464 + Interview Questions in ATOMS IN GENERAL AWARENESS Page 2 InterviewSolution

51.	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)c)/(2pih)`C. ` (e^(2)h)/(2pihc)`D. `(e^(2))/(2epsi_(0)hc)`
Answer» Correct Answer - D Speed of electron in first orbit of hydrogen atom is given by `" " v=(e^(2))/(2epsi_(0)h)` ` :. " " (v)/(c)=(e^(2))/(2epsi_(0)hc)`

Discussion

52.	the wavelength limit present in the pfund series is `(R=1-097xx10^7m^-1)`A. 1572 nmB. 1898 nmC. 2278 nmD. 2535 nm
Answer» Correct Answer - C The wavelength for Pfund series is given by `(1)/(lambda)=R[1/(5^2)-1/(n^2)]` for series limit, `n=prop` therefore lambda `=(25)/(R)=(25)/(1.097xx10^7)=2278nm`.

Discussion

53.	An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be proportional toA. 1/mB. `1//v^(2)`C. 1/zeD. `v^(2)`
Answer» Correct Answer - A Distamnce of closet apporach , ` r_(0)=(Ze^(2))/(2mv^(2)piepsi_(0))` ` rArr " " r_(0)prop (1)/(m)`

Discussion

54.	The shoretest wavelength in Lyman series is 91.2 nm. The longest wavelength of the series isA. 121.6 nmB. 182.4 nmC. 243. 4nmD. 364.8 nm
Answer» Correct Answer - A The wavelength `(lambda)` of lines is given by `" " (1)/(lambda)=R((1)/(1^(2))-(1)/(n^(2)))` For Lyman seris, the shortest wavelength is for `n=oo` and longest is for n=2 `:. " " (1)/(lambda_(s))=R((1)/(1^(2)))" "....(i)` `" " (1)/(lambda_(L))=R((1)/(1^(2))-(1)/(2^(2)))=(3)/(4)R" "....(ii)` On dividing Eq. (ii) by Eq. (ii), we get `" " (lambda_(L))/(lambda_(s))=(4)/(3)` Given `" " lambda_(s)=91.2 nm rArr lambda_(L)=91.2xx(4)/(3)=121.6 nm`

Discussion

55.	What is the energy required to remove an electron from second orbit of hydrogen atom ?
Answer» Energy of second orbit `E_2= -13.6/2^2=-3.4` eV `therefore` Ionisation energy =0-(-3.4) =3.4 eV

Discussion

56.	Assertion Energy `E_(1)` is required to remove first electron from helium atom and energy `E_(2)` is to required to remove the second electron . Them `E_(1)ltE_(2).` Reason Ionisation energy of single electron of `He^(+)` is 54.4 eV.A. If both Assertion and Reason ar true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not correct explanation of AssertionC. If Assertion is true by Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - B `E_(H)=13.6eV` (inoisation energy ) `E_(He^(+))=13.6(Z)^(2)=(13.6)(Z)^(2)=54.4eV`

Discussion

57.	Energy of 24. eV is required to remove one of the electron from a neutral helium atom. The energy ( in e V ) reauired to remove both the electorns from a netural helium aotm isA. 38.2B. 49.2C. 51.8D. 78.6
Answer» Correct Answer - D After reamining one electron from helium atom it will become hydrogen like atom `:. E=24+(13.6) (2)^(2) =78.6 eV`

Discussion

58.	Determine the radius of the first orbit of the hydrogen atom. What would be the velocity and frequency of the electron in the first orbit ? Given : `h=6.62xx10^(-34) J s, m =9.1xx10^(-31) kg, e=1.6xx10^(-19) C, k = 9xx10^(9)m^(2)C.`
Answer» Given `h=6.62xx10^(-34) J-s,` `m=9.1xx10^(-31) kg` `e=1.6xx10^(-19)C,` `k=9xx10^(9)Nm^(2)C^(-2), n=1` i) `r_(1)=(n^(2)h^(2))/(4pi^(2)" mke"^(2))` `=((1)^(2)xx(6.62xx10^(-34))^(2))/(4xx(3.14)^(2)xx9.1xx10^(-31)xx9xx10^(9)(1.6xx10^(-19))^(2))` `therefore r_(1)=0.529 overset(0)A cong 0.53 overset(0)A` ii) `V=sqrt((KZ)/(mr))xxe` `=sqrt((9xx10^(9)xx1)/(9.1xx10^(-31)xx0.53)xx1.6xx10^(-19)` `therefore V=2.19xx10^(6) ms^(-1)` iii) `v=(KZe^(2))/(nhr)` `=(9xx10^(9)xx1xx(1.6xx10^(-19))^(2))/(1xx6.62xx10^(-34)xx0.529xx10^(-10))` `therefore v=6.6xx10^(15)Hz.`

Discussion

59.	The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is `5.30xx10^(-11)m` and `22xx10^(6)m s^(-1)` respectively, then the orbital peroid of this electron in second excited state will beA. `1.21xx10^(-14)s`B. `1.21xx10^(-12)s`C. `1.21xx10^(-10)s`D. `1.21xx10^(-15)s`
Answer» Correct Answer - D Here, `r_1=5.30xx10^(-11)m,v_1=2.2xx10^6ms^(-1)` In the second excited state, `r_n=n^2r_1,v_n=(v_1)/(n)` `therefore r_2=4r_1=4xx5.30xx10^(-11)m=2.12xx10^(-10)m` and `v_2=(v_1)/(2)=(2.2xx10^(6))/(2)ms^(-1)=1.1xx10^(6)ms^(-1)`

Discussion

60.	Statement-1 : In X-rays diffraction, the wavelength of the scattered X-ray is same as that of the incident wavelength . Statement-2 : The compton wavelength for scattering of X-rays by bound atoms is very smallA. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - D

Discussion

61.	The radius of the first electron orbit of a hydrogen atom is `5.3 xx 10^(-11)m`. What is the radius of the second orbit ?
Answer» `r_(n)prop n^(2)` `(r_(2))/(r_(1))=(2^(2))/(1^(2))=(4)/(1) rArr r_(2)=4r_(1)` `therefore r_(2)=4xx5.3xx10^(-11)=2.12xx10^(-10)m`.

Discussion

62.	Statement-1 : The Bohr model of the hydrogen atom does not explain the fine structure of spectral lines. and Statement-2 : The Bohr model does not take into account the spin of the electronA. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False, Statement-2 is True
Answer» Correct Answer - A

Discussion

63.	How is impact parameter related to the scattering angle?
Answer» 1) Impact parameter (b) : Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom. 2) Scattering angle `(theta)` : The scattering angle `(theta)` is the angle between the asymtotic direction of approach of the `alpha` - particle and the asymtotic direction in which it receeds. 3) The relation between b and `theta` is b `=(1)/(4 pi epsi_(0))(Ze^(2))/(E)cot""(theta)/(2)" where "E= K.E" of "alpha-" particle "=(1)/(2) mv^(2)`.

Discussion

64.	Distinguish between excitation potential and ionization potential.
Answer» Excitation Potential : 1) When the electron jumps from lower orbit to higher orbit by absorbing energy is called excited electron and the process is known as excitation. The minimum accelerating potential which provides an electron energy sufficient to jump from the inner most orbit(ground state) to one of the outer orbits is called excitation potential or resonance potential. 2) a) For example, in case of hydrogen atom, `E_(1)=-13.6 eV.` `E_(2)=-3.4 eV, E_(3)=-1.51 eV and" soon, "E_(oo)=0` `therefore` Energy required to raise an electron from ground state (n=1) to first excited state (n=2) is `E=E_(2)-E_(1)=-3.4-(-13.6)=10.2 eV`. The corresponding excitation potential = 10.2 Volt. b) Similarly, energy required to raise an electron from ground state (n=1) to second excited state (n=3) is `E=E_(3)-E_(1)=-1.51-(-13.6)=-1.51+13.6=12.09 eV` The corresponding excitation potential =12.09 Volt and so on. 3) The excitation potential of an atom is not one. It can have many values, depending on the state to which the atom is excited. Ionisation potential : 1) The energy supplied is so large that it can remove an electron from the outer most orbit of an atom, the process is called Ionisation. Thus ionisation is the phenomenon of removal of an electron from the outer most orbit of an atom. 2) The minimum accelerating potential which would provide an electron energy sufficient just to remove it from the atom is called Ionisation potential. 3) For example, total energy of electron in ground state of hydrogen atom, +13.6 eV energy is required. `therefore` Ionisation energy of hydrogen atom = 13.6 eV. Ionisation potential of hydrogen atom = 13.6 Volts 4) The general expression for ionisation potential of an atom is `V=(13.6 Z^(2))/(n^(2))` Volt. where Z is the charge number of the atom and n is number of orbit from which electron is to be removed. 5) For a given element, ionisation potential is fixed, but for different elements, ionisation potentials are different.

Discussion

65.	the excitation energy of lyman last lines isA. the same as ionisation energyB. the same as the last absorption line in lyman seriesC. both (a) and (b)D. different from (a) and (b)
Answer» Correct Answer - C

Discussion

66.	the number of de broglie wavelength contained in the second bohr orbit of hydrogen atom isA. 1B. 2C. 3D. 4
Answer» Correct Answer - B `because 2pir_n=nlambda` `therefore n=2,2pir_2=2lambda=2xx`de broglie wavelength.

Discussion

67.	The de-Broglie wavelength of an electron in the first Bohr orbit isA. equal to one- fourth the circumference of the first orbitB. equal to half the circumference of first orbitC. equal to twice the circumference of first orbit.D. equal to the circumference of the first orbit.
Answer» Correct Answer - D Angular momentum `=(nh)/(2pi)` `rArr` moment of momentum `=pxxr_n` `rArrpxxr_n=(nh)/(2pi)rArrr(h)/(lambda)r_nrArrlambda =(2pir_n)/(n)` For `1^st` orbit, `n=1,lambda =2pir_1` `rArr =` circumference of `1^st` orbit.

Discussion

68.	Half life of a radioactive substance A is 4 day.s The probability that a nucleus will decay in two half lives is(1) 1/4(2) 3/4(3) 1/16(4) 1
Answer» Correct option : (2) 3/4 Explanation : After two half lives (1/4)^th fraction of nuclei will remain unchanged. Hence (3/4)^th fraction will decay. Hence the probability that a nucleus decays in two half lives is 3/4.

68.

Half life of a radioactive substance A is 4 day.s The probability that a nucleus will decay in two half lives is(1) 1/4(2) 3/4(3) 1/16(4) 1

Answer»

Correct option : (2) 3/4

Explanation :

After two half lives (1/4)^th fraction of nuclei will remain unchanged. Hence (3/4)^th fraction will decay. Hence the probability that a nucleus decays in two half lives is 3/4.

Discussion

69.	The ionization enegry of the electron in the hydrogen atom in its ground state is `13.6 ev`. The atoms are excited to higher energy levels to emit radiations of `6` wavelengths. Maximum wavelength of emitted radiation corresponds to the transition betweenA. n=3 and n=2 statesB. n=3 and n=1 stateC. n=2 and n=1 stateD. n=4 and n=3 states
Answer» Correct Answer - D We know that, the wavelength of emited radiation is givne by `(1)/(lambda)R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=RxxK"where,"K=(1)/(n_(1)^(2))-(1)/(n_(2)^(2))` For `lambda_("max")` K should be minimum. Which corresponds to ` n_(1)=3,n_(2)=4` So, transition takes place between n=4 and n=3 states.

Discussion

70.	The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of photon with the most enegy ? A. IIIB. IVC. ID. II
Answer» Correct Answer - A Energy of emitted radition ,E= Rhc `((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `rArrR_(4to3)=Rhc((1)/(3^(2))-(1)/(4^(2)))=Rhc((7)/(9xx16))=0.05Rhc` `E_(4to2)=Rhc((1)/(2^(2))-(1)/(4^(2)))=Rhc((3)/(16))=0.2Rhc` `E_(2to1)=Rhc[(1)/((1)^(2))-(1)/((2)^(2))]=Rhc((3)/(4))=0.75Rhc` and `E_(1to4)=Rhc[(1)/((4)^(2))(1)/((1)^(2))]=-(8)/(9)Rhc=-0.9Rhc` Thus, transition III gives largest amout of energy

Discussion

71.	The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of photon with the most enegy ? A. IB. IIC. IIID. IV
Answer» Correct Answer - C `I^st` transition is showing absorption of a photon. From rest of three transitions , III is having maximum energy from level `n=2` to `n=1` `Delta Eprop((1)/(n_1^2)-(1)/(n_2^2))`

Discussion

72.	The ionization enegry of the electron in the hydrogen atom in its ground state is `13.6 ev`. The atoms are excited to higher energy levels to emit radiations of `6` wavelengths. Maximum wavelength of emitted radiation corresponds to the transition betweenA. n=3 to n=1 statesB. n=4 to n=3 statesC. n=3 to n=2 statesD. n=2 to n=1 states
Answer» Correct Answer - B Number of wavelelngth =`(n(n-b))/(2)` where, n=number of orbit from which transition place `:. " " 6=(n(n-1))/(2)rArrn=4` `:.` The wavelength of emitted radiations will be maximum for transition n=4 to n=3

Discussion

73.	The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-11 m. What are the radii of the n = 2 and m = 3 orbits?
Answer» Given r₁ = 5.3 × 10^-11m n₁ = 1, n₂ = 2, r₂ = ? n₃ = 3, r₃ = ? Since r ∝ n² ∴ r₁ ∝ n₁² And r₂ ∝ n₂² ∴ \(\frac{r_2}{r_1}=\frac{n^2_2}{n_1^2}\) = \(\frac{2^2}{1}=\frac{4}{1}\) or r₂ = 4 × r₁ = 4 × 5.3 × 10^-11 or e₂ = 2.12 × 10^-10m Also r₃ ∝ n₃² \(\frac{r_3}{r_1}=\frac{n^2_3}{n_1^2}\)= \(\frac{3^2}{1}=\frac{9}{1}\) or r₃ = 9r₁ = 9 × 5.3 × 10^-11 or r₃ = 4.77 × 10^-10m.

73.

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-11 m. What are the radii of the n = 2 and m = 3 orbits?

Answer»

Given r₁ = 5.3 × 10^-11m
n₁ = 1, n₂ = 2, r₂ = ?
n₃ = 3, r₃ = ?
Since r ∝ n²
∴ r₁ ∝ n₁²
And r₂ ∝ n₂²

∴ \(\frac{r_2}{r_1}=\frac{n^2_2}{n_1^2}\) = \(\frac{2^2}{1}=\frac{4}{1}\)

or r₂ = 4 × r₁ = 4 × 5.3 × 10^-11
or e₂ = 2.12 × 10^-10m
Also r₃ ∝ n₃²

\(\frac{r_3}{r_1}=\frac{n^2_3}{n_1^2}\)= \(\frac{3^2}{1}=\frac{9}{1}\)

or r₃ = 9r₁ = 9 × 5.3 × 10^-11
or r₃ = 4.77 × 10^-10m.

Discussion

74.	If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is(a) 10.2 eV(b) 0 eV(c) 3.4 eV(d) 6.8 eV
Answer» (c) 3.4 eV Explanation: E_n = \(\frac{13.6}{n^2}\) ⇒ E₂ = \(\frac{13.6}{(2)^2}\) = 3.4 eV

74.

If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is(a) 10.2 eV(b) 0 eV(c) 3.4 eV(d) 6.8 eV

Answer»

(c) 3.4 eV

Explanation:

E_n = \(\frac{13.6}{n^2}\)

⇒ E₂ = \(\frac{13.6}{(2)^2}\)

= 3.4 eV

Discussion

75.	In Bohr model of the hydrogen atom, let R,v and E represent the radius of the orbit, speed of the electron and the total energy respectively. Which of the following quantities are directly proportional to the quantum number n?A. VRB. REC. `R//E`D. none of these
Answer» Correct Answer - A `Rpropn^2`,`Vpropn^(-1)`,`EpropZ^2`

Discussion

76.	Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de Broglie wavelength λ of that electron as(a) (0.529) nλ (b) \(\sqrt{nλ}\)(c) (13.6)λ (d) nλ
Answer» Correct option is (d) nλ

Discussion

77.	When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the forms of electromagnetic radiation. Why cannot it be emitted as other forms of energy?
Answer» The transition of an electron from a higher energy to a lower energy level can appear in the form of electromagnetic radiation because electrons interact only electromagnetically.

Discussion

78.	In the Bohr model of hydrogen atom, the electron is pictured to rotate in a circular orbit of radius `5xx 10^-11 m`, at a speed `2.2 xx 10^6` m/ s. What is the current associated with electron motion?
Answer» Frequecncy of revolution of electron can be wirtten as, `f=(v)/(2pir)=(2.2xx10^(6))/(2pi(5xx10^(-11)))=7.1xx10^(15)Hz` `:.` Currently associated, `I=q//t=qf=(1.6xx10^(-19))(7xx10^(15))( :. (1)/(t)=f)` `=11.2xx10^(-4)A=1.12mA`

Discussion

79.	The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because (a) of the electrons not being subject to a central force. (b) of the electrons colliding with each other (c) of screening effects (d) the force between the nucleus and an electron will no longer be given by Coulomb’s law.
Answer» (a) of the electrons not being subject to a central force. option (a) of the electrons not being subject to a central force. The electrostatic force of attraction between electron and nucleus is a central force which provides necessary centripetal force for the circular motion of the electron. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.

79.

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because (a) of the electrons not being subject to a central force. (b) of the electrons colliding with each other (c) of screening effects (d) the force between the nucleus and an electron will no longer be given by Coulomb’s law.

Answer»

(a) of the electrons not being subject to a central force.

option (a) of the electrons not being subject to a central force.

The electrostatic force of attraction between electron and nucleus is a central force which provides necessary centripetal force for the circular motion of the electron.

The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.

Discussion

80.	Imagine removing one electron from `He^(4)` and `He^(3)`. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why ?
Answer» On removing one electron from `He^(4)` and `He^(3)`, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass. Also after removing one electron from `He^(4)` and `He^(3)` atoms contain one electron and are hydrogen like atoms.

Discussion

81.	Imagine removing one electron from He4 and He3. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.
Answer» Because both the nuclei are very heavy as compared to electron mass.

Discussion

82.	An ionised H-molecules consists of an electron and wo protons. The protons are seperated by a small distance of the order of angstrom. In the ground state,A. the electrons would not move iin circular orbitsB. the energy would be `(2)^(4)` times that of a H-atomC. the electrons, orbits would go around the protonsD. the molecule will soon decay in a proton and a H-atom
Answer» Correct Answer - A::C The protons are separated by a small distance of the orderf of angstrom. In the ground the electron would not move in circular orbits the electrons, orbit would go around protons.

Discussion

83.	An ionised H-molecules consists of an electron and wo protons. The protons are seperated by a small distance of the order of angstrom. In the ground state,A. the electron would not move in circular orbitsB. the energy would be `(2)^4` times that of a H-atomC. the molecule will soon decay in to a proton and a H-atomD. none of these
Answer» Correct Answer - A

Discussion

84.	The simple Bohr modle is not applicable to He4 atom because(a) He4 is an inert gas.(b) He4 has neutrons in the nucleus.(c) He4 has one more electron.(d) electrons are not subject to central forces.
Answer» (c), (d) (c) He⁴ has one more electron. (d) electrons are not subject to central forces.

84.

The simple Bohr modle is not applicable to He4 atom because(a) He4 is an inert gas.(b) He4 has neutrons in the nucleus.(c) He4 has one more electron.(d) electrons are not subject to central forces.

Answer»

(c), (d)

(c) He⁴ has one more electron.

(d) electrons are not subject to central forces.

Discussion

85.	An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state, (a) the electron would not move in circular orbits. (b) the energy would be (2)4 times that of a H-atom. (c) the electrons, orbit would go arround the protons. (d) the molecule will soon decay in a proton and a H-atom.
Answer» (a) the electron would not move in circular orbits. (c) the electrons, orbit would go arround the protons.

85.

An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state, (a) the electron would not move in circular orbits. (b) the energy would be (2)4 times that of a H-atom. (c) the electrons, orbit would go arround the protons. (d) the molecule will soon decay in a proton and a H-atom.

Answer»

(a) the electron would not move in circular orbits.

(c) the electrons, orbit would go arround the protons.

Discussion

86.	The Bohr model for the spectra of a H-atom(a) will not be applicable to hydrogen in the molecular from.(b) will not be applicable as it is for a He-atom.(c) is valid only at room temperature.(d) predicts continuous as well as discrete spectral lines.
Answer» (a), (b) (a) will not be applicable to hydrogen in the molecular from. (b) will not be applicable as it is for a He-atom.

86.

The Bohr model for the spectra of a H-atom(a) will not be applicable to hydrogen in the molecular from.(b) will not be applicable as it is for a He-atom.(c) is valid only at room temperature.(d) predicts continuous as well as discrete spectral lines.

Answer»

(a), (b)

(a) will not be applicable to hydrogen in the molecular from.
(b) will not be applicable as it is for a He-atom.

Discussion

87.	Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom, (a) because of energy conservation. (b) without simultaneously releasing energy in the from of radiation. (c) because of momentum conservation. (d) because of angular momentum conservation.
Answer» (a), (b) (a) because of energy conservation. (b) without simultaneously releasing energy in the from of radiation.

87.

Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom, (a) because of energy conservation. (b) without simultaneously releasing energy in the from of radiation. (c) because of momentum conservation. (d) because of angular momentum conservation.

Answer»

(a), (b)

(a) because of energy conservation.
(b) without simultaneously releasing energy in the from of radiation.

Discussion

88.	hydrogen atom emits light when it changes from `n=5` energy level to `n=2` energy level. Which colour of light would the atom emit ?A. redB. yellowC. greenD. voilet.
Answer» Correct Answer - D `(1)/(lambda)=R[1/(2^2)-1/(5^2)]=R[1/(4)-1/(25)]=(21R)/(100)` `therefore lambda=(100)/(21R)=(100)/(21xx1.1xx10^7)=(100)/(23.1)xx10^-7m=4346Å` This corresponds to voilet.

Discussion

89.	The kinetic energy of the orbital electron in a hydrogen atom in the excited state corresponding to n = 2 is (A) 3.4 eV (B) 6.8 eV (C) 13.6 eV(D) 27.2 eV.
Answer» Correct option is (A) 3.4 eV

Discussion

90.	when an atomic gas or vapour is excited at low pressure, by passing an electric current through it thenA. emission spectrum is observedB. absorption spectrum is observedC. band spectrum is observedD. both a and c
Answer» Correct Answer - A

Discussion

91.	Calculate the energy generated in kWh, when `100g` of `._(3)Li^(7)`. are converted into `._(2)He^(4)` by proton bombardment. Given mass of `._(3)Li^(7)=7.0183a.m.u ,` mass of `._(2)He^(4)=4.0040a.m.u`, mass of `._(1)H^(1)=1.0081a.m.u.`
Answer» Correct Answer - `2.36 xx 10^(13)` J

Discussion

92.	One gram of radium is reduced by `2` miligram in `5` yers by `alpha`-decay. Calculate the half-life of radium.
Answer» Correct Answer - 1730.8 years

Discussion

93.	The life-life of `Bi^210` is `5` days. What time is taken by `(7//8)^th` part of the sample of decay ?
Answer» Correct Answer - 15 days

Discussion

94.	Figure shows the enegry levels `P, Q, R, S` and `G` of an atom where `G` is the ground state. A red line in the emission spectrum of the atom can be obtaned by an energy level change from `Q` so `S`. A blue line can be obtained by following energy level change A. P to QB. Q to PC. R to SD. R to G
Answer» Correct Answer - D If E is the energy radiated in transition , then form figure `" " E_(RtoG)gtE_(Qtos)gtE_(RtoS)gtE_(QtoR)gtE_(PtoQ)` For obtaining blue line energy radiated should be be more. Hence, correct option is (d)

Discussion

95.	The number of `alpha`-particless scattered per unit area N `(theta)` at scattering angle `theta` varies inversely asA. `cos^(4)((theta)/(2))`B. `sin^(4)((theta)/(2))`C. `tan^(4)((theta)/(2))`D. `cot^(4)((theta)/(2))`
Answer» Correct Answer - B

Discussion

96.	Assertion: For the scattering of `alpha`-particles at a large angles, only the nucleus of the atom is responsible. Reason: Nucleus is very heavy in comparison to electrons.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 both assertion and reason are false.
Answer» Correct Answer - A we know that an electron is very light particle as compared to an `alpha`-particle. Hence electron cannot scatter the `alpha`-particle at large angles, according to law of conservation of momentum. On the other hand, mass of nucleus is comparable with the mass of `alpha`-particle, hence only the nucleus of atom is responsible for scattering of `alpha`-particles.

Discussion

97.	The diagram shown the path of four `alpha`-particles of the same energy being scattered by the nucleus of an atom simutaneously. Which of these are/is not physically possible ? A. Both 3 and 4B. Both 2 and 3C. Both 1 and 4D. Only 4
Answer» Correct Answer - D `alpha` particles cannot be attracted by the nucleus

Discussion

98.	The graph of the total number of `alpha`-particles scattered at different angles in a given interval of time for `alpha`- particles scattering in the geiger- marsden experiment is given byA. B. C. D.
Answer» Correct Answer - A

Discussion

99.	(A) most of the mass of the atom is concentrated in its nucleus. (R) all alpha particles striking a gold sheet are scattered in different directions.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 both assertion and reason are false.
Answer» Correct Answer - C The results of `alpha`- particle scattering experiment showed that most of the mass of the atom is concentrated in its nucleus. Not all but only high envery `alpha`-particles are scattered when thet strike an extremely thin foil of metal of passing through nucleus.

Discussion

100.	An `alpha`-particle colliding with one of the electrons in a gold atom loosesA. Most of its momentumB. About `1/3`rd of its momentumC. Little of its energyD. Most of its energy
Answer» Correct Answer - C

Discussion

Explore topic-wise InterviewSolutions in .

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)c)/(2pih)`C. ` (e^(2)h)/(2pihc)`D. `(e^(2))/(2epsi_(0)hc)`

the wavelength limit present in the pfund series is `(R=1-097xx10^7m^-1)`A. 1572 nmB. 1898 nmC. 2278 nmD. 2535 nm

An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be proportional toA. 1/mB. `1//v^(2)`C. 1/zeD. `v^(2)`

The shoretest wavelength in Lyman series is 91.2 nm. The longest wavelength of the series isA. 121.6 nmB. 182.4 nmC. 243. 4nmD. 364.8 nm

What is the energy required to remove an electron from second orbit of hydrogen atom ?

Energy of 24. eV is required to remove one of the electron from a neutral helium atom. The energy ( in e V ) reauired to remove both the electorns from a netural helium aotm isA. 38.2B. 49.2C. 51.8D. 78.6

Determine the radius of the first orbit of the hydrogen atom. What would be the velocity and frequency of the electron in the first orbit ? Given : `h=6.62xx10^(-34) J s, m =9.1xx10^(-31) kg, e=1.6xx10^(-19) C, k = 9xx10^(9)m^(2)C.`

The radius of the first electron orbit of a hydrogen atom is `5.3 xx 10^(-11)m`. What is the radius of the second orbit ?

How is impact parameter related to the scattering angle?

Distinguish between excitation potential and ionization potential.

the excitation energy of lyman last lines isA. the same as ionisation energyB. the same as the last absorption line in lyman seriesC. both (a) and (b)D. different from (a) and (b)

the number of de broglie wavelength contained in the second bohr orbit of hydrogen atom isA. 1B. 2C. 3D. 4

The de-Broglie wavelength of an electron in the first Bohr orbit isA. equal to one- fourth the circumference of the first orbitB. equal to half the circumference of first orbitC. equal to twice the circumference of first orbit.D. equal to the circumference of the first orbit.

Half life of a radioactive substance A is 4 day.s The probability that a nucleus will decay in two half lives is(1) 1/4(2) 3/4(3) 1/16(4) 1

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of photon with the most enegy ? A. IIIB. IVC. ID. II

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of photon with the most enegy ? A. IB. IIC. IIID. IV

The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-11 m. What are the radii of the n = 2 and m = 3 orbits?

If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is(a) 10.2 eV(b) 0 eV(c) 3.4 eV(d) 6.8 eV

In Bohr model of the hydrogen atom, let R,v and E represent the radius of the orbit, speed of the electron and the total energy respectively. Which of the following quantities are directly proportional to the quantum number n?A. VRB. REC. `R//E`D. none of these

Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de Broglie wavelength λ of that electron as(a) (0.529) nλ (b) \(\sqrt{nλ}\)(c) (13.6)λ (d) nλ

When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the forms of electromagnetic radiation. Why cannot it be emitted as other forms of energy?

In the Bohr model of hydrogen atom, the electron is pictured to rotate in a circular orbit of radius `5xx 10^-11 m`, at a speed `2.2 xx 10^6` m/ s. What is the current associated with electron motion?

Imagine removing one electron from `He^(4)` and `He^(3)`. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why ?

Imagine removing one electron from He4 and He3. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

The simple Bohr modle is not applicable to He4 atom because(a) He4 is an inert gas.(b) He4 has neutrons in the nucleus.(c) He4 has one more electron.(d) electrons are not subject to central forces.

The Bohr model for the spectra of a H-atom(a) will not be applicable to hydrogen in the molecular from.(b) will not be applicable as it is for a He-atom.(c) is valid only at room temperature.(d) predicts continuous as well as discrete spectral lines.

hydrogen atom emits light when it changes from `n=5` energy level to `n=2` energy level. Which colour of light would the atom emit ?A. redB. yellowC. greenD. voilet.

The kinetic energy of the orbital electron in a hydrogen atom in the excited state corresponding to n = 2 is (A) 3.4 eV (B) 6.8 eV (C) 13.6 eV(D) 27.2 eV.

when an atomic gas or vapour is excited at low pressure, by passing an electric current through it thenA. emission spectrum is observedB. absorption spectrum is observedC. band spectrum is observedD. both a and c

Calculate the energy generated in kWh, when `100g` of `._(3)Li^(7)`. are converted into `._(2)He^(4)` by proton bombardment. Given mass of `._(3)Li^(7)=7.0183a.m.u ,` mass of `._(2)He^(4)=4.0040a.m.u`, mass of `._(1)H^(1)=1.0081a.m.u.`

One gram of radium is reduced by `2` miligram in `5` yers by `alpha`-decay. Calculate the half-life of radium.

The life-life of `Bi^210` is `5` days. What time is taken by `(7//8)^th` part of the sample of decay ?

Figure shows the enegry levels `P, Q, R, S` and `G` of an atom where `G` is the ground state. A red line in the emission spectrum of the atom can be obtaned by an energy level change from `Q` so `S`. A blue line can be obtained by following energy level change A. P to QB. Q to PC. R to SD. R to G

The number of `alpha`-particless scattered per unit area N `(theta)` at scattering angle `theta` varies inversely asA. `cos^(4)((theta)/(2))`B. `sin^(4)((theta)/(2))`C. `tan^(4)((theta)/(2))`D. `cot^(4)((theta)/(2))`

The diagram shown the path of four `alpha`-particles of the same energy being scattered by the nucleus of an atom simutaneously. Which of these are/is not physically possible ? A. Both 3 and 4B. Both 2 and 3C. Both 1 and 4D. Only 4

The graph of the total number of `alpha`-particles scattered at different angles in a given interval of time for `alpha`- particles scattering in the geiger- marsden experiment is given byA. B. C. D.

An `alpha`-particle colliding with one of the electrons in a gold atom loosesA. Most of its momentumB. About `1/3`rd of its momentumC. Little of its energyD. Most of its energy