Structure of Atom · Question 47

Consider the following sets of quantum numbers: array|c|c|c|c|c| & n & l & m & s \ (i) & 3 & 0 & 0 & +1/2 \ (ii) & 2 & 2 & 1 & +1/2 \ (iii) & 4 & 3 & -2 & -1/2 \ (iv) & 1 & 0 & -1 & -1/2 \ (v) & 3 & 2 & 3 & +1/2 \ array Which of the following sets of quantum numbers are not possible?

1. (i), (ii), (iii) and (iv)
2. (ii), (iv) and (v)
3. (i) and (iii)
4. (ii), (iii) and (iv)

The orientation of an atomic orbital is governed by

1. principal quantum number
2. azimuthal quantum number
3. spin quantum number
4. magnetic quantum number.

Given : The mass of electron is 9.11 10-31 kg, Planck constant is 6.626 10-34 J s, the uncertainty involved in the measurement of velocity within a distance of 0.1 A is

1. 5.79 105 m s-1
2. 5.79 106 m s-1
3. 5.79 107 m s-1
4. 5.79 108 m s-1

The energy of second Bohr orbit of the hydrogen atom is -328 kJ mol-1; hence the energy of fourth Bohr orbit would be

1. -41 kJ mol-1
2. -82 kJ mol-1
3. -164 kJ mol-1
4. -1312 kJ mol-1

The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in hydrogen atom will be (Given ionization energy of H = 2.18 10-18 J atom-1 and h = 6.625 10-34 J s)

1. 1.54 1015 s-1
2. 1.03 1015 s-1
3. 3.08 1015 s-1
4. 2.00 1015 s-1

The radioactive isotope 6027Co which is used in the treatment of cancer can be made by (n, p) reaction. For this reaction the target nucleus is

1. 5928Ni
2. 5927Co
3. 6028Ni
4. 6027Co

The value of Planck's constant is 6.63 10-34 J s. The velocity of light is 3.0 108 m s-1. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8 1015 s-1 ?

1. 2 10-25
2. 5 10-18
3. 4 101
4. 3 107

In hydrogen atom, energy of first excited state is -3.4 eV. Then find out K.E. of same orbit of hydrogen atom

1. +3.4 eV
2. +6.8 eV
3. -13.6 eV
4. +13.6 eV

23592U, nucleus absorbs a neutron and disintegrates into 13954Xe, 9438Sr and x so, what will be product x?

1. 3-neutrons
2. 2-neutrons
3. -particle
4. -particle

The following quantum numbers are possible for how many orbitals : n = 3, l = 2, m = +2 ?

1. 1
2. 2
3. 3
4. 4