclide formed ?A. `3,7`B. `4,8`C. `3,6`D. `4,7` |
|
Answer» Correct Answer - A `_(4)He^(7) + _(-1)e^(0) rarr _(3)LI^(7)` So atomic number `= 3`mass number `= 7` |
|
| 453. |
A bulb emits light of ` 4500 Å `. The bulb is rated as ` 150` watt and ` 9 %` of the energy is emited as light . How many photons are emitted by the bulb per second ? |
|
Answer» We know , Energy one photon `= ( hc)/( lambda) ` ` = ( 6. 6625 xx 10 ^(-34) xx 3xx 10^8 )/( 4500 xx 10^(-10)) J = 4 . 42 xx 10^(-19) J` Energy emitted by bulb`= 15 0 xx 8/( 100 ) J //sec` Energy emitted by bulb = ` 150xx 8/( 100) J//sec` `(watt = J//sec )` `:. n xx 4.44 xx 10 10^(19) = 150 xx 8/( 100) ` ( where (n) is no. of photons ) `:. n= 27 . 2 xx 10 ^(18)`. |
|
| 454. |
Photons of minimum energy `496k,J. mol^(-1)` are needed to an atoms. Calculate the lowest frequency of light that will ionize a sodium atom.A. `1.24xx10^(14)s^(-1)`B. `1.24xx10^(15)s^(-1)`C. `2.48xx10^(15)s^(-1)`D. `2.48xx10^(14)s^(-1)` |
|
Answer» Correct Answer - 2 `Na(g)+496kJ mol^(-1)rarr Na^(+)(g)` For one photon `:E=(496kJ mol^(-1))/(N_(A)mol^(-1))=hv` `v=((496kJ mol^(-1))((10^(3)J)/(1kJ)))/((6.63xx10^(-34)Js)(6.022xx10^(23)mol^(-1)))=1.24xx10^(15)s^(-1)(Hz)` |
|
| 455. |
`Be^(3+)` and a proton are accelerated by the same potential, their de`-` Broglie wavelengths have the ratio ( assume mass of proton `=` mass of neutron `) :`A. `1:2`B. `1:4`C. `1:1`D. `1:3sqrt(3)` |
|
Answer» Correct Answer - 4 `lambda_(Be^(+3))=(h)/(sqrt(2xx3eVm_(Be^(3+))))=(h)/(sqrt(2xx3eV9m_(p)))` `lambda_(p)=(h)/(sqrt(2eVm_(p)))` Hence `(lambda_(Be^(+3)))/(lambda_(p))=sqrt((2eVm_(p))/(2xx3eVxx9m_(p)))=(1)/(3sqrt(3))` |
|
| 456. |
If the ionization energy of `He^(+)` is `19.6xx10^(-18) J` per atom then the energy of `Be^(3+)` ion in the second stationary state is `:`A. `-4.9xx10^(-18)J`B. `-44.1xx10^(-18)J`C. `-11.025xx10^(-18)J`D. `-19.4xx10^(-18)J` |
|
Answer» Correct Answer - 4 `E_(H)` in first orbit `=(-19.6xx10^(-18))/(4)J` `E_(Be^(3+))` in second orbit `=-((19.4xx10^(-18))/(4))xx(16)/(4)=-19.4xx10^(-18)J` |
|
| 457. |
A pronton captures a free electron whose K.E. is zero & from a hydrogen atom of lowest energy-level (n=1) . If a photons is emitted in the process, what will be the wavelenght of radiation ? In which region of electromagetic spectrum, will this radiation fall ? (Inoisation potential of hydrogen =13.6volt, `(h=6.6xx10^(-34)K//S,C=3.0xx10^(8)m//s)` |
|
Answer» Correct Answer - `910Å;U.V` `DeltaE=13.6=(Z^(2))/(n^(2))=13.6 eV` `DeltaE=(hc)/(lambda)=13.6xx1.6xx10^(-19)` `(lambda)=910Å` |
|
| 458. |
A blube emits light of `lambda = 4500 Å`. The bulb is rated as 150 watt and `8%` of the energy is emitted as light. Number of photons emitted by bulb per second is : `["Take" hc = 12400 eVÅ]` |
|
Answer» Energy emitted per second by the bulb `=150xx(8)/(100)J` Energy of 1 photon `=(hc)/(lamda)=(6.626xx10^(-34)xx3xx10^(8))/(4500xx10^(-10))` Let n photons be evolved per second. `thereforenxx4.42xx10^(-19)=150xx(8)/(100)` `n=27.2xx10^(18)` |
|
| 459. |
Find the maximum number of photons number of photons emitted by an H-atom, if atom is excited to atates with principal quantum number four.A. `4`B. `3`C. `2`D. `1` |
|
Answer» Correct Answer - B For maximum number, N=n-1=4-1=3 |
|
| 460. |
In (Q.18)problem, the minimum number of photons emitted by the H-atom isA. `1`B. `2`C. `3`D. `4` |
|
Answer» Correct Answer - A When transition takes place from n=4to n=1. One photon will be emitted. |
|
| 461. |
The kinetic energy of an electron in hydrogen atom is `3.40 eV`.The minimum energy required to ionise the hydrogen atom isA. `-3.40eV`B. `6.40 eV`C. `-6.80 eV`D. `3.40 e V` |
|
Answer» Correct Answer - D Given,`KE=3.40 eV` So, total energy=`-3.40 eV` When H-atom is ionised, its minimum total energy will become zero. So,ionisation energy =`-(-3.40)=3.40eV` |
|
| 462. |
Two H atoms in the ground state collide in elastically. The maximum amount by which their combined kinetic energy is reduced isA. `10.20 eV`B. `20.40 eV`C. `13.6 eV`D. `27.2 eV` |
| Answer» Correct Answer - A | |
| 463. |
The correct ground state electronic configuration of chromium atom isA. `[Ar]3d^(5)4s^(1)`B. `[Ar]3d^(4)4s^(2)`C. `[Ar]3d^(6)4s^(0)`D. `[Ar]3d^(5)4s^(2)` |
|
Answer» Correct Answer - A `Cr(Z = 24)` `[Ar]3d^(6)4s^(1)` This is because `d^(5)` is a more half -filled configuration Reason for the stablity of half-filled and fallty filled ortained are symmetry and excharge energy |
|
| 464. |
The outer electronic configuration of the ground state chromium atom is `3d^(2),4s^(2)` |
|
Answer» `Cr(Z = 24)` `[Ar]3d^(6)4s^(1)` This is because `d^(5)` is a more half -filled configuration Reason for the stablity of half-filled and fallty filled ortained are symmetry and excharge energy |
|
| 465. |
The electronic configuration of chromium (z=24) isA. 2,8,14B. 2,8,8,6C. 2,8,12,2D. 2,8,13,1 |
|
Answer» Correct Answer - D `Cr=24=(1s^2)/2,(2s^2 2p^6)/8, (3s^2 3p^6)/13 , (4s^1)/1, 3d^5` |
|
| 466. |
Calculate the ratio of the frequencies of the long wavelength limits of the Balmer and Lyman series of hydrogen.A. `27:5`B. `5:27`C. `4:1`D. `1:4` |
| Answer» Correct Answer - A | |
| 467. |
Which of the following are isotones ?A. `._(18)Ar^(40)`B. `._(20)Ca^(42)`C. `._(21)Se^(43)`D. `._(21)^(Se)^(41)` |
|
Answer» Correct Answer - A::B::C Isotopes: Electrons that contain same atomic number |
|
| 468. |
Assertion(A): P-orbital can accomadate 6 electrons Reason (R): No two electrons in atom can have same set of four quantum numbersA. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
| Answer» Correct Answer - D | |
| 469. |
If the uncertainties in the measurement of position and momentum of an electron are equal, calculate the uncertainty in measuring the velocity.A. ` 1/m sqrt h/(pi)`B. ` sqrt h/(pi)`C. `1/( 22m) sqrt h/(pi)`D. `sqrt h/(2pi)` |
|
Answer» Correct Answer - C According to Heisenbegr uncetainty principle ` Delta p. Delta x le h/( 4 pi)` or ` Delta u . Delta x. m le h/( 4 pi)` ` or ( m delta u) ^2 le h/( 4 pi) (:. Delta x =Delta p) ` or ` Delta u = 1/( 2m) sqrt h/(pi)`. |
|
| 470. |
The correct ground state electronic configuration of chromium atom(Z=24) is :A. `[Ar] 3d^5 4s^1`B. `[Ar] 3d^4 4s^2`C. `[Ar] 3d^6 4s^0`D. `[Ar] 4d^5 4s^1` |
|
Answer» Correct Answer - A 3d subshell filled with 5 electrons (half-filled) is more stable than that filled with 4 electrons. 1,4s electron jump into 3d subshell for more stability |
|
| 471. |
Which is not an atomic orbital?A. 2dB. 5pC. 3pD. 4d |
|
Answer» Correct Answer - A 2d orbital is not possible |
|
| 472. |
For a certain atom, there are energy levels A,B,C corresponds to energy values `E_(A)ltE_(B)ltE_(C)`. Choose the correct option if `lambda_(1),lambda_(2),lambda_(3) `are the wavelength of rediations corresponding to the transition from C to B,B to A and C to A respectively.A. `lambda_(3)=lambda_(1)+lambda_(2)`B. `lambda_(3)=(lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2))`C. `lambda_(1)+lambda_(2)+lambda_(3)=0`D. `3lambda_(2)=lambda_(3)+2lambda_(2)` |
|
Answer» Correct Answer - B `E_(B)-E_(A)=(hc)/(lambda^(2))Rightarrow E_(0)-E_(B)(hc)/(lambda2)` or `1/lambda_(3)=1/lambda_(1)+1/lambda_(2)Rightarrowlambda_(3)=(lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2))` |
|
| 473. |
Assertion (A) : On increasing the internsity of incident radiation, the photoelectrons eject and then KE increases Reason (R ) : Greater the intensity means greater the energy which in turn means greater the frequency of the radiation.A. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If both (A) and (R ) are incorrect |
| Answer» Correct Answer - Both are false | |
| 474. |
The quantum number which cannot say any thing about an orbital isA. nB. lC. mD. s |
|
Answer» Correct Answer - D spin quantum number represent `e^(-)` spin onlu |
|
| 475. |
The uncertainties in the velocities of two particles, A and B are 0.05 and 0.02 `ms^(-1)`, respectively. The mass of B is five times to that of the mass A. What is the ratio of uncertainties `((Delta_(X_A))/(Delta_(X_B)))` in their positionsA. 2B. 0.25C. 4D. 1 |
|
Answer» Correct Answer - A According to Heisenberg `Deltax xx mDeltav = h/(4pi)` For particle A `Deltax=Deltax_A , m=m , Deltav=0.05` So, `Deltax_Axxmxx0.05=h/(4pi)` .....(i) For particle B `Deltax=Deltax_B , m=5m , Deltav=0.02` So, `Deltax_B xx 5 m xx 0.02 =h/(4pi)` .....(ii) Eq. (i) (ii) , we get `(Deltax_A)/(Deltax_B)=2` |
|
| 476. |
Calculate the energy required to excite an electron in hydrogen atom from the ground state to the next higher state, if the ionsation energy for the hydrogen atom is `13.6 eV`.A. `3.4 eV`B. `10.2 eV`C. `12.1 eV`D. `1.3 eV` |
| Answer» Correct Answer - B | |
| 477. |
Any radiation in the ultraviolet region of hydrogen spectrum is able to eject photoelectrons from a metal. What should be the maximum value of threshold frequency for the metal?A. `3.288times10^(15)Hz`B. `2.466times10^(15)Hz`C. `4.594times10^(14)Hz`D. `8.220times10^(14)Hz` |
|
Answer» Correct Answer - B For minimum frequency in Lyman series of H -atom, transition n=2to n=1 takes place. Now , hv=`DeltaE=E_(2)-E_(1)` or hv=`[(-(13.6)/(2^(2)))-(-(13.6)/(1^(2)))]eV` `v=(10.2times1.6times10^(-19))/(6.63times10^(-34))` ` therefore v=2.466times 10^(15)Hz` If the threshold frequency of metal is v or less, photoelectron will come out. |
|
| 478. |
Assertion (A) : An orbital cannot have more than two electron Reason (R ) : The two electrons is an orbital create opposite magnetic fieldA. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
|
Answer» Correct Answer - C The two lobes are oriented along x-axis. |
|
| 479. |
Statement-1 : The groundstate configuration of Cr is [Ar] `3d^(5) 4s^(1)` Statement-2 : The energy of atom is lesser in `3d^(5)4s^(1)` configuration compared to `3d^(4)4s^(2)` configuration.A. Both (A) and (R) are true and (R) is the correct explanation of (A)B. Both (A) and (R) are true and (R) is not the correct explanation of (A)C. (A) is true but (R) is falseD. (A) is false but (R) is true |
|
Answer» Correct Answer - A `psi^(2)` is the probability function `psi` is the wave function. |
|
| 480. |
The number of waves in the fourth bohr orbit of hydrogen isA. 3B. 4C. 9D. 12 |
|
Answer» Correct Answer - B |
|
| 481. |
The wavelngth fo a spectrl line for an electronic transition is inversely related to :A. Velocity of electron undergoing transitionB. Number of electrons undergoing transactionC. The difference in energy levels involved in the transitionD. None of these |
|
Answer» Correct Answer - C `DeltaE = (hc)/(lambda)` |
|
| 482. |
Satement-1: Emitted radiation will fall in visible range when an electron jumps from higher level ot `n=2 "in"Li^(+2) "ion".` Statement-2: Balmer series radiations belong to visible range in all H-atoms. |
|
Answer» Correct Answer - d |
|
| 483. |
Find the wavelength of the emitted radiation,if electron in hydrogen atom jumps from third orbit to second orbit.A. `lambda=(36)/(5R)`B. `lambda=(5R)/(36)`C. `lambda=5/R`D. `lambda=R/6` |
| Answer» Correct Answer - A | |
| 484. |
The correct set of quantum number for the unpaired electron of chlorine atom is `{:(n,l,m_(1)):}` " " `{:(n,l,m_(1)):}`A. `{:(2,1,0):}`B. `{:(2,1,1):}`C. `{:(,1,1):}`D. `{:(3,0,0):}` |
| Answer» Correct Answer - C | |
| 485. |
Statement-1: The orbital angular momentum of d-electron in orbitals is `sqrt(6)(h)/(2pi)` Statement-2: Angular momentum of `e^(-)` in orbit is `mvr =(nh)/(2pi)`A. If both the statement are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-2B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-2C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
|
Answer» Correct Answer - B `L = sqrt(l(l+1)) .(h)/(2pi) rArr mvr = (nh)/(2pi)` |
|
| 486. |
The wavelngth fo a spectrl line for an electronic transition is inversely related to :A. the number of electrons undergoing the transitionB. the nuclear charge of the atomC. the difference in the energy of the energy levels involved in the transitionD. the velocity of the electron undergoing the transition. |
|
Answer» Correct Answer - C ( c) Difference in the energy of the energy levels involved in the transition. |
|
| 487. |
When an electron jumps from higher orbit to the second orbit in `He^(+)` ion,the radiation emitted out will be in `(R=1.09times10^(7)m^(-1))`A. ultraviolet regionB. visible regionC. infrared regionD. X-ray region |
| Answer» Correct Answer - B | |
| 488. |
The quantum numbers `n = 3, l = 1, m = +1` and `s = +1//2` represent the unpaired electron present inA. Sodium atomB. Aluminium atomC. Fluorine atomD. Potassium atom |
|
Answer» Correct Answer - B 3p orbital |
|
| 489. |
The wavelngth fo a spectrl line for an electronic transition is inversely related to :A. number of electrons undergoing transitionB. the nuclear charge of the atomC. the velocity of an electron undergoing transitionD. the difference in the energy levels involved in the transition |
|
Answer» Correct Answer - D |
|
| 490. |
When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as `E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2))` Where, `m_(e)=` rest mass of electron `DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV` per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum `bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1)` Where `R_(H) = 1.1 xx 10^(7)m^(-1)` (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of `n_(1) = 1,2,3,4,5` respectively and `n_(2) =oo` for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be `.^(n)C_(2)`. Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass `(mu)`. `(1)/(mu) = (1)/(m_(N))+ (1)/(m_(e))` Here, `m_(N)=` mass of nucleus `m_(e)=` mass of electron Answer the following questions The ratio of the wavelength of the first line to that of second line of Paschen series of H-atom isA. `256:175`B. `175:256`C. `15:16`D. `16:15` |
|
Answer» Correct Answer - A `bar(v) = (1)/(lambda) = R_(H)z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` For 1st line: `n_(1) = 3, n_(2) = 4` For 2nd line: `n_(1) = 3, n_(2) = 4` |
|
| 491. |
When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as `E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2))` Where, `m_(e)=` rest mass of electron `DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV` per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum `bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1)` Where `R_(H) = 1.1 xx 10^(7)m^(-1)` (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of `n_(1) = 1,2,3,4,5` respectively and `n_(2) =oo` for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be `.^(n)C_(2)`. Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass `(mu)`. `(1)/(mu) = (1)/(m_(N))+ (1)/(m_(e))` Here, `m_(N)=` mass of nucleus `m_(e)=` mass of electron Answer the following questions The emission spectrum of `He^(+)` involves transition of electron from `n_(2) rarr n_(1)` such that `n_(2)+n_(1) = 8` and `n_(2) -n_(1) = 4`. what whill be the total number of lines in the spectrum?A. 10B. 15C. 20D. 21 |
|
Answer» Correct Answer - A `n_(2) = 6, n_(1) = 2`, no of spectral lines `=Deltan ((Delta n+1))/(2)` |
|
| 492. |
When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as `E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2))` Where, `m_(e)=` rest mass of electron `DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV` per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum `bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1)` Where `R_(H) = 1.1 xx 10^(7)m^(-1)` (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of `n_(1) = 1,2,3,4,5` respectively and `n_(2) =oo` for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be `.^(n)C_(2)`. Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass `(mu)`. `(1)/(mu) = (1)/(m_(N))+ (1)/(m_(e))` Here, `m_(N)=` mass of nucleus `m_(e)=` mass of electron Answer the following questions In which of the following region the spectrum of `He^(+)` will be observed in above transition?A. UltravioletB. VisibleC. InfraredD. far infrared |
|
Answer» Correct Answer - B `n_(1) = 2` i.e. balmer series |
|
| 493. |
`A` near ultra violet photon of wavelength `300nm` is absorbed by a gas and then emitted as two photons. One photons is of red light with wavelength `760nm`. What would be the wave length of the second photon? |
|
Answer» It may noted that energy of photon which absorbed is emitted as sum of the energy of two photons. Energy absorbed `hv=(hc)/(lambda)` According to available information, `(hc)/(lambda)=(hc)/(lambda_(1))+(hc)/(lambda_(2)),(1)/(lambda)=(1)/(lambda_(1))+(1)/(lambda_(2)),(1)/(lambda_(2))=[(1)/(lambda)-(1)/(lambda_(1))]` Now, `lambda = 300 nm , lambda_(1)=760nm,lambda_(2)` can be calculated as : `(1)/(lambda_(2))=[(1)/(300)-(1)/(760)]=(760-300)/(300xx760)(nm^(-1))` `(1)/(lambda_(2))=(460)/(760xx300)(nm^(-1))` `lambda_(2)=(760)/(460)(nm)=496nm` |
|
| 494. |
In an atom when an electron jumps from higher energy level to lower energy level it amits energy in form of elecromagnetic radiations. When these elecromagnetic radiations are passed through a prism and received on a photographic film some lines are observed on that film and those lines are called spectral lines. For hydrogen like species when jump takes from any excited state to ground state `(n=1)`, line produced in that case is called a Lyman series line. If transition occurs from 3rd or above level to second level then corresponding lines produced are called Balmer lines. Similarly, for next level it is called Paschen series line. If there are 3 atoms of a hydrogen like species one in 2nd one in 3rd and one in 4th excited state, then how many maximum total different Balmer and Paschen lines can be produced ?A. 2B. 3C. 4D. 5 |
|
Answer» Correct Answer - d |
|
| 495. |
In an atom when an electron jumps from higher energy level to lower energy level it amits energy in form of elecromagnetic radiations. When these elecromagnetic radiations are passed through a prism and received on a photographic film some lines are observed on that film and those lines are called spectral lines. For hydrogen like species when jump takes from any excited state to ground state `(n=1)`, line produced in that case is called a Lyman series line. If transition occurs from 3rd or above level to second level then corresponding lines produced are called Balmer lines. Similarly, for next level it is called Paschen series line. The wavelength of photon corresponding ot second least energy of Lyman series in H-atom is :A. 102 nmB. 121 nmC. 91 nmD. 486 nm |
|
Answer» Correct Answer - a |
|
| 496. |
Wavelengths of different radiations are given below: (A) 300nm (B) `300 mu m` (C) 3nm (`30A^(@)` Arrange these radiations in the increasing order of their energies.A. `B gt A gt C gt D`B. `B lt A lt C lt D`C. `B lt A lt C =D`D. `B gt A gt C = D` |
|
Answer» Correct Answer - C `Ex (1)/(lambda)` |
|
| 497. |
Which of the following will not show deflection from the path on passing through an electric field? Proton,cathode rays, electron,neutron.A. protonB. cathode raysC. electronD. neutron |
|
Answer» Correct Answer - D neutrons are natural |
|
| 498. |
`{:(,"Column-I",,"Column-II",),((A),"Cathode rays",(p),"Helium nuclei",),((B),"dumb-bell",(q),"Uncertainty principle",),((C),"Alpha particles",(r),"Electromagnetic radiation" ,),((D),"Moseley",(s),"p-orbital",),((E),"Heisenberg",(t),"Atomic number",),((F),"X-ray",(u),"electrons",):}` |
| Answer» Correct Answer - `(A) rarr u ; (B) rarr s ; (C ) rarr p ; (D) rarr t ; (E) rarr q ; (F) rarr r` | |
| 499. |
Cathode rays have:A. mass onlyB. charge onlyC. no mass and no chargeD. mass and charge both |
| Answer» Correct Answer - D | |
| 500. |
What happens to the cathode rays under a strong magnetic field or an electric field ? What is conclusion made from this ? |
|
Answer» Electric field-move towards positive plate, negatively charged. Magnetic field-deflect perpendicular to the applied magnetic field. |
|