In a hydrogen atom, if energy of an electron in ground state is 13.6. ev, then that in the 2^{nd} excited state is [2002]
2^{nd} excited state will be the 3^{rd }energy level.
Uncertainty in position of a minute particle of mass 25 g in space is 10^{–5} m. What is the uncertainty in its velocity (in ms^{–1}) ? (h = 6.6 x 10^{34} Js) [2002]
TIPS/Formulae :
= 2.1 x 10^{28} ms^{1}
The number of delectrons retained in Fe^{2+} [2003] (At. no. of Fe = 26) ion is
Fe^{++} (26 – 2 = 24) = 1s^{2} 2s^{2} 2p^{6 }3s^{2} 3p^{6} 4s^{0} 3d^{6} hence no. of d electrons retained is 6. [Two 4s electron are removed]
The orbital angular momentum for an electron revolving in an orbit is given by This momentum for an selectron will be given by [2003]
TIPS/Formulae : For selectron, ℓ = 0
∴Orbital angular momentum =
Which one of the following groupings represents a collection of isoelectronic species ?(At. nos. : Cs : 55, Br : 35)[2003]
N^{3–}, F^{–} and Na^{+} contain 10 electrons each.
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following interorbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]
Planck’s constant, h = 6.63 × 10^{–34} Js
The lines falling in the visible region comprise Balmer series. Hence the third line from red would be n_{1 }=2,
n_{2} = 5 i.e. 5 → 2.
The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately [2003]
Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? [2004]
The possible quan tum numbers for 4f electron are n = 4, ℓ = 3, m = – 3, –2 –1, 0, 1 , 2 , 3 and s =
Of various possiblities only option (a) is possible.
Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, ℓ = 1 and 2 are, respectively [2004]
Electronic configuration of Cr atom (z = 24) = 1s^{2}, 2s^{2} 2p^{6}, 3s^{2} 3p^{6} 3d^{5}, 4s^{1}
when ℓ = 1, p  subshell,
Numbers of electrons = 12
when ℓ = 2, d  subshell,
Numbers of electrons = 5
The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×10^{7} m^{–1}) [2004]
TIPS/Formulae :
λ = 91.15 x 10 ^{9} m ≈ 91nm
Which one of the followin g sets of ion s represents the collection of isoelectronic species? [2004]
(Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)
each contains 18 electrons.
In a multielectron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields? [2005]
(A) n = 1, l = 0, m = 0
(B) n = 2, l = 0, m = 0
(C) n = 2, l = 1, m = 1
(D) n = 3, l = 2, m = 1
(E) n = 3, l = 2, m = 0
The energy of an orbital is given by (n + l) in (d) and (c). (n + l) value is (3 + 2) = 5 hence they will have same energy, since there n values are also same.
Of the following sets which one does NOT contain isoelectronic species? [2005]
Calculating number of electrons
Hence the species in option (b) are not isoelectronic.
According to Bohr's theory, the angular momentum of an electron in 5^{th} orbit is [2006]
Angular momentum of an electron in nth orbital is given by ,
For n = 5, we have
Angular momentum of electron
Uncertainty in the position of an electron (mass = 9.1 × 10^{–31} kg) moving with a velocity 300 ms^{–1}, accurate upto 0.001% will be [2006]
(h = 6.63 × 10^{–34} Js)
Given m = 9.1 × 10^{–31}kg , h = 6.6 × 10^{–34}Js
= 0.003ms^{–1}
From Heisenberg's uncertainity principle
= 1.92 x 10^{2} m
Which one of the following sets of ions represents a collection of isoelectronic species? [2006]
(a)
(b) Li^{+} = 3+1= 4e^{–}, Na^{+} = 11–1 = 10e^{–},
Mg^{++ } = 12–2=10e^{–}
Ca^{++} = 20 – 2 = 18e^{– } (not isoelectronic)
(c) K^{+} = 19 – 1= 18e^{–}, Cl^{–} =17 + 1 = 18e^{–},
Ca^{++} = 20 – 2 =18e, Sc3^{+ }= 21–3 = 18e^{–}
(isoelectronic)
(d) Ba^{++} 56 – 2 = 54e, Sr^{++} 38–2 = 36e^{– }
K^{+}= 9–1 = 18e^{–}, Ca^{++}= 20–2 = 18e^{–}
(not isoelectronic)
Which of the following sets of quantum numbers represents the highest energy of an atom? [2007]
(a) n = 3, ℓ = 0 means 3sorbital and n + ℓ = 3
(b) n = 3, ℓ = 1 means 3porbital n + ℓ = 4
(c) n = 3, ℓ = 2 means 3dorbital n + ℓ = 5
(d) n = 4, ℓ = 0 means 4sorbital n + ℓ = 4
Increasing order of energy among these orbitals is 3s < 3p < 4s < 3d
∴ 3d has highest energy..
Which one of the following constitutes a group of the isoelectronic species? [2008]
Species having same n umber of el ectr on s ar e isoelectronic calculating the number of electrons in each species given here, we get.
CN^{–} (6 + 7 + 1 = 14); N_{2} (7 + 7 = 14);
O_{2}^{2–}(8 + 8 +2 = 18) ; C_{2}^{2–} (6 + 6 + 2 = 14);
O_{2}^{–} (8 + 8 + 1 = 17) ; NO^{+} (7 + 8 – 1 = 14)
CO (6 + 8 = 14) ; NO (7 + 8 = 15)
From the above calculation we find that all the species listed in choice (b) have 14 electrons each so it is the correct answer.
The ionization enthalpy of hydrogen atom is 1.312 × 10^{6} J mol^{–1}. The energy required to excite the electron in the atom from n = 1 to n = 2 is [2008]
(ΔE), The energy required to excite an electron in an atom of hydrogen from n = 1 to n = 2 is ΔE (difference in energy E_{2 }and E_{1}) Values of E_{2} and E_{1} are, E_{1 }= – 1.312 × 10^{6} J mol^{–1}
= –3.28 × 10^{5} J mol^{–1}
ΔE is given by the relation,
∴ ΔE = E_{2 }– E_{1} = [–3.28 × 10^{5}]– [–1.312 × 10^{6} ] J mol^{–1 }
= (–3.28 × 10^{5} + 1.312 × 10^{6}) J mol^{–1}
= 9.84 × 10^{5} J mol^{–1}
Thus the correct answer is (d)
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 10^{3} ms^{ –1}. (Mass of proton = 1.67 × 10^{–27 }kg and h = 6.63 × 10^{–34} Js)
= 3.97 × 10^{–10 }meter = 0.397 nanometer
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is ( h = 6.6 × 10^{–34} kg m^{2}s^{–1}, mass of electron, em = 9.1 × 10^{–31} kg) : [2009]
According to Heisenberg uncertainty principle.
Here = 0.03
So,
= 1.92 × 10^{–3} meter
The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol^{–1}. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 × 10^{8} ms^{–1} and NA = 6.02 × 10^{23} mol^{–1}). [2010]
Energy required to break one mole of Cl – Cl bonds in Cl_{2}
= 0.4947 × 10^{–6} m = 494.7 nm
Ionisation energy of He^{+} is 19.6 × 10^{–18 }J atom^{–1}. The energy of the first stationary state (n = 1) of Li^{2+} is [2010]
Given I_{1} = – 19.6 × 10^{–18} , Z_{1 }= 2, n_{1 }= 1 , Z_{2} = 3 and n_{2} = 1
Substituting these values in equation (ii).
= – 4.41 × 10^{–17 }J/atom
A gas a bsor bs a ph oton of 355 n m an d emits at two wavelengths. If one of the emissions is at 680 nm, the other is at:[2011]
Energy of absorbed photon = Sum of the energies of emitted photon
= 1.346 x10^{6}
or λ_{2}= 1/1.346 × 10^{6} = 743 × 10^{–9}m = 743 nm
The electrons identified by quantum numbers n and l :
(A) n = 4, l = 1
(B) n = 4, l = 0
(C) n = 3, l = 2
(D) n = 3, l = 1
can be placed in order of increasing energy as : [2012]
(a) 4 p (b) 4 s (c) 3 d (d) 3 p
Accroding to Bohr Bury's (n + l) rule, increasing order of energy (D) < (B) < (C) < (A).
Note : If the two orbitals have same value of (n + l) then the orbital with lower value of n will be filled first.
Energy of an electron is given by E = – 2.178 ×
Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be : (h = 6.62 × 10 ^{–34} Js and c = 3.0 × 10^{8} ms^{–1}) [JEE M 2013]
= 1.214 × 10^{–7}m
The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is: [JEE M 2014]
The electronic configuration of Rubidium (Rb = 37) is
1s^{2}2s^{2} 2 p^{6} 3s^{2}3 p^{6} 3d^{10} 4s^{2} 4 p^{6} 5s^{1 }
Since last electron enters in 5s orbital Hence n = 5, l = 0, m = 0, s =
Which of the following is the energy of a possible excited state of hydrogen ? [JEE M 2015]
Total energy =
where n = 2, 3, 4 ....
Putting n = 2
= 3.4eV
A stream of electrons from a heated filaments was passed two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by: [JEE M 2016]
As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron will be eV
Wavelength of electron wave
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 




