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601.

The discovery of _______ proved that atom in divisible.

Answer» Correct Answer - Electron
602.

Mass spectrograph helps in the detection of isotopes because they:A. hae different atomic massesB. have same number of electronsC. have sae atomic numberD. have same atomic masses.

Answer» Correct Answer - A
603.

The speed of an electron having its wavelength being equal to speed can vbegiven by :A. `sqrt h/m`B. `sqrt m/h`C. ` sqrt h/p`D. `sqrt h/( 2KE)`

Answer» Correct Answer - A
`lambda= u ` the from ` lambda= h/( mu) , lambda^2 = h/m`.
604.

A green ball weighs ` 75 g` and comes travelling towards you at ` 400 cm //sec `. A photon of light emitted form green ball has wavelength of ` 5xx 10^(-5)` cm . Assuming that the error in the position of ball is the same as the wavelength of itself calculate the error in momentum of the green ball .

Answer» ` 1.055 xx 10^(23)` ,
605.

When `alpha particle are sent through a this metal foil mass of then go straight through the foil becauseA. `alpha` particle are much he avier than electronB. `alpgha`particle are positively chargedC. Most part of the atom is empty spaceD. `alpha` particle move with light speed

Answer» Correct Answer - A::C
606.

Which of the following sets of quantum number is //are not perrmitted ?A. `n = 3,l = 3 ,m = +1 , s= +(1)/(2)`B. `n = 3,l = 2 ,m = +2 , s= -(1)/(2)`C. `n = 3,l = 1 ,m = +2 , s= -(1)/(2)`D. `n = 3,l = 0 ,m = 0 , s= +(1)/(2)`

Answer» Correct Answer - A::B::C
607.

A certain transition to H spectrum from an excited state to ground state in one more steps given rise to total `10` lines .How many of these belong to UV spectrum ?A. 3B. 4C. 5D. 6

Answer» Correct Answer - B
4
608.

The uncertaintuy invelved in the measurement fo velocity within a distance of ` 0.1 Å ` is :A. `5.79xx10^(8)ms^(-1)`B. `5.79xx10^(5)ms^(-1)`C. `5.79xx10^(6)ms^(-1)`D. `5.79xx10^(7)ms^(-1)`

Answer» Correct Answer - C
`Delta u. Delta x = h/(4 pi m)`
` :. Delta u = h/( 4 pi m. Delta x)`
` = ( 6.626 xx 10^(-34))/( 4 xx 3.14 xx 9. 11 xx 10^(-31) xx 0.1 xx 10^(-10))`
` = 5. 8 xx 10^6 ms^(-1)`.
609.

The orientiation of an atomic orbital is governed by :A. Magnetic quantum numbreB. Principal quantum bumberC. Aximumthal quantum numberD. Spin quantum number

Answer» Correct Answer - A
Magnetic quantum number signifies the possile number of orientarions of an orbital .
610.

An electron in a hydrogen like atom is in an excited state3 . It has a total energy of -3 4 eV. Calculate : (a) The kinetic energy of electron . (b) The de-Broglie wavelength of electron . ( `h = 6.6 xx 10^(023) , m_e = 9 .108 xx 10^(-31) kg)`

Answer» (a) ` 3.4e V, (b) 6.618 xx 10^(10) m,`
611.

The total energy of an electron in the second excited state of the hydrogen atom is about -1.5 eV. The kinetic energy and potential energy of the electron in this state are:A. 1.5 eV and -3 eVB. `-1.5eV` and -1.5eVC. 3eV and -4.5 eVD. `-0.75` eV and `-0.75eV`

Answer» Correct Answer - A
`KE=-TE,PE=2TE`
612.

What should be the number of electrons presents in `X^(2+)` on the basis of electronic configuiration if the ion `X^(3+)` has `14` protons ?

Answer» `X^(3-)` has `14` proton so X also has `14` proton .Therefore it the `14` electron
`X = 14 = 1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(2)`
`X^(2+) = 1s^(2) 2s^(2) 2p^(6)3s^(2) = 12` electrons
613.

The potential energy of an electron in the hydrogen atom is -6.8 eV. Indicate in which excited state, the electron is present?A. firstB. secondC. thirdD. fourth

Answer» Correct Answer - A
`PE = 2(T.E)`
614.

Identify the paris which are not of isotopes?A. `._(6)^(12)X,._(6)^(13)Y`B. `._(17)^(35)X,._(17)^(37)Y`C. `._(6)^(14)X,._(7)^(14)Y`D. `._(4)^(8)X,._(4)^(9)Y`

Answer» Correct Answer - D
Isotopes have same atomic number but different mass number
615.

What is a particle or matter wave and how it is different from electro-megnetic wave?

Answer» `|{:(,"Particle/Matter wave",,"e/m wave",),((i),"Made of microscopic particle eg":[e^(-),p,n],(i),"Made of quanta",),((ii),"Veloctiy"ltlt3xx10^(8)m//s,(ii),"Velocity"=3xx10^(8)m//s,),((iii),"Very small wavelength {Very high frequency}",(iii),"Very large wavelength [small frequency]",),((iv),"No electric or magnetic field associated",(iv),"Electric and magnetic field lie at right angles",),((v),"Matter waves remain localized particles and are not emitted",(v),"They are emitted by source and leave the source They propogate in all direcions",),((vi),"Pass through a medium but not through vacuum",(vi),"Can pass through vacuum as well as through medium",):}|`
616.

If change in energy `(DeltaE)=3xx10^(-8)` J, `h=6.64xx10^(-34)` J-s and `c=3xx10^8`m/s, then wavelength of the light isA. `6.36xx10^3` ÅB. `6.36xx10^5` ÅC. `6.64xx10^(-8)` ÅD. `6.36xx10^(18)` Å

Answer» Correct Answer - C
`DeltaE="hc"/lambda "or " lambda="hc"/(DeltaE)`
`=(6.64xx10^(-34) xx 3xx10^8)/(3xx10^(-8))=6.64xx10^(-8) Å`
617.

What is an `alpha`-particle ?

Answer» Doubly ionised He atom
618.

The atoms of different elements having same mass number but different atomic number are known as isobars. The sum of protons and neutrons, in the isobars is always different.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false

Answer» Correct Answer - C
The assertion that the isobars are the atoms of different elements having same mass number but different atomic number, is correct but reason is false because atomic mass is sum of number of neutron and protons which should be same for isobars.
619.

The atoms of different elements having same mass number but different atomic number are known as isobars. The sum of protons and neutrons, in the isobars is always different.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true.

Answer» Correct Answer - C
( c) The assertion that the isobars are the atoms is different elements having same mass number but different atomic number, is correct but reason is false because atomic mass is sum of number of neutron and protons which should be same for isobars.
620.

Which of the Following is not H-Like speciesA. `He^+`B. `Li^(2+)`C. `B^(4+)`D. `C^(4+)`

Answer» Correct Answer - D
621.

Which of the following series belong to Visible regionA. Lyman seriesB. Balmer SeriesC. Paschen SeriesD. Pfund Series

Answer» Correct Answer - B
622.

Aufbay principle fails to explain the configuration of element with atomic numberA. 18B. 21C. 24D. 27

Answer» Correct Answer - C
Chronium `(z = 24)` and copper `(z=29)`
623.

Total number of electron in any energy level isA. `sum_(l = 0)^(l = n-1)2 (2l +1)`B. `sum_(l=0)^(l=n)2(2l +1)`C. `sum_(l=0)^(l=n+1)2(2l+1)`D. `sum_(l=1)^(l=n-1)2 (2l+1)`

Answer» Correct Answer - A
no. of `e^(-)` in a sublevel `=4l +2`
624.

Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also true. this is summed up in what we now call the Heisenberg uncertainty principal. The equation is `Deltax.Delta(mv) ge (h)/(4pi)` The uncertainty is the position or in the momentum of a macroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electron is small enough for the uncertainty to be relatively large and significant. If the uncertainties in position and momentum are equal, the uncertainty in the velocity is:A. `sqrt((h)/(pi))`B. `sqrt((h)/(2pi))`C. `(1)/(2m)sqrt((h)/(pi))`D. `(1)/(2)sqrt((h)/(pim))`

Answer» Correct Answer - C
`Delta x.Deltav = (h)/(4pim)`
625.

Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also true. this is summed up in what we now call the Heisenberg uncertainty principal. The equation is `Deltax.Delta(mv) ge (h)/(4pi)` The uncertainty is the position or in the momentum of a macroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electron is small enough for the uncertainty to be relatively large and significant. If the uncertainty in velocity and position is same, then the uncertainty in momentum will beA. `sqrt((hm)/(4pi))`B. `msqrt((h)/(4pi))`C. `sqrt((h)/(4pim))`D. `(1)/(m)sqrt((h)/(4pi))`

Answer» Correct Answer - A
`Delta x.Deltap = (h)/(4pi)`
626.

Statement : The 3p-orbital has higher energy level than 3s in `He^(+)` ion. Explanation: The energy of an orbital depends upon n and l.A. S is correct but E is wrongB. S is wrong but E is correct.C. Both S and E are correct and E is correct explanation of SD. Both S and E are correct but E is not correct explanation of S

Answer» Correct Answer - C
Higher is the value of ( n+1) more is the energy level of orbital.
627.

Statement : The 3p-orbital has higher energy level than 3s in `He^(+)` ion. Explanation: The energy of an orbital depends upon n and l.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true.

Answer» Correct Answer - D
(d) Higher is the value of `(n + 1)` more is the energy level of orbital for one electron system energy of `3 s = 3 p`.
628.

Orbital angular momentum of 3s and 3p are :A. `(h)/(2pi) , (h)/(2pi)`B. `(h)/(sqrt2 pi) , (h)/(sqrt2 pi)`C. `0 , (sqrt(2h))/(pi)`D. `0 , (h)/(sqrt2 pi)`

Answer» Correct Answer - D
629.

Choose the correct statement(s) .A. The orbital wave function or `Psi` for an electron has no physical meaningB. Square of wave function `(Psi^(2))` at a point gives the probability density of the electron at that point .C. Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals .D. All of the above

Answer» Correct Answer - D
630.

The series limit for Balmer series of H-spectra isA. 3800B. 4200C. 3646D. 4000

Answer» Correct Answer - C
The last time in any series is called series limit. Series limit for Balmer series in 3646 Å
631.

The greater the energy of a photon , the :A. longer the wavelength and the higher the frequency .B. longer the wavelength and the lower the frequency .C. shorter the wavelength and the higher the frequency .D. shorter the wavelength and the lower the frequency .

Answer» Correct Answer - C
632.

The de Broglie wavelength associated with a ball of mass `1kg` having kinetic enegry `0.5J` isA. `6.626xx10^(-34)` mB. `13.20xx10^(-34)` mC. `10.38xx10^(-21)` mD. `6.626xx10^(-34)` Å

Answer» Correct Answer - A
`lambda=h/(sqrt2m(KE))=(6.626xx10^(-34)) / sqrt(2xx1xx0.5) =6.626xx10^(-34)` m
633.

For an electron whose x-positional uncertainty is `1.0 xx 10^(-10)` m . The uncertainty in the x - component of the velocity in `ms^(-1)` will be the order of :A. `10^(6)`B. `10^(9)`C. `10^(12)`D. `10^(15)`

Answer» Correct Answer - A
634.

The velocity of an electron with de Broglie wavelength of `1.0 xx 10^(2)` nm is:A. `7.2 xx 10^(5)cm//sec`B. `72 xx 10^(5)cm//sec`C. `7.2 xx 10^(4)cm//sec`D. `3.6 xx 10^(5)cm//sec`

Answer» Correct Answer - A
`lambda = (h)/(mv)`
635.

What is the velocity of an electron (m = `9.11 xx 10^(-28)` g) that exhibits a de Broglie wavelength of `10.0` nm ? [`1 J = 1kg dot" m^(2)* s^(2)`]A. `72.7 mdot"s^(-1)`B. `270 mdot"s^(-1)`C. `7.27 xx 10^(4) m dot"s^(-1)`D. `7.27 xx 10^(6) m dot "s^(-1)`

Answer» Correct Answer - C
636.

All the energy levels are called excied state when the val,ue of the principal quantum number isA. `n = 1`B. `n gt 1`C. `n lt 1`D. `n gt - 1`

Answer» All the energy state in which n is greater than `1` are called excited state
637.

In sample of hydrogen atoms in ground state, electron make transition from ground state to particular excited state, where path length is four times of debroglie wavelenght . Electrons make back transition to the ground state prouducing all possible photons.A. Wavelenght of photon used for this transition is 121.5 nmB. Final excitd state of `Li^(+2)` is 6C. kinetic energy of electron in intaial excited state of `Li^(+2)` is 13.6 eVD. All are correct

Answer» Correct Answer - D
`3^(rd)` highest energetic photons `rArr(13.6-3.4)eV=10.2 eV`, transition is `2 to 1`
`lambda=(hc)/(E)=(1240)/(10.2)nm=121.5nm`
For H atoms =gor `Li^(+)`ion
`13.6xx1[1-(1)/(2^(2))]=13.6xx9[(1)/(3^(2))-(1)/(6^(2))]`
`therefore` for `Li^(+)`ion transition is `3 to 6`
Initial excited state=3
Final excited state =6
K.E. of electron in intial excited state of `Li^(+2)`=`13.6xx(3^(3))/(3^(3))`=`13.6 eV`.
638.

If the average life time of an excited state of hydrogen is of the order of `10^(-8) s`, estimate how many whits an alectron makes when it is in the state `n = 2` and before it suffers a transition to state` n = 1 (Bohrredius a_(0) = 5.3 xx 10^(-11)m)`?

Answer» Correct Answer - `8xx10^(6)`
`V_(2)=V_(0)xx(1)/(2)=(V_(0))/(2)`
`x=vxxt`
`x=(V_(0))/(2)xx10^(-8)sec=((V_(0)xx10^(-8))/(2))m`
`2pir rarr 1` round
`(V_(0)xx10^(-8))/(2)=(V_(0)xx10^(-8))/(2)xx(1)/(2pir)`
`r_(2)=r_(0)xxn^(2)=4r_(0)`
so, no. of revolutions
`=(V_(0)//2xx10^(-8))/(2pixx4r_(0))=(V_(0)xx10^(-8)xx1)/(2xx2pixx4r_(0))`
`=(2.18xx10^(6)xx10^(-18))/(2xx2xx3.14xx4xx0.529)`
`(2.18xx10^(-12))/(2.6xx10^(-21))=0.838xx10^(9)=8xx10^(6)`
639.

For silver metal `mu_(0)` is `1.13 xx 10^(17)s^(-1)` .What is the maximum energy of the photoelectron produced by shiniong atraviolet light wavelength `1.5` nm on the metal by shining liught wavelength `1.5 nm` on the metal .

Answer» `hv = hv_(0) + KE`
`KE = hv - hv_(0)`
`hc/(lambda) -hv_(0)`
`:. KE = ((6.626 xx 10^(-34) Js) xx 3 xx 10^(8) ms^(-1))/(1.5 xx 10^(-9) m)`
`-(6.626 xx 10^(-34) Js)(1.13 xx 10^(17) s^(-1))`

`= 1.325 xx 10^(-16) J - 7.487 xx 10^(-17) J`
`= 0.576 xx 10^(-16) J`
640.

Calculate the mass of a photon of sodium light wavelength `600` and velocity `3 xx 10^(8)m s^(-1)`.

Answer» Wavelength of photon
`lambda = 600 xx 10^(-9) m = 6.0 xx 10^(-7)m`
Velocity of photon `(v) = 3 xx 10^(8) m s^(-1)`
Using relationship
`m = (h)/(c lambda)`
we get
Mass of photon `= (6.626 xx 10^(-34) J s)/( 3 xx 10^(8) ms^(-1) xx 6.0 xx 10^(-7) m)`
`= 3.68 xx 10^(--36) kg`
641.

What is the potential energy of an electron present in `N-` shell of the `Be^(3+)` ion ?A. `-3.4eV`B. `-27.2 eV`C. `-13.6 eV`D. `-6.8 eV`

Answer» Correct Answer - 2
Energy of `N-` shell `=(-13.6xx(4)^(2))/((4)^(2))=-13.6eV`
`:. P.E.=2xxErArr2xx-13.6=-27.2eV`
642.

The angular speed of an electron revolving around the H-nucleus is proportional toA. `1//r`B. `1//r^(3//2)`C. `1//r^(2)`D. `r^(3//2)`

Answer» Correct Answer - B
We know that, therefore `1/(4pivarepsilon_(0))cdot(Ze^(2))/r^(2)=(mv^(2))/r`
therefore`v=sqrt((1/(4pivarepsilon_(0))cdot(Ze^(2))/(mr)))=k/sqrtr`
(Let `k=(Ze^(2))/(4pivarepsilon_(0)m)` )
Angular speed ,`omega=v/r=k/(sqrtrcdotr)=k/(r^(3//2))`
643.

Chlorine atom differs from chloride ion in the number ofA. ProtonB. NeutronC. ElectronD. protons and electrons

Answer» Correct Answer - C
Cl and `Cl^-` differs in number of eletrons. Cl has `17e^-` while `Cl^-` has `18e^-`
644.

A particle of charge equal to that of an electron and mass `208` times the mass of the electron moves in a circular orbit around a nucleus of charge `+3e`. Assuming that the Bohr model of the atom is applicable to this system, (a) derive an expression for the radius of the `n^(th)` bohr orbit, (b) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom, and (c ) find the wavelength of the radiation emitted when the revolving particle jumps from the third orbit to the first.

Answer» Correct Answer - `r_(n)=(n^(2)h^(2))/(4Kpi^(2)xx3e^(2)xx208m_(e));n=25 ; 55.2 "pm"`
(a) `(1)/(lambda)=((1)/(lambda_(1))+(1)/(lambda_(2)))=rxx4[(1)/(1^(2))-(1)/(n^(2))]`
(b) `DeltaE_(2rarr4)=2.7=IE[(1)/(4)-(1)/(16)]`
`IE=2.7xx(16)/(3)eV`
(c ) `Deltaoverset("max")E_(4rarr1)=IE[(1)/(k)-(1)/(1)]`
`DeltaE_(4rarr3)=IE[(1)/(16)-(1)/(9)]`
645.

Calculate the binding energy per mole when threshold wavelength of photon is `240nm`

Answer» Correct Answer - `497kJ//mol`
`E=(1240)/(240)eV " "E=5.167eV`
`E=497.9kJ//"mol"`
646.

The mass number of an anion, `X^(3-)`, is 14. if there are ten electrons in the anion, the number of neutrons in the nucleus of atom, `X_2` of the element will beA. 10B. 14C. 7D. 5

Answer» Correct Answer - C
`._7X^14, n=14-7=7`
647.

Bohr advanced the idea of :A. Stationary electronsB. statioinary nucleusC. statioary orbitsD. elliptical orbits

Answer» Correct Answer - C
648.

A monoatomic (X) ion has a charge of +3 the nucleus of the ion has a mass number of 45. the number of neutrons in the nucleus is 1.14 times that of number protons. Find out : (a) Number of electrons in atom x. (b) Number of electrons in `X^(3+)` ion. (c) Configurations of X, `X^(3+)` and `X^(1+)` ion. (d) Suggest which of the these `(X,X^(3+)` and `X^(+)`) are paramagentic (e) the Total magnetic moment of `X^(+)` ion.

Answer» (a) 21, (b) 18, (d) `X,X^+`, (e ) `sqrt(8)`. B.M,, .
649.

If `((0.51times10^(-10))/4)`meter is the radius of smallest electron orbit in hydrogen like atom,then this atom isA. hydrogen atomB. `He^(+)`C. `Li^(2+)`D. `Be^(3+)`

Answer» Correct Answer - D
650.

If an electron is moving around a nucleus of charge `3theta`in a circular orbit of radius `10^(-10)`m,then calculate the initial frequency of light emitted by the electron.A. `4.2times10^(15)Hz`B. `0.36times10^(15)Hz`C. `3.6times10^(15)Hz`D. `4.2times10^(15)Hz`

Answer» Correct Answer - C
We know that, `(mv^(2))/r=(2theta^(2))/(5pivarepsilon_(0)r^(2))or v=((2theta^(2))/(4pivarepsilon_(0)mr))^(1//2)`
Also,frequency=f=`v/(2pir)= ((2theta^(2))/(4pivarepsilon_(0)mr))^(1//2)times1/2pir`
therefore`f=((1.414)(9times10^(9))^(1//2)times1.6times10^(-19))/((9.1times10^(-31))^(1//2)times2pi(10^(-10))^(3//2))`
cong`3.6times10^(15)Hz`