708 + Interview Questions in STRUCTURE-OF-ATOM IN GENERAL AWARENESS Page 13 InterviewSolution

601.	Assertion : `Cu^(2+)` ion is a coloured ion . Reason : Every ion with unpaired electron is coloured .A. if both assertion and reason are correct and reason is correct explanation for assertion .B. If both assertion and reason are correct but reason in not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect.
Answer» Correct Answer - C Correct reason. An ion with unpaired electron in the d- or f-sub shell is coloured.

Discussion

602.	Assertion (A) : `Fe^(3+)` (g) ion is more stable than `Fe^(2+)(g)` ion. Reason (R) : `Fe^(3+)` ion has more number of unpaired electrons than `Fe^(2+)` ion.A. if both assertion and reason are correct and reason is correct explanation for assertion .B. If both assertion and reason are correct but reason in not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect.
Answer» Correct Answer - C Correct reason. In `Fe^(3+)`ion. all the five 3d orbitals are half filled while they are not in `Fe^(2+)`ion.

Discussion

603.	The unpaired electrons in `Al` and `Si` are present in `3p` orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Answer» Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital. (ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus. (iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

Discussion

604.	The nuclear radius is of the order of `10^(-13)` cm while atomic radius is of the order `10^(-8)` cm. Assuming the nucleus and the atom to be spherical, what fraction of the atomic volume occupied by the nucleus ?
Answer» The volume of a sphere `= 4pi r^(3)//3` where r is the radius of the sphere `:.` Volume of the nucleus `= 4pi r^(3)//3 = 4pi (10^(-13))^(3)//3 cm^(3)` Similarly, Volume of the atom `= 4pi r^(3)//3 = 4pi (10^(-8))^(3)//3 cm^(3)` `:.` Fraction of the volume of atom occupied by the nucleus `= (4pi (10^(-13))^(3)//3 cm^(3))/(4pi (10^(-8))^(3)//3) cm^(3) = 10^(-15)`

Discussion

605.	Calculate (a) wave number and (b) frequency of yellow radiations having wavelength of `5800Å`
Answer» Calculate of wave number: Wave number `bar(v) = (1)/(lambda)` But `lamda = 5800Å = 5800 xx 10^(-10)m " (Given)" :. bar(v) = (1)/(5800 xx 10^(-10)m) = 1.72 xx 10^(6) m^(-1)` (b) Calculate of Frequency: Frequency `v = (c)/(lamda)` Substituting `c = 3 xx 10^(8) m//sec and lamda = 5800 xx 10^(-10)m`, we get `v = (3 xx 10^(8) ms^(-1))/(5800 xx 10^(-10)m) = 5.172 xx 10^(14) s^(-1)` or cycles/sec or Hz

Discussion

606.	Calculate the wavelength (in nanometer) associated with a proton moving at `1.0 xx 10^(3) ms^(-1)`A. 0.032 nmB. 0.40 nmC. 2.5 nmD. 14.0 nm
Answer» Correct Answer - B `lamda = (h)/(mv) = (6.62 xx 10^(-34) kg m^(2) s^(-1))/((1.67 xx 10^(-27) kg) (10^(3) ms^(-1)))` `= 3.96 xx 10^(-10) m = 0.40 nm`

Discussion

607.	The two extra -nuclear electrons in the orbital of helium have antiparallel spins `(uarr darr)`. Why not they have parallel spins `(uarr uarr)` ?
Answer» If they had parallel spins, the repulsion will increase. As a result, energy will increase and the stability will decrease.

Discussion

608.	Which of the following orbitals are possible ? `1pm, 2s, 3f, 3d`
Answer» 2s and 3d (l never be eual to n or greater than n )

Discussion

609.	The 3d-orbitals having electron density in all the three axes isA. `3d_(xy)`B. `3d_(z^(2))`C. `3d_(yz)`D. `3d_(zx)`
Answer» Correct Answer - B

Discussion

610.	An electron is in one of the `3d` orbitals. Give the possible values of `n, l,` and `m` for this electron.
Answer» for the 3d orbital Principal quantum number (n) = 3 Azimuthal quantum number (l) = 2 Magnetic quantum number `(m_(l)) = – 2, – 1, 0, 1, 2`

Discussion

611.	Give the number of electrons in the species `H_(2)^(+), H_(2)` and `O_(2)^(o+)`
Answer» Number of electrons present in hydrogen molecule `(H_(2))` = 1 + 1 = 2 number of electrons in `H_(2)^(+) = 2-1= 1 ` `H_(2):` number of electorn in `H_(2) = 1+1 =2 ` `O_(2)^(+)` number of electrons present in oxygen molecule `(O_(2))` = 8+8 = 16 number of electrons in `O_(2)^(+)` = 16-1=15

Discussion

612.	An atom of an element contains `29` electrons and `35` neutrons. Deduce a. The number of protons and b. The elctonic configuration of the element.
Answer» No. of protons in a neutral atom = No. of electrons = 29 Electronic configuration = `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(10) 4s^(1)`.

Discussion

613.	An atom of an element contains `29` electrons and `35` neutrons. Deduce a. The number of protons and b. The electronic configuration of the element.
Answer» (i) For an atom to be neutral, the number of protons is equal to the number of electrons. Number of protons in the atom of the given element = 29 (ii) The electronic configuration of the atom is `1s^(2) 2s^(2) 2p^(2)3s^(6) 4s^(2) 3d^(10)`

Discussion

614.	An element with mass number `81` contains `31.7%` more neutrons as compared to protons. Assign the atomic symbol.
Answer» An element can be identified by its atomic number only. Let us find the atomic number. Let the number of protons = x `:.` Number of neutrons `=x+(x xx31.7 )/(100) =(x + 0.317 x)` Now, Mass no. of element = No. of protons + No. of neutrons `81 = x+x+0.317 x = 2.317 x or x=(81)/(2.317)=35` `:. " "` No. of protons = 35, No. of neutrons = 81 -35 = 46 Atomic number of element (Z) = No. of protons = 35 The element with atomic number (Z) 35 is bromine `._(35)^(81)Br`.

Discussion

615.	Write down the quantum numbers n, l and m for the following orbitals: (i) `3d_(x^(2) - y^(2))` (ii) `4d_(z^(2))` (iii) `3d_(xy)` (iv) `4d_(xz)` (v) `2p_(z)` (vi) `3p_(x)`
Answer» (i) `n = 3, l = 2, m = +2` (ii) `n = 4, l = 2, m = 0` (iii) `n = 3, l = 2, m = -2` (iv) `n =4, l = 2, m = +1` (v) `n = 2, l = 1, m = 0` (vi) `n =3, l =1, m = +1`

Discussion

616.	Match the following `:` `{:(,"Column-I (Orbitals)",,"Column-II (Nature and No. of Nodes)"),((A),3d_(z^(2)),(p),"One sperhical node"),((B),4d_(xy),(q), "Two nodal planes"),((C ) , 2s,(r ),"Node in yz plane"),((D) ,3p_(x),(s) , "Two angular nodes"):}`
Answer» Correct Answer - `A(s),B(p,q,r,s),C(p),D(p,r)`

Discussion

617.	Which of the following orbitals are degenerate ? `3d_(xy), 4d_(xy) 3d_(z^(2)), 3d_(yz), 4d_(yz), 4d_(z^(2))`
Answer» Degenerate orbitals are the orbitals of the same subshell of the same main shell. Hence, these are `(3d_(xy), 3d_(z^(2)), 3d_(yz)) and (4d_(xy), 4d_(yz), 4d_(z^(2))`

Discussion

618.	Match the following `:` `{:(,"Column-I",,"Column-II"),((A),3d_(xy),(p),Psi=0[zx "plan"]),((B),2p_(y),(q),Psi=0[yz "plane"]),((C ) , 2s,(r ),Psi= "max"["yz plane"]),((D),2s,(s),Psi="max[zx plane]"),(,,(t),Psi"versus r graph has wo maximas"):}`
Answer» Correct Answer - `A(p,q),B(q,s),C(r,s,t),D(r,s)`

Discussion

619.	A student while heating solid lead nitrate taken in a test tube would observe :A. white residue of `PbO_(2)`B. green residue of `NO_(2)`C. yellow residue of PbOD. brown residue of NO
Answer» Correct Answer - C

Discussion

620.	A student took some lead nitrate compound in a boiling tube and heated it strongly. The gas/gases evolved on heating this compound is/are :A. `NO_(2)`B. `NO_(2)+CO_(2)`C. `NO_(2)+O_(2)`D. `N_(2)+O_(2)`
Answer» Correct Answer - C

Discussion

621.	The colour of residue left behind on heating lead nitrate when it is still hot is :A. greenB. yellowC. blackD. reddish-brown
Answer» Correct Answer - D

Discussion

622.	In which of the following orbitals , the sum of `( n+1)` value is lowest ? `4s, 3d, 4p`
Answer» `4s rarr n = 4, l = 0, ( n+l) =4` `3d rarr n=3,l= 2, ( n+l)= 5` `4p rarr n = 4 , l= 1, ( n+1) = 5` Therefore, 4s- orbital is lowest in energy .

Discussion

623.	Which of the following combinations of quantum numbers is possible for a `4p` orbital?A. `n = 4,l = 1, m_(l) = 0, +-1`B. `m= 4, l = 1, m_(l) =- 1`C. `n = 4, l = 1,m_(l) = 0`D. `n = 4, l = 1, m_(l) = +1`
Answer» Correct Answer - A The designation `4p` indicates that the orbital has a principal quantum number `n = 4` and an angular-momentum quantum number `l = 1`. The magnetic quantum number can have any of the values `-1, 0`, or `+1`.

Discussion

624.	Radial wave function depends on the quantum numbers (i) `n` (ii) l` `(iii) m_(l)` (iv) m_(s)`A. `(ii),(iii)`B. `(i),(ii)`C. `(i),(ii),(iii)`D. `(i),(ii),(iii),(iv)`
Answer» Correct Answer - B Radial wave function, `R(r )`, that depends obn the distance `(r )` form the nucleus is depends on the quantum numbers `n` and `l`. On the other hand, angular wave function, `A(theta, psi)`, that depends on the direction form the nucleus is dependent on the quantum numbers `l` and `m_(l)`.

Discussion

625.	An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. If `lamda_(1)` is the wavelength of the line for jump of the electron from 4th to 3rd orbit and `lamda_(2)` that of the line for jump from 3rd to 2nd orbit, then wavelength of the line for the jump from 4th orbit to 2nd orbit will beA. `lamda_(1) + lamda_(2)`B. `(lamda_(1) lamda_(2))/(lamda_(1) + lamda_(2))`C. `(lamda_(1) lamda_(2))/(lamda_(1) - lamda_(2))`D. `(lamda_(1) + lamda_(2))/(lamda_(1) lamda_(2))`
Answer» Correct Answer - D `bat(v)_(1) = (1)/(lamda_(1)) = R ((1)/(3^(2)) - (1)/(4^(2)))` `bar(v)_(2) = (1)/(lamda_(2)) = R ((1)/(2^(2)) - (1)/(3^(2)))` `bar(v) = (1)/(lamda) = R ((1)/(2^(2)) - (1)/(4^(2)))` Hence, `(1)/(lamda) = (1)/(lamda_(1)) + (1)/(lamda_(2)) = (lamda_(1) + lamda_(2))/(lamda_(1) lamda_(2)) or lamda = (lamda_(1) lamda_(2))/(lamda_(1) + lamda_(2))`

Discussion

626.	An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. The total number of lines produced in the spectrum will beA. 3B. 6C. 9D. 12
Answer» Correct Answer - B Total no. of lines produced when the electron returns to the ground state `Sigma (n_(2) - n_(1)) = Sigma (4 -1) = Sigma 3 = 3 + 2 + 1 = 6`

Discussion

627.	A student placed a clean iron nail in blue coloured copper sulphate solution for a considerable time. He observes that :A. iron nail gets green coatingB. iron nail gets brown coatingC. iron nail gets no coatingD. iron nail gets blue coating
Answer» Correct Answer - B

Discussion

628.	A student was asked to carry out a chemical reaction by placing four different metal strips in `CuSO_(4)` solution for a considerable time, one by one. Which of the following metal strip will turn the blue `CuSO_(4)` solution to a light green solution in due course of time ?A. FeB. AuC. MgD. Ag
Answer» Correct Answer - A

Discussion

629.	A student was asked by her science teacher to prepare a true solution. She dissolved the solute in water but forget to record its name. What may be the correct name of the solute ?A. barium sulphateB. sulphur powderC. alumD. egg albumin
Answer» Correct Answer - C

Discussion

630.	For the electrons of oxygen atom, which of the following statemetns correct?A. `Z_("eff")` for an electron in a 2s orbital is the same as `Z_("eff")` for an electron in a 2p orbital.B. An electron in the 2s orbital has the same energy as an electron in the 2p orbital.C. `Z_("eff")` for an electron in 1s orbital is the same as `Z_("eff")`for an electron in a 2s orbital.D. The two electrons present in the 2s orbital have spin quantum numbers `m_(s)` but of opposite sign.
Answer» Correct Answer - D is the correct answer.

Discussion

631.	`P` is the probability of finding the Is electron of hydrogen atom in a spherical shell of infitesimal thickness, dr, at a distance `r` from the nucleus. The volume of this shell is `4pi r^(2)dr`. The qualitative sketch of the dependence of `P` on r isA. B. C. D.
Answer» Correct Answer - C Probabilty of finding 1s electron is maximum near the nucleus and goes on increasing till it reaches a maximum value at a distance 52.9 pm and then begins to decrease abruptly. Even at large distance from the nucleus, there is a finite through small probability of finding an electron of a given energy.

Discussion

632.	A certain particle carries `2.5 xx 10^(-16) C` of static electric charge. Calculate the number of electrons present in it.
Answer» Charge carried by one electron `= 1.6022 xx 10^(-19) C` `:.` Electrons present in particle carrying `2.5 xx 10^(-16) C " charge" = (2.5 xx 10^(-16))/(1.6022 xx 10^(-19)) = 1560`

Discussion

633.	If `n = 3, l = 0, m = 0` then atomic number isA. 12, 13B. 13, 14C. 10, 11D. 11, 12
Answer» Correct Answer - D `n = 3, l = 0` means last shell is 3s `:. E.C. " will be " 1s^(2) 2s^(2) 2p^(6) 3s^(1 -2)` Atomic no. is 11 or 12

Discussion

634.	How many neutrons are there in `""_(38)^(88)Sr` ?A. 38B. 50C. 126D. 88
Answer» Correct Answer - B In `._(38)^(88)Sr` Atomic number = No. of protons = No. of electrons = 38 Atomic mass = 88 Number of neutrons = 88 - 38 = 50

Discussion

635.	How many number of electrons are present in a particle which carries a charge of `5.5xx10^(-16)C?`A. 3432B. 1560C. 8240D. 2432
Answer» Correct Answer - A Charge carried by one electron = `1.6022xx10^(-19)C` Electrons present in `5.5xx10^(-16)C =(5.5xx10^(-16))/(1.6022xx10^(-19))=3432`

Discussion

636.	The best way to think about a wave function is to regard it as an expression whose______defines the probability of finding the electron within a given volume of space around the nucleus.A. cubeB. squareC. square rootD. cube root
Answer» Correct Answer - B In the quantum mechanical description of the atom, the idea of an electron particles is replaced by the concept of the electron wave, i.e., associating wave packets. The value of the wave function `(psi)` used to describe the wave packet of the wave at any in space is related to the amplitude of the wave at that point. its intensity of vibration at that point in space is proportional to the square of the wavefunction `(\|psi\|^(2))` at that point.

Discussion

637.	The number of nodes in `3p` orbitalA. 3B. 4C. 2D. 1
Answer» Correct Answer - D The number of radial nodes for 3p orbitals : (n-l-1)=3-1-1=1

Discussion

638.	The probability of finding out an electron at a point within an atom is proportional to theA. square of the orbital wave function i.e., `Psi^(2)`B. orbital wave function i.e., `Psi`C. Hamiltonian operator i.e., HD. principal quantum number i.e., n
Answer» Correct Answer - A

Discussion

639.	Calculate the energy required for the process `He^(+) (g) rarr He^(2+) (g) + e^(-)`. The ionization energy for the H atom in the ground state is `2.18 xx 10^(-18) J "atom"^(-1)`
Answer» For H-like particles, `E_(n) = - (2pi^(2) m Z^(2) e^(4))/(n^(2) h^(2))` For H-atom, I.E. `= E_(oo) - E_(1) = 0 - (-(2pi^(2) me^(4))/(1^(2) xx h^(2))) = (2pi^(2) me^(4))/(h^(2)) = 2.18 xx 10^9-18) J "atom"^(-1)` (Given) For the given process, Energy required `= E_(oo) - E_(1) = 0 = (-(2pi^(2) m xx 2^(2) xx e^(4))/(1^(2) xx h^(2))) = 4 xx (2pi^(2) me^(4))/(h^(2)) = 4 xx 2.18 xx 10^(-18) J` `= 8.72 xx 10^(-18) J`

Discussion

640.	The diameter of zinc atom is `2.6 A`. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of `1.6 cm` if the zinc atoms are arranged side by side lengthwise.
Answer» (a) Radius of zinc atom = `("Diameter")/(2)` = `(2.6 Å)/(2)` = `1.3 xx 10^(-10) m ` = `130 xx 10^(-12) m = 130 `pm (b) Length of the arrangement = `1.6` cm = `1.6 xx 10^(-2)` m Diameter of zinc atom = `2.6 xx 10^(-10) m ` `therefore` Number of zinc atoms present in the arrangement = `(1.6 xx 10^(-2) m)/(2.6 xx 10^(-10 m )` = `0.6153 xx 10^(8) m ` = `6.153 xx 10^(7) `

Discussion

641.	Calculate the energy required for the process , ` He^+ (g) rarr He^(2+) (g) + e` The ionization energy for the H-atom in the grounds state is ` 2. 18 xx 10 ^(-18) J "atom"^(-1)`.
Answer» Energy associated with hydrogen-like species is given by, `E_(a) = -2.18 xx 10^(-18) ((Z^(2))/(n^(2))) J` For ground state of hydrogen atom, `Delta E = E_(oo) - E_(1)` = `0 - [-2.18 xx 10^(-18) {((1)^(2))/((1)^(2)) }]J` `Delta E = 2.18 xx 10^(-18) J` For the given process , `He_((g))^(+) to He_((g))^(2+) + e^(-)` An electron is removed from n = 1 to n = `oo` `Delta E = E_(oo) - E_(1)` = `0 - [-2.18 xx 10^(-18) {((2)^(2))/((1)^(2)) }]` `Delta E = 8.72 xx 10^(-18) J` `therefore` The energy required for the process is `8.72 xx 10^(-18) J`.

Discussion

642.	A certain metal when irradiated to light `(v = 3.2 xx 10^(16) Hz)` emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiation by light `( v = 2.0 xx 10^(16)Hz)`. The `v_(0)` Threshold frequency ) of the metal isA. `1.2xx10^(14)`HzB. `8xx10^(15)` HzC. `1.2xx10^(16)` HzD. `4xx10^(12)` Hz
Answer» Correct Answer - B `(K.E)_(1)=hv_(1)-hv_(0) (K.E)_(2)=hv_(2)-hv_(0) As" "(K.E)_(1)=2xx(K.E)_(2) therefore" "(hv_(1)-hv_(0))=2(hv_(2)-hv_(0)) or" "v_(0)=2v_(2)-v_(1)=2xx(2xx10^(16))-(3.2xx10^(16)) =0.8xx10^(16)Hz or 8xx10^(15)Hz`

Discussion

643.	When a certain metal was irradiated with light of frequency `1.6xx10^(16)Hz`, the photoelectrons emitted had twice the kinetic energy as photoelectrons emitted when the same metal was irradiated with light of frequency `1.0xx10^(6)Hz`. Calculate the threshold frequency `(v_(0))` for the metal.
Answer» Correct Answer - `4xx10^(15)Hz` According to available information : `KE_(1)=h(v_(1)-v_(0))=h(1.6xx10^(16)-v_(0))` `KE_(2)=h(v_(2)-v_(0))=h(1.0xx10^(16)-v_(0))` Dividing eqn. (ii) by (i), we have `(KE_(2))/(KE_(1))=((1.0xx10^(16)-v_(0)))/((1.6xx10^(16)-v_(0)))` or `(KE_(1))/(2xxKE_(1))=((1.0xx10^(16)-v_(0)))/((1.6xx10^(16)-v_(0)))(KE_(2)=1//2KE_(1))` or` ((1.0xx10^(16)-v_(0)))/((1.6xx10^(16)-v_(0)))=(1)/(2) , 2.0xx10^(16)-2v_(0)=1.6xx10^(16)-v_(0)` or ` v_(0)=(2.0-1.6)xx10^(16)=0.4xx10^(16)=0.4xx10^(16)=4xx10^(15)`Hz.

Discussion

644.	Calcualte the maximum kinetic energy of photoelectrons emitted where as light of freqency `2 xx 10^(16)Hz` is irradiated on a metal surface with threshold frequency `(v_(0))` equal to`8.68 xx10^(15)Hz`.
Answer» `hv= hv_(0)+KE` Threshold frequency `(v_(0)) = 8.68 xx 10^(15) Hz ( s^(-1))` Frequency of light `(v) = 2 xx 10^(16) Hz` KE `= h(v-v_(0))` `= 6.626 xx 10^(-34) ( 2 xx 10^(16) - 8.68 xx 10^(-15))` `KE = 7.5 xx 10^(-18)J` Maximum kinetic energy of photoelectrons `= 7.5 xx 10^(-18) J`

Discussion

645.	A beam of `Ne^(+)` ions in the mass spectrometer is split into________segments.A. twoB. threeC. fourD. no splitting at all
Answer» Correct Answer - B Neon occurs in nature as three isotopes: `._(10)^(20)Ne,_(10)^(21)Ne`, and `._(10)^(22)Ne` Neon was first element to be separated into isotopes.

Discussion

646.	Which electronic level would allow the hydrogen atom to absorbs a photon but not to emit a photonA. 1s and 2s orbitals are spherical in shape.B. 2sC. 3sD. 4s
Answer» Correct Answer - A

Discussion

647.	A 600 W mercury lamp emits monochromatic radiation of wavelength 313.3 nm. How many photons are emitted from the lamp per second ? (`h = 6.626 xx 10^(-34) Js`, velocity of light `= 3xx 10^(8) ms^(-1)`)A. `1 xx 10^(19)`B. `1 xx 10^(20)`C. `1 xx 10^(21)`D. `1 xx 10^(23)`
Answer» Correct Answer - C `600 W = 600 J s^(-1) (1 W = 1 Js^(-1))` i.e., Energy emitted per sec = 600 J Energy of one photon, `E = hv = h (c)/(lamda)` `= (6.626 xx 10^(-34) Js xx 3 xx 10^(8) ms^(-1))/(313.3 xx 10^(-9) m)` `= 6.344 xx 10^(-19) J` `:.` Photons emitted per sec `= (600)/(6.344 xx 10^(-19))` `~= 1 xx 10^(21)`

Discussion

648.	Calculate the energy of one mole of photons of radiation whose frequency is `5 xx 10^(14) Hz`.A. 199.51 kJB. `3.3xx10^(-19)J`C. `6.626xx10^(-34)J`D. `2.31xx10^(5)J`
Answer» Correct Answer - B Energy of one photon is given by the expression, E = hv E = `6.626xx10^(-34)Jsxx5xx10^(14)s^(-1)=3.313xx10^(19)J`

Discussion

649.	Find the energy of each of the photons which (i) corresponds to light of frequency `3 xx 10^(15)Hz` (ii) have wavelength of `0.50 Å`
Answer» (i) `v = 3 xx 10^(15) Hz, E = hv = (6.626 xx 10^(-34) Js) (3 xx 10^(15) s^(-1)) = 1.988 xx 10^(-18) J` (ii) `lamda = 0.50 xx 10^(-10) m, E = hv = h (c)/(lamda) = ((6.626 xx 10^(-34) Js) (3 xx 10^(15)s^(-1)))/(0.50 xx 10^(-10) s) J = 3.98 xx 10^(-15) J`

Discussion

650.	Which of the following properties of atom could be explained correctly by Thomson model of atom?A. Overall neutrality of atom.B. Spectra of hydrogen atom.C. Position of electrons, protons and neutrons in atom.D. Stability of atom.
Answer» Correct Answer - A Only overall neutrality of an atom could be explained correctly by Thomson model of atom.

Discussion

Explore topic-wise InterviewSolutions in .

The unpaired electrons in `Al` and `Si` are present in `3p` orbital. Which electrons will experience more effective nuclear charge from the nucleus?

The nuclear radius is of the order of `10^(-13)` cm while atomic radius is of the order `10^(-8)` cm. Assuming the nucleus and the atom to be spherical, what fraction of the atomic volume occupied by the nucleus ?

Calculate (a) wave number and (b) frequency of yellow radiations having wavelength of `5800Å`

Calculate the wavelength (in nanometer) associated with a proton moving at `1.0 xx 10^(3) ms^(-1)`A. 0.032 nmB. 0.40 nmC. 2.5 nmD. 14.0 nm

The two extra -nuclear electrons in the orbital of helium have antiparallel spins `(uarr darr)`. Why not they have parallel spins `(uarr uarr)` ?

Which of the following orbitals are possible ? `1pm, 2s, 3f, 3d`

The 3d-orbitals having electron density in all the three axes isA. `3d_(xy)`B. `3d_(z^(2))`C. `3d_(yz)`D. `3d_(zx)`

An electron is in one of the `3d` orbitals. Give the possible values of `n, l,` and `m` for this electron.

Give the number of electrons in the species `H_(2)^(+), H_(2)` and `O_(2)^(o+)`

An atom of an element contains `29` electrons and `35` neutrons. Deduce a. The number of protons and b. The elctonic configuration of the element.

An atom of an element contains `29` electrons and `35` neutrons. Deduce a. The number of protons and b. The electronic configuration of the element.

An element with mass number `81` contains `31.7%` more neutrons as compared to protons. Assign the atomic symbol.

Write down the quantum numbers n, l and m for the following orbitals: (i) `3d_(x^(2) - y^(2))` (ii) `4d_(z^(2))` (iii) `3d_(xy)` (iv) `4d_(xz)` (v) `2p_(z)` (vi) `3p_(x)`

Match the following `:` `{:(,"Column-I (Orbitals)",,"Column-II (Nature and No. of Nodes)"),((A),3d_(z^(2)),(p),"One sperhical node"),((B),4d_(xy),(q), "Two nodal planes"),((C ) , 2s,(r ),"Node in yz plane"),((D) ,3p_(x),(s) , "Two angular nodes"):}`

Which of the following orbitals are degenerate ? `3d_(xy), 4d_(xy) 3d_(z^(2)), 3d_(yz), 4d_(yz), 4d_(z^(2))`

Match the following `:` `{:(,"Column-I",,"Column-II"),((A),3d_(xy),(p),Psi=0[zx "plan"]),((B),2p_(y),(q),Psi=0[yz "plane"]),((C ) , 2s,(r ),Psi= "max"["yz plane"]),((D),2s,(s),Psi="max[zx plane]"),(,,(t),Psi"versus r graph has wo maximas"):}`

A student while heating solid lead nitrate taken in a test tube would observe :A. white residue of `PbO_(2)`B. green residue of `NO_(2)`C. yellow residue of PbOD. brown residue of NO

A student took some lead nitrate compound in a boiling tube and heated it strongly. The gas/gases evolved on heating this compound is/are :A. `NO_(2)`B. `NO_(2)+CO_(2)`C. `NO_(2)+O_(2)`D. `N_(2)+O_(2)`

The colour of residue left behind on heating lead nitrate when it is still hot is :A. greenB. yellowC. blackD. reddish-brown

In which of the following orbitals , the sum of `( n+1)` value is lowest ? `4s, 3d, 4p`

Which of the following combinations of quantum numbers is possible for a `4p` orbital?A. `n = 4,l = 1, m_(l) = 0, +-1`B. `m= 4, l = 1, m_(l) =- 1`C. `n = 4, l = 1,m_(l) = 0`D. `n = 4, l = 1, m_(l) = +1`

Radial wave function depends on the quantum numbers (i) `n` (ii) l` `(iii) m_(l)` (iv) m_(s)`A. `(ii),(iii)`B. `(i),(ii)`C. `(i),(ii),(iii)`D. `(i),(ii),(iii),(iv)`

An electron in the hydrogen atom absorbs energy and jumps to the 4th orbit. Then it jumps back to the ground state in steps, i.e., from 4th to 3rd orbit, then from 3rd to 2nd orbit and finally to the ground state. The total number of lines produced in the spectrum will beA. 3B. 6C. 9D. 12

A student placed a clean iron nail in blue coloured copper sulphate solution for a considerable time. He observes that :A. iron nail gets green coatingB. iron nail gets brown coatingC. iron nail gets no coatingD. iron nail gets blue coating

A student was asked to carry out a chemical reaction by placing four different metal strips in `CuSO_(4)` solution for a considerable time, one by one. Which of the following metal strip will turn the blue `CuSO_(4)` solution to a light green solution in due course of time ?A. FeB. AuC. MgD. Ag

A student was asked by her science teacher to prepare a true solution. She dissolved the solute in water but forget to record its name. What may be the correct name of the solute ?A. barium sulphateB. sulphur powderC. alumD. egg albumin

`P` is the probability of finding the Is electron of hydrogen atom in a spherical shell of infitesimal thickness, dr, at a distance `r` from the nucleus. The volume of this shell is `4pi r^(2)dr`. The qualitative sketch of the dependence of `P` on r isA. B. C. D.

A certain particle carries `2.5 xx 10^(-16) C` of static electric charge. Calculate the number of electrons present in it.

If `n = 3, l = 0, m = 0` then atomic number isA. 12, 13B. 13, 14C. 10, 11D. 11, 12

How many neutrons are there in `""_(38)^(88)Sr` ?A. 38B. 50C. 126D. 88

How many number of electrons are present in a particle which carries a charge of `5.5xx10^(-16)C?`A. 3432B. 1560C. 8240D. 2432

The best way to think about a wave function is to regard it as an expression whose______defines the probability of finding the electron within a given volume of space around the nucleus.A. cubeB. squareC. square rootD. cube root

The number of nodes in `3p` orbitalA. 3B. 4C. 2D. 1

The probability of finding out an electron at a point within an atom is proportional to theA. square of the orbital wave function i.e., `Psi^(2)`B. orbital wave function i.e., `Psi`C. Hamiltonian operator i.e., HD. principal quantum number i.e., n

Calculate the energy required for the process `He^(+) (g) rarr He^(2+) (g) + e^(-)`. The ionization energy for the H atom in the ground state is `2.18 xx 10^(-18) J "atom"^(-1)`

The diameter of zinc atom is `2.6 A`. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of `1.6 cm` if the zinc atoms are arranged side by side lengthwise.

Calculate the energy required for the process , ` He^+ (g) rarr He^(2+) (g) + e` The ionization energy for the H-atom in the grounds state is ` 2. 18 xx 10 ^(-18) J "atom"^(-1)`.

When a certain metal was irradiated with light of frequency `1.6xx10^(16)Hz`, the photoelectrons emitted had twice the kinetic energy as photoelectrons emitted when the same metal was irradiated with light of frequency `1.0xx10^(6)Hz`. Calculate the threshold frequency `(v_(0))` for the metal.

Calcualte the maximum kinetic energy of photoelectrons emitted where as light of freqency `2 xx 10^(16)Hz` is irradiated on a metal surface with threshold frequency `(v_(0))` equal to`8.68 xx10^(15)Hz`.

A beam of `Ne^(+)` ions in the mass spectrometer is split into________segments.A. twoB. threeC. fourD. no splitting at all

Which electronic level would allow the hydrogen atom to absorbs a photon but not to emit a photonA. 1s and 2s orbitals are spherical in shape.B. 2sC. 3sD. 4s

A 600 W mercury lamp emits monochromatic radiation of wavelength 313.3 nm. How many photons are emitted from the lamp per second ? (`h = 6.626 xx 10^(-34) Js`, velocity of light `= 3xx 10^(8) ms^(-1)`)A. `1 xx 10^(19)`B. `1 xx 10^(20)`C. `1 xx 10^(21)`D. `1 xx 10^(23)`

Calculate the energy of one mole of photons of radiation whose frequency is `5 xx 10^(14) Hz`.A. 199.51 kJB. `3.3xx10^(-19)J`C. `6.626xx10^(-34)J`D. `2.31xx10^(5)J`

Find the energy of each of the photons which (i) corresponds to light of frequency `3 xx 10^(15)Hz` (ii) have wavelength of `0.50 Å`

Which of the following properties of atom could be explained correctly by Thomson model of atom?A. Overall neutrality of atom.B. Spectra of hydrogen atom.C. Position of electrons, protons and neutrons in atom.D. Stability of atom.