708 + Interview Questions in STRUCTURE-OF-ATOM IN GENERAL AWARENESS Page 10 InterviewSolution

451.	Assertion : For l = 2, `m_(l)` can be -2, -1, 0, +1 and +2. Reason : For a given value of l, (2l + 1) values of `m_(l)` are possible.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

Discussion

452.	An orbital has `l=4` , what are the possible values of `m_(l)` ? Hint `: m_(l) = -1,"…...",0"…......," +l`
Answer» `m_(l) = -4 ,-3,-2,-1,0,+1,+2,+3,+4`

Discussion

453.	In multi-electron atom 4s-orbital is lower in energy thanA. 3d-orbitalB. 3p-oribitalC. 2s-orbitalD. 2p-orbital
Answer» Correct Answer - 1

Discussion

454.	The possible valuesof `m_(l)`in a givensubshell are given by the formulaA. `n^(20`B. `2n^(2)`C. `2l +1`D. `4l+1`
Answer» Correct Answer - 3

Discussion

455.	The main energy shell in which the electron is present is given byA. Principal quantum numberB. Azimuthal quantum numberC. Spin quantum numberD. Magnetic quantum number
Answer» Correct Answer - 1

Discussion

456.	Shape of an orbital is given byA. Principal quantum numberB. Spin quantum numberC. Azimuthal quantum numberD. Magnetic quantum number
Answer» Correct Answer - 3

Discussion

457.	The shape of `rho-` orbital isA. Dumb-bellB. Double dumb -bellC. SphericalD. Conical
Answer» Correct Answer - 1

Discussion

458.	The correct set of quantum numbers for 4d-electrons isA. 4, 3, 2, +1/2B. 4, 2, 1, 0C. 4, 3, -2, +1/2D. 4, 2, 1, -1/2.
Answer» Correct Answer - D

Discussion

459.	The shape of the orbital is determined byA. spin quantum numberB. magnetic quantum numberC. azimuthal quantum numberD. principal quantum number.
Answer» Correct Answer - C

Discussion

460.	Two electrons occupying the same orbital are distinguished by :A. spin quantum numberB. Principal quantum numberC. Magnetic quantum numberD. Azimuthal quantum number.
Answer» Correct Answer - A The electrons present in the same orbital have always opposite spins (+1/2 and -1/2). An orbital can always have a maximum of two electrons only.

Discussion

461.	It travelling at same speeds, whichof the following mater waves have the shortest wavelength?A. ElectronB. Alpha partice `(He^(2+))`C. NeutronD. Proton
Answer» From de-Brogile equation, wavelenght `lamdba=(h)/(mv)`. For same speed of different particles i.e., electron, proton, neutron and a-particle. `lamdba=(1)/(m)` As h is constnat, Greater the mass of matter waves. Lesser is wavelength and vice-versa. in these matter waves. alpha particle `(He^(2+))` has higher mass , therefore, shortest wavelength

Discussion

462.	What is the correct order of the methods you would apply to separate the components of a mixture of ammonium chloride, common salt and sand ?A. dissolving in water, filtration, evaporation and sublimationB. dissolving the water, evaporation and sublimationC. sublimation, dissolving in water, filtration and evaporationD. moving a magnet, dissolving in water and sublimation
Answer» Correct Answer - C

Discussion

463.	The white of an egg, common salt, sugar and fine sand are added to water separately in beakers as shown below. The mixture is stirred well. A suspension will be formed in the beaker A. IB. IIC. IIID. IV
Answer» Correct Answer - D

Discussion

464.	Which of the following is the correct set of apparatus to separate common salt and sand by filtration process : A. IB. IIC. IIID. IV
Answer» Correct Answer - A

Discussion

465.	A student added the following substances to water kept in four separate beakers. He stirred the mixture well and filtered each one of them through a filter paper. He would obtain a solid residue on the filter paper in the case of :A. egg albuminB. common saltC. chalk powderD. alum
Answer» Correct Answer - C

Discussion

466.	In an experiment, carbon disulphide was added to a test-tube containing a mixture of iron filings and sulphur powder as shown in the given diagrams : The correct observation is represented in diagram :A. IB. IIC. IIID. IV
Answer» Correct Answer - C

Discussion

467.	A student, by mistake, mixed sulphur powder with iron filings. The following techniques were suggested to separate the sulphur from the mixture out of which he has to choose one : A. dissolving in carbon disulphide filtration, evaporation B. dissolving in water at room temperature and filtration C. dissolving in hot waterm filtration and evaporation D. dissolving in ice cold water and filtration The correct technique is :A. AB. BC. CD. D
Answer» Correct Answer - A

Discussion

468.	Composition of the nuclei of two atomic species X and Y is given as under : `{:(,"X","Y"),("Protons :",6,6),("Neutrons :",6,8):}` Give the mass number of X and Y. What is the relation between the two species and which element or elements they represent ?
Answer» We know that : Mass numbers = No. of protons + No. of protons + No. of neutrons So, Mass number of X = 6 + 6 = 12 Mass number of Y = 6 + 8 = 14 Thus, the mass number of X is 12 and that of Y is 14 Now, X contain 6 protons, therefore, the atomic number of X is 6. Y contains 6 protons, therefore, the atomic number of Y is also 6. Since X and Y have the same atomic number (of 6) but different mass numbers (of 12 and 14), they are a pair of isotopes. Atomic number 6 is of carbon element. So, both X and Y represent carbon element.

Discussion

469.	Arrange s, p and d sub-shells of a shell in the increasing order of effective nuclear charge `(Z_("eff"))` experienced by the electron present in them.
Answer» s-orbital, being spherical in shape, shields the electrons from the nucleus more effectively than p-orbital which in turn shields more effectively than d-orbital. Hence, the increasing order of effective nuclear charge `(Z_("eff"))` experienced by the electrons present in them is: `d lt p lt s`

Discussion

470.	Which of the following has non-spherical sub-shell of electron ?A. HeB. BC. BeD. Li
Answer» Correct Answer - B B `(1s^(2) 2s^(2) 2p^(1))` has non-spherical 2p orbital with dump-bell shape.

Discussion

471.	In a hydorgen atom, enegry of the first excited state is `-3.4 eV`. Find out the kinetic enegry of the same orbit of `H` atom.A. `+3.4eV`B. `+6.8 eV`C. `-13.6 eV`D. `+13.6 eV`
Answer» Correct Answer - A As discussed in `KE =` Total enegry `=- (-3.4 eV)` `=+3.4 eV`

Discussion

472.	Photoelectrons are removed with kinetic energy `1.864 xx 10^(-21)J`, when photons of light with energy ` 4.23 xx 10^(-19) J` fall on the metal. What is the minimum energy required per mole to remove an electron potassium metal? Hing`: hv_(0)= hv-KE` Energy required per mole `=hv_(0) xx 6.022 xx 10^(23)`
Answer» Correct Answer - `253.6 kJ mol^(-1)`

Discussion

473.	A proton is moving with kinetic energy `5 xx 10^(-27) J`. What is the wavelength of the de Broglie wave associated with it ?
Answer» Correct Answer - `1.62 xx 10^(-7) m` Mass of proton `= (1.008 xx 10^(-3))/(6.02 xx 10^(23)) kg = 1.67 xx 10^(-27) kg` Further proceed as in Solved Problem 3.

Discussion

474.	Light of wavelength `5000 Å` fall on a metal surface of work function `1.9` eV. Find a. The energy of photon b. The kinetic energy of photoelectrons
Answer» Wavelength of light `(lambda) = 5000 Ã… = 5000 xx 10^(-10)m =5 xx 10^(-7)m`. Work function `(v^(0))=1.9eV=1.9 xx 1.6 xx 10^(-19) J`. Step I. Energy of photons `E=hv=(hc)/(lambda)=((6.626 xx 10^(-34)Js)xx (3 xx 10^(8) ms^(-1)))/((5xx 10^(-7)m))=3.97 xx 10^(-19)J`. Step II . Kinetic energy of photoelectrons ltbr. `KE(1//2 mv^(2))=hv-hv^(0)=(3.97 xx 10^(-19)J)-(1.9 xx 1.6 xx 10^(-19)J) ` `=(3.97 xx 10^(19)J)-(3.04 xx 10^(-19)J)=9.3 xx 10^(-20)J`.

Discussion

475.	What is the wavelength for the electron accelerated by ` 1.0 xx 10^(4) `volts?
Answer» Step I. Calculating of the velocity of electron Energy (kinetic energy ) of electron `=1.0 xx 10^(4)` volts. `=1.0 xx 10^(4) xx 1.6 xx 10^(-19)J=1.6 xx 10^(-15)J` `=1.6 xx 10^(-15)kg m^(2)s^(-2)` or `" " 1//2 m upsilon^(2)=1.6 xx 10^(-15)kg m^(2)s^(-2)` or `" " upsilon =((2xx1.6 xx10^(-15)kg m^(2)s^(-2))/(9.1 xx 10^(-31)kg))^(1//2)=5.93 xx 10^(7) ms^(-1)` Step II. calculation of the wavelength of electron According to de Broglie equation, `lambda =(h)/(m upsilon) , lambda = ((6.626 xx 10^(-34) kg m ^(2) s^(-1)))/((9.1 xx 10^(-31 ) kg) xx (5.93 xx 10^(7) ms^(-1)))=1.22 xx 10^(-11)m`.

Discussion

476.	The kinetic energy of an electron is `5 xx 10^(5) eV` (electron volts). Calculate the wavelength of the wave associated with the electron. The mass of the electron may be taken as `10^(-30) kg`
Answer» Correct Answer - `1.66 xx 10^(-7) m` `1 eV = 1.6 xx 10^(-19)J, KE = (1)/(2) mv^(2) or (5 xx 10^(-5)) (1.6 xx 10^(-19)) = (1)/(2) xx 10^(-30) xx v^(2)` or `v^(2) = 16 xx 10^(6) or v = 4 xx 10^(3) m s^(-1)` Hence, `lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(10^(-30) kg xx 4 xx 10^(3) ms^(-1)) = 1.66 xx 10^(-7) m`

Discussion

477.	Out of cosumic rays, `lamda`-rays, X-rays and radiowaves,....have minimum wavelength and .....have maximum wavelength
Answer» cosmic rays, radiowaves

Discussion

478.	The probability density plots of 1 s and 2s orbitals are given in figure. The density of dots in a region represetns the probability density of finding electrons in the region. On the basis of above diagram which of the following statements is incorrect?A. 1s and 2s orbitals are spherical in shapeB. The probability of finding the electron is maximum near the nucleusC. The probability of finding the electron at a given distance is equal in all directionsD. The probability density of electrons for 2s orbital dec reases uniformly as distance from the nucleus increases.
Answer» The probability density of electrons in 2s orbital first increases then decreases and after that it begins to increases again as distance increases from nucleus.

Discussion

479.	The probability density plots of 1 s and 2s orbitals are given in figure. The density of dots in a region represetns the probability density of finding electrons in the region. On the basis of above diagram which of the following statements is incorrect?A. 1s and 2s orbitals are spherical in shape.B. The probability of finding the electron is maximum near the nucleus.C. The probability of finding the electron at a given distance is equal in all directions.D. The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases.
Answer» Correct Answer - D The statement is incorrect. The probability density of electrons for 2s orbital first increases, then decreases and after that it begins to increases again.

Discussion

480.	The probability density plots of 1s and 2s atomic orbitals are given in figures. The density of dots in a region represents the probability density of finding electrons in the region. On the basis of the above diagram, which of the following statements is incorect?A. 1s and 2s orbitals are spherical in shape.B. The probability of finding the electron is maximum near the nucleus.C. The probability of finding the electron at a given distance is equal in all directions.D. The probability density of electron for 2s orbital decreases uniformly as distance from the nucleus increases.
Answer» Correct Answer - D For 1s orbital, the probability density is maximum at the nucleus ant it decreases sharply as we move away from it. On the other hand, for 2s orbital, the probability density first decreases sharply to zero and again starts increasing.

Discussion

481.	What is the packet of enegry called?A. ElectronB. PhotonC. PositronD. Proton
Answer» Correct Answer - B The packet of any electromagnetic enegry is called quantum but the packet of enegry fo visible light is called photon.

Discussion

482.	The number of `d` electrons in `Ni` (at.no `= 28`) is equal to that of theA. `s` and `p` electrons in `F^(-)`B. `p` electrons in `Ar`(at.no `= 18)`C. `d` electrons in `Ni^(2+)`D. total electrons in `N`
Answer» Correct Answer - C `Ni (Z = 28)`is `[Ar]3d^(8)4s^(2)`. After the removal of `2` electrons, w eget `Ni^(2+)` with the configuration `[Ar]3d^(8)`. Both `Ni` and `Ni^(2+)` ions have the same number of `d` electrons (i.e, `8`).

Discussion

483.	It is possible to take an `X` ray photograph (radiograph) of the bones of a living person becauseA. bones are white in colorB. bones are very hardC. bones are more opaque than the surrounding fleshD. bones are flexible
Answer» Correct Answer - C The lower the density, the more transparent is the material to `X` rays.

Discussion

484.	The number of radial nodes in 3s and 2p respectively areA. 2 and 0B. 1 and 2C. 0 and 2D. 2 and 1
Answer» Correct Answer - A Radial nodes `= n - l -1`

Discussion

485.	The number of radial nodes in `3s` and `2p`, respectively, areA. `0`and`2`B. `1`and`2`C. `2`and`0`D. `2`and`1`
Answer» Correct Answer - C The number of radial nodes or spherical nodes is given by the formula `n - l-1` where `n` is principle quantum number and `l` is azimuthal quantum number. For `3s` orbital, `n - l-1 = 3-0-1=2` ? For `2p` orbital, `n - l - 1=2 - 1- 1=0`

Discussion

486.	Statement-1. 2p orbitals do not have any spherical node. Statement.2 The number of nodes in p-orbitals is given by `(n -2)` where n is the principal quantum numberA. 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 Statement -2 is the correct explanation of statement -1. This is because number spherical nodes in any orbital `= (n -l -1)`. For 2p-orbital, `n = 2, l = 1`

Discussion

487.	Which of the following is the right electronic confiuration of the element palladium `(Z = 46)`?A. `[Kr] 4d^(10)`B. `[Kr] 4d^(9)5s^(1)`C. `[kr] 4d^(8)5s^(2)`D. `[Kr]4d^(7)5s^(3)`
Answer» Correct Answer - A Palladium has `46` electrons. We need `36` electrons `(2+8 +8 +18)` to complete the fourth period, leaving us with `10` more electrons to distribute among the `5s` (maximum `2e^(-)` and `4d` (maximum `10e^(-)s`) subshell. The three choice are `(1),(2)` and `(3)`. Since atomic palladium is diamagnetic, its electron configuration must be `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)4p^(6)4d^(10)` or simply `[Kr] 4d^(10)` Note that a diamagnetic substance is a substance that is not attracted by a magnetic field or is very slightly repelled by such a fiels. This property generally means that the sunstance has only paired electrons.

Discussion

488.	Humphrey series in the hydrogen spectrum is obtainned as a result of the jump of electrons from `n_(2) ge ....." to " n_(1) =`......
Answer» Correct Answer - 7, 6

Discussion

489.	Give the electronic configurations of the elements : `._(19)K, ._(25)Mn, ._(20)Ca`
Answer» `._(19)K= 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(1) : ._(25)Mn=1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5) 4s^(2), ._(20)Ca=1s^(2) 2s^(2) 2p^(6) 3s^(2)3p^(6) 4s^(2)`

Discussion

490.	The correct ground state electronic configuration of chromium atom(Z=24) is :
Answer» `Cr(Z=24), 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)`.

Discussion

491.	Which of the following set of quantum numbers represents the highest energy of an atom ?A. `n = 4,l = 0,m_(l) = 0,s =+1//2`B. `n = 3,l = 1, m_(l) = 1, s =+1//2`C. `n = 3,l = 0, m_(l)=0, s =+1//2`D. `n = 3,l = 2, m_(l) = 1, s = +1//2`
Answer» Correct Answer - D This can be decided by means of the `(n +l)` rule which states that in a neutral isolated atom, the lower the value of `(n+l)` for an orbital electron, the lower is its enegry. However, if the two orbital electrons have the same value of `(n+l)`, the orbital electron with lower value of `n` has lower energy. The fourth set of quantum numbers provides the highest `(n+l)` value and hence, the corresponding orbital electron has the highest enegry.

Discussion

492.	According to Louis de Broglie, a French physicist, every moving material particle has a dual nature i.e., wave and particle nature. The two characters are co-related by de Broglie relation `lambda=(h)/(m upsilon(p)).` Here `lambda` represents wave nature while p or `m upsilon` accounts for the particle nature. Since h is constant , the two characters are inversely proportional to each other. This relationship or equation is valid mainly for microscopic particles such as electrons, protons, atoms, ions, molecules etc. It does not apply to semi-micro or micro particles. For particles having same kinetic energy , the de Broglie wavelength is :A. directly proportional to velocityB. inversely proportional to velocityC. independent of velocity and massD. meaningless
Answer» Correct Answer - A `lambda=(h)/(m upsilon)=(h upsilon)/(2xxKE)` If KE is fixed then `lambda prop upsilon`.

Discussion

493.	Assertion. Electronic configurations of `Cr^(3+)` (containing 21 electrons) is same as that of `Sc (Z = 21)`, i.e., isoelectronic species have the same electronic configuration Reason. Orbitals of atoms as well as ions are filled in order of increasing energy following aufbau principleA. If both assertion and reason are true, and reason is the true explanation of the arrertionB. If both assertion and reason are true, but reason is not the true explanation of the assertionalC. If assertion is true, but reason is falseD. If both assertion and reason are false
Answer» Correct Answer - D Correct A. Atomic and cations containing the same number of electrons do not have the same electronic configuration Correct R. Orbitals of atoms are filled in order of increasing energy.

Discussion

494.	The de Broglie wavelength of a ball of mass 10g moving with a velocity of `10 ms^(-1)` is `(h = 6.626 xx 10^(-34) Js)`A. `6.626 xx 10^(-33) m`B. `6.626 xx 10^(-29) m`C. `6.626 xx 10^(-31) m`D. `6.626 xx 10^(-36) m`
Answer» Correct Answer - A

Discussion

495.	A body of mass `10 mg` is moving with a velocity of `100 ms^(-1)`. The wavelength of the de Broglie wave associated with it would beA. `6.63 xx 10^(-31)m`B. `6.63 xx10^(-35)m`C. `6.63 xx10^(-34) m`D. `6.63 xx10^(-7) m`
Answer» Correct Answer - A The de Broglie equation allows the calculation of wavelength of any perticle or object of mass `m` moving at velocity `v`: `lambda = (h)/(mv)` `=((6.63xx10^(-34)kgm^(2)s^(-1)))/((10xx10^(-6)kg)(100ms^(-1)))` `= 6.63 xx 10^(-31)m` Note that `1mg = 10^(-3)g` and `1g = 10^(-3)kg`. Thus, `1mg = 10^(-6)kg`

Discussion

496.	A body of mass `10 mg` is moving with a velocity of `100 ms^(-1)`. The wavelength of the de Broglie wave associated with it would beA. `6.63xx10^(-35)m`B. `6.63xx10^(-31)m`C. `6.63xx10^(-37)m`D. `6.63xx10^(-34)m`
Answer» Correct Answer - B `lambda=(h)/(m upsilon)=((6.626xx10^(-34)kgm^(2)s^(-1)))/((10xx10^(-6)kg)xx(100ms^(-1)))` `=6.63xx10^(-31)m`.

Discussion

497.	Calculate the energy associated with the first orbitof `He^(+)`.What is the radius of this orbit ? Hint `: E_(n) = - 2.18 xx10^(-18) ((Z^(2))/( n^(2))) J //`atom `He^(+) ( Z=2)` `r_(n) = ( 52.9(n^(2)))/(Z) ` pm
Answer» `E_(1) = - 8.72 xx 10^(-18)J` `r_(10 = 0.0 264 nm`

Discussion

498.	What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from an energy level with `n=4` to an energy level with `n=2` ? What is the colour corresponding to this wavelength ? (Rydberg constant `=109, 677 cm^(-1)`)
Answer» `bar(v)=R(1/n_(1)^(2)-1/n_(2)^(2))=109677 (1/2^(2)-1/4^(2)) cm^(-1) =20564.4 cm^(-1)` `lambda =1/v=1/(20564.4 cm^(-1))=486xx10^(-7) cm=486xx10^(-9) m=486 nm` The colour corresponding to this wavelength is blue.

Discussion

499.	The wavelength associated with an electron moving with velocity `10^(10) ms^(-1)` isA. `6.62 x10^(-10) m`B. ` 7.27 xx 10^(-14) m`C. ` 3.69 xx 10^(-12) m`D. ` 4.92 xx 10^(-11) m`
Answer» Correct Answer - 2

Discussion

500.	A base ball of mass 200g is moving with a velocity of `30 xx 10^(2) m s^(-1)`. If we can locate the base ball with an error equal to the wavelength of the light used `(5000 Å)`, how will the uncertainty in momentum be compared with the total momentum of the base ball ?
Answer» Correct Answer - `Delta p//p = 1.75 xx 10^(-31)` `m = 200//1000 = 0.2 kg, v = 30 xx 10^(2) m s^(-1), Delta x = 5000 Å = 5000 xx 10^(-10) m = 5 xx 10^(-7) m` `Delta x. Delta p = h//4p pi`. Calculate `Deltap`. Also `p - mv`. Calculate `Delta p//p`

Discussion

Explore topic-wise InterviewSolutions in .

An orbital has `l=4` , what are the possible values of `m_(l)` ? Hint `: m_(l) = -1,"…...",0"…......," +l`

In multi-electron atom 4s-orbital is lower in energy thanA. 3d-orbitalB. 3p-oribitalC. 2s-orbitalD. 2p-orbital

The possible valuesof `m_(l)`in a givensubshell are given by the formulaA. `n^(20`B. `2n^(2)`C. `2l +1`D. `4l+1`

The main energy shell in which the electron is present is given byA. Principal quantum numberB. Azimuthal quantum numberC. Spin quantum numberD. Magnetic quantum number

Shape of an orbital is given byA. Principal quantum numberB. Spin quantum numberC. Azimuthal quantum numberD. Magnetic quantum number

The shape of `rho-` orbital isA. Dumb-bellB. Double dumb -bellC. SphericalD. Conical

The correct set of quantum numbers for 4d-electrons isA. 4, 3, 2, +1/2B. 4, 2, 1, 0C. 4, 3, -2, +1/2D. 4, 2, 1, -1/2.

The shape of the orbital is determined byA. spin quantum numberB. magnetic quantum numberC. azimuthal quantum numberD. principal quantum number.

Two electrons occupying the same orbital are distinguished by :A. spin quantum numberB. Principal quantum numberC. Magnetic quantum numberD. Azimuthal quantum number.

It travelling at same speeds, whichof the following mater waves have the shortest wavelength?A. ElectronB. Alpha partice `(He^(2+))`C. NeutronD. Proton

The white of an egg, common salt, sugar and fine sand are added to water separately in beakers as shown below. The mixture is stirred well. A suspension will be formed in the beaker A. IB. IIC. IIID. IV

Which of the following is the correct set of apparatus to separate common salt and sand by filtration process : A. IB. IIC. IIID. IV

A student added the following substances to water kept in four separate beakers. He stirred the mixture well and filtered each one of them through a filter paper. He would obtain a solid residue on the filter paper in the case of :A. egg albuminB. common saltC. chalk powderD. alum

In an experiment, carbon disulphide was added to a test-tube containing a mixture of iron filings and sulphur powder as shown in the given diagrams : The correct observation is represented in diagram :A. IB. IIC. IIID. IV

Composition of the nuclei of two atomic species X and Y is given as under : `{:(,"X","Y"),("Protons :",6,6),("Neutrons :",6,8):}` Give the mass number of X and Y. What is the relation between the two species and which element or elements they represent ?

Arrange s, p and d sub-shells of a shell in the increasing order of effective nuclear charge `(Z_("eff"))` experienced by the electron present in them.

Which of the following has non-spherical sub-shell of electron ?A. HeB. BC. BeD. Li

In a hydorgen atom, enegry of the first excited state is `-3.4 eV`. Find out the kinetic enegry of the same orbit of `H` atom.A. `+3.4eV`B. `+6.8 eV`C. `-13.6 eV`D. `+13.6 eV`

A proton is moving with kinetic energy `5 xx 10^(-27) J`. What is the wavelength of the de Broglie wave associated with it ?

Light of wavelength `5000 Å` fall on a metal surface of work function `1.9` eV. Find a. The energy of photon b. The kinetic energy of photoelectrons

What is the wavelength for the electron accelerated by ` 1.0 xx 10^(4) `volts?

The kinetic energy of an electron is `5 xx 10^(5) eV` (electron volts). Calculate the wavelength of the wave associated with the electron. The mass of the electron may be taken as `10^(-30) kg`

Out of cosumic rays, `lamda`-rays, X-rays and radiowaves,....have minimum wavelength and .....have maximum wavelength

What is the packet of enegry called?A. ElectronB. PhotonC. PositronD. Proton

The number of `d` electrons in `Ni` (at.no `= 28`) is equal to that of theA. `s` and `p` electrons in `F^(-)`B. `p` electrons in `Ar`(at.no `= 18)`C. `d` electrons in `Ni^(2+)`D. total electrons in `N`

It is possible to take an `X` ray photograph (radiograph) of the bones of a living person becauseA. bones are white in colorB. bones are very hardC. bones are more opaque than the surrounding fleshD. bones are flexible

The number of radial nodes in 3s and 2p respectively areA. 2 and 0B. 1 and 2C. 0 and 2D. 2 and 1

The number of radial nodes in `3s` and `2p`, respectively, areA. `0`and`2`B. `1`and`2`C. `2`and`0`D. `2`and`1`

Which of the following is the right electronic confiuration of the element palladium `(Z = 46)`?A. `[Kr] 4d^(10)`B. `[Kr] 4d^(9)5s^(1)`C. `[kr] 4d^(8)5s^(2)`D. `[Kr]4d^(7)5s^(3)`

Humphrey series in the hydrogen spectrum is obtainned as a result of the jump of electrons from `n_(2) ge ....." to " n_(1) =`......

Give the electronic configurations of the elements : `._(19)K, ._(25)Mn, ._(20)Ca`

The correct ground state electronic configuration of chromium atom(Z=24) is :

Which of the following set of quantum numbers represents the highest energy of an atom ?A. `n = 4,l = 0,m_(l) = 0,s =+1//2`B. `n = 3,l = 1, m_(l) = 1, s =+1//2`C. `n = 3,l = 0, m_(l)=0, s =+1//2`D. `n = 3,l = 2, m_(l) = 1, s = +1//2`

The de Broglie wavelength of a ball of mass 10g moving with a velocity of `10 ms^(-1)` is `(h = 6.626 xx 10^(-34) Js)`A. `6.626 xx 10^(-33) m`B. `6.626 xx 10^(-29) m`C. `6.626 xx 10^(-31) m`D. `6.626 xx 10^(-36) m`

A body of mass `10 mg` is moving with a velocity of `100 ms^(-1)`. The wavelength of the de Broglie wave associated with it would beA. `6.63 xx 10^(-31)m`B. `6.63 xx10^(-35)m`C. `6.63 xx10^(-34) m`D. `6.63 xx10^(-7) m`

A body of mass `10 mg` is moving with a velocity of `100 ms^(-1)`. The wavelength of the de Broglie wave associated with it would beA. `6.63xx10^(-35)m`B. `6.63xx10^(-31)m`C. `6.63xx10^(-37)m`D. `6.63xx10^(-34)m`

Calculate the energy associated with the first orbitof `He^(+)`.What is the radius of this orbit ? Hint `: E_(n) = - 2.18 xx10^(-18) ((Z^(2))/( n^(2))) J //`atom `He^(+) ( Z=2)` `r_(n) = ( 52.9(n^(2)))/(Z) ` pm

What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from an energy level with `n=4` to an energy level with `n=2` ? What is the colour corresponding to this wavelength ? (Rydberg constant `=109, 677 cm^(-1)`)

The wavelength associated with an electron moving with velocity `10^(10) ms^(-1)` isA. `6.62 x10^(-10) m`B. ` 7.27 xx 10^(-14) m`C. ` 3.69 xx 10^(-12) m`D. ` 4.92 xx 10^(-11) m`

A base ball of mass 200g is moving with a velocity of `30 xx 10^(2) m s^(-1)`. If we can locate the base ball with an error equal to the wavelength of the light used `(5000 Å)`, how will the uncertainty in momentum be compared with the total momentum of the base ball ?