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The electrons are more likely to be found: 

  • Option 1)

    in the region a and c

  • Option 2)

    in the region a and b 

  • Option 3)

    only in the region a 

  • Option 4)

    only in the region c

  The energy (E) of a quantum of radiation - Where h is plank’s constant and  is frequency -     Bohr's model - 1. Force of attraction between the nucleus and an electron is equal to centripetal force. 2.      n = principal quantum number. 3. Energy can be absorbed or emitted when electron transfer orbit   -   We knew that probability of finding an electron is proportional  Now on converting...

The graph between \left | \psi \right |^{2} and  r ( radial distance ) is shown below.

This represents :

  • Option 1)

    3s orbital

  • Option 2)

    2s orbital

  • Option 3)

    1s orbital

  • Option 4)

    2p orbital

For 's' , graph always starts from above. For every other orbital, graph starts from origin.  this graph is for 's'-orbital. => Thus , l=0 => In this graph, only one radial node is present ( point A). We know that, radial node is given by (n-l-1)  n-l-1 = 1( as only one radial node is present) n-1 = 1 (  l = 0) => n = 2  graph is for 2s So, option (2) is correct.Option 1)3s orbitalOption 2)2s...

Among the following, the energy of 2s orbital is lowest in:

 

  • Option 1)

    K

  • Option 2)

    H

  • Option 3)

    Li

  • Option 4)

    Na

  (hydrogen-like species)  (for other species) for minimum 2S orbital energy  Zeff more E less Zeff of K is the highest  Zeff more means nuclear charge more on K    Option 1)KOption 2)HOption 3)LiOption 4)Na

The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9 . The spectral series are :

  • Option 1)

    Lymen and Paschen

  • Option 2)

    Balmer and Brackett

  • Option 3)

     Brackett and Pfund 

  • Option 4)

    Paschen and Pfund

for shortest wavelength  For Lymen =          Balmer=          Paschen=         Brackett=           Pfund=          Option 1)Lymen and PaschenOption 2)Balmer and BrackettOption 3) Brackett and Pfund Option 4)Paschen and Pfund

Which one of the following about an electron occupying the 1s orbitals in a hydrogen atom is incorrect ? ( The Bohr radius is represented by a_{o} )

  • Option 1)

    The probability density of finding the electron is maximum at the nucleus.

  • Option 2)

    The electron can be found at a distance 2a_{o} from the nucleus.

  • Option 3)

    The magnitude of the potential energy is double that of its kinetic energy on an average .

  • Option 4)

    The total energy of the electron is maximum when it is at a distance a_{o} from the nucleus.

  Total energy of elctron in nth orbit - Where z is atomic number -       a. , so probability density of finding the electron is maximum at the nucleus .   b. True, the electron can be found at any distance from the nucleus. c.    on average. d.                   is minimum not maximum is incorrect.  Option 1)The probability density of finding the electron is maximum at the nucleus.Option...

For any given series of spectral lines of atomic hydrogen, let \Delta \overline{v}=\overline{v}_{max}-\overline{v}_{min} be the difference in maximum and minimum frequencies in cm^{-1}.

The ratio \Delta \overline{v}_{Lyman}/\Delta \overline{v}_{Balmer} is :

  • Option 1)

    4:1

  • Option 2)

    9:4

  • Option 3)

    5:4

  • Option 4)

    27:5

 
    Lyman Series :     (electron jump)                                                                                                                                                       Balmen Series :                                                                                                                                                         Option 1) Option 2) Option 3) Option 4)

If P is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength \lambda,  then for 1.5 p momentum of the photoelectron, the wavelength of the light should be:

(Assume kinetic energy of elected photoelectron to be very high in comparison to work function):

  • Option 1)

    \frac{3}{4}\lambda

  • Option 2)

    \frac{1}{2}\lambda

  • Option 3)

    \frac{2}{3}\lambda

  • Option 4)

    \frac{4}{9}\lambda

given                    (K.E)            (work function)  Option 1)Option 2)Option 3)Option 4)

The quantum number of four electrons are given below:

I.n=4,l=2,m_{l}=-2,m_{s}=-1/2

I\! I.n=3,l=2,m_{l}=1,m_{s}=+1/2

I\!I \! I.n=4,l=1,m_{l}=0,m_{s}=+1/2

IV.n=3,l=1,m_{l}=1,m_{s}=-1/2

The correct order of their increasing energies will be :

 

  • Option 1)

    I<II<III<IV

  • Option 2)

    IV<III<II<I

  • Option 3)

    IV<II<III<I

     

  • Option 4)

    I<III<II<IV

 
    I 4 2 6 4d II 3 2 5 3d III 4 1 5 4p IV 3 1 4 3p Higher the value of () energy. If two orbitals have same value of () , the orbital with value of n will have higher energy. Option 1) Option 2) Option 3)   Option 4)
According to Bohr's model - 1. Force of attraction between the nucleus and an electron is equal to centripetal force. 2.      n = principal quantum number. 3. Energy can be absorbed or emitted when electron transfers between the orbits    The orbital angular momentum is given by  where l is the azimuthal quantum number for p orbital,   orbital angular momentum =                                    

8 in oxygen and two electrons in hydrogen

goodnight sir have a nice day thankyou so much
Effective nuclear depends upon 2 things: (i) Nuclear charge: Total charge present in the nucleus that attracts the electrons towards itself. (ii) Number of shielding electrons: These are the electrons present inside in the atom that repel the valence electron and thus decrease the nuclear attraction to the electrons. Effective nuclear charge is always less than the actual nuclear charge.

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm . Which spectral line of H-atom is suitable for this purpose ?

\left [ R_{H}= 1\times 10^{5}cm^{-1},h= 6.6\times 10^{-34}Js,c= 3\times 10^{8}ms^{-1} \right ]

  • Option 1)

    Paschen , 5\rightarrow 3

     

  • Option 2)

    Lyman,\: \: \infty \rightarrow 1

  • Option 3)

    \; \; \; Balmer,\: \: \infty \rightarrow 2\: \:

     

  • Option 4)

    Paschen,\: \: \infty \rightarrow 3

  Paschen , Bracket and Pfund Series spectrums - Infrared Region -     Line Spectrum of Hydrogen like atoms -   - wherein Where R is called Rhydberg constant, R = 1.097 X 107 , Z is atomic number n1= 1,2 ,3…. n2= n1+1, n1+2 ……     Balmer Series Spectrum - Where It lies in visible region As we know that Paschen, 3 -is correct    Option 1)Paschen ,   Option 2)Option 3)  Option 4)

The de Broglie wavelength (\lambda) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency] :

  • Option 1)

    \lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{1}{4}}}

     

     

     

  • Option 2)

    \lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{3}{2}}}

     

  • Option 3)

    \lambda \: \alpha \frac{1}{(v-v_{0})}

  • Option 4)

    \lambda \: \alpha \frac{1}{(v-v_{0})^{\frac{1}{2}}}

  De-broglie wavelength - - wherein where m is the mass of the particle v its velocity  p its momentum   Orbital frequency - - wherein m & e are mass and charge of electron  h = planck's constant z = atomic number n = principal quantum number Formula  ???????   Option 1)      Option 2)  Option 3)Option 4)

If the de Broglie wavelength of the electron in n th Bohr orbit in a hydrogenic atom is equal to 1.5 \pi a_{0}(a0 is Bohr radius), then the value of n/z is :

 

  • Option 1)

    0.75

  • Option 2)

    1.0

  • Option 3)

    1.50

  • Option 4)

    0.40

  De-broglie wavelength - - wherein where m is the mass of the particle v its velocity  p its momentum       Atomic number ( Z ) - Z = number of protons in the nucleus of an atom As we have learned in debroglics wavelength we know -    Option 1)0.75Option 2)1.0Option 3)1.50Option 4)0.40

THe 71s electron of an element X with an atomic  number of 71 enters into the orbital:

  • Option 1)

    6p

  • Option 2)

    5d

  • Option 3)

    4f

  • Option 4)

    6s

  Aufbau Principle - In the ground state of the atoms, the orbitals are filled in order of their increasing energies. - wherein     Pauli Exclusion Principle - “No two electrons in an atom can have the same set of four quantum number" . -     Hund’s Rule of Maximum Multiplicity - "pairing of electrons in the orbitals belonging to the same subshell ( p,d or f ) does not take place until each...

Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface ?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Photoelectric Effect - The electrons are ejected from the metal surface as soon as the beam of light of a particular frequency strikes the surface. -   As we learned in Atomic sttructure Kinetic energy                          Option 1)Option 2)Option 3)Option 4)

L = 2 means d orbitals

electronic configuration of Fe2+ = [Ar] 3d6

six electrons in d orbitals 4 will be unpaired 2 will be paired

electronic configuration of Co2+ = [Ar] 3d7

three electrons will be unpaired four will be paired

electronic configuration of Ni2+ = [Ar] 3d8 
two electrons will be unpaired and six will be paired

So the sum of paired electrons is 2+4+6=12.

The ground state energy of hydrogen atom is 13.6 eV. The energy of second excited state of He^{+} ion in eV is:

  • Option 1)

    -54.4

  • Option 2)

    -3.4

  • Option 3)

    -27.2

  • Option 4)

    -6.04

  Total energy of elctron in nth orbit - Where z is atomic number -   As we have learned in energy of atom Option 1)-54.4Option 2)-3.4Option 3)-27.2Option 4)-6.04

What is the work function of the metal if the light of wavelength 400\AA generates photoelectrons of velocity 6\times 10^{5}ms^{-1} from it ?

(Mass of electron = 9\times 10^{-31}kg

Velocity of light = 3\times 10^{8}ms^{-1}

Planck's constant = 6.626\times 10^{-34}Js

Charge of electron = 1.6\times 10^{-19}JeV^{-1})

  • Option 1)

    0.9eV

  • Option 2)

    3.1eV

  • Option 3)

    2.1eV

  • Option 4)

    4.0eV

  De-broglie wavelength - - wherein where m is the mass of the particle v its velocity  p its momentum     Total energy of elctron in nth orbit - Where z is atomic number As we have learned in work function -    Option 1)0.9eVOption 2)3.1eVOption 3)2.1eVOption 4)4.0eV

Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ?

(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in orbital of lower angular momentum .

(b) For a given value of the principal quantum number , the size of the orbit is inversely propotional to the azimuthal quantum number .

(c) According to wave mechanics , the ground state angular momentum is equal to \frac{h}{2\pi } .

(d) The plot of   \psi \; \; Vs\: \: r    for various azimuthal quantum numbers , shows peak shifting towards higher r value .

  • Option 1)

    (a) ,(c)

  • Option 2)

    (b) ,(c)

  • Option 3)

    (a) ,(d)

  • Option 4)

    (a) ,(b)

  Planck’s Quantum Theory - Atom and molecules could emit (or absorb) energy only in discrete quantities known as quanta and not in a continuous manner. -     Principal Quantum Number - The principal quantum number determines the size and too large extent the energy of the orbital. -     Principal Quantum Number (n) -  It is a positive integer with value of n = 1,2,3....... -     Azimuthal...
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