The complete symbol for the atom with the given atomic number (Z) and atomic mass (A) would be, respectively :

i.	Z = 17 , A = 35.
ii.	Z = 92 , A = 233.
iii.	Z = 4 , A = 9.

1.  ${}_4{}^9\mathrm{Be}$   ${}_{92}{}^{233}\mathrm{U}$     ${}_4{}^9\mathrm{Be}$

2.  ${}_{17}{}^{35}\mathrm{Cl}$    ${}_{92}{}^{233}\mathrm{U}$    ${}_4{}^9\mathrm{Be}$

3.  ${}_4{}^9\mathrm{Be}$    ${}_{17}{}^{35}\mathrm{Cl}$     ${}_{92}{}^{233}\mathrm{U}$

4. ${}_{92}{}^{233}\mathrm{U}$    ${}_4{}^9\mathrm{Be}$    ${}_{17}{}^{35}\mathrm{Cl}$

Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. The frequency (ν) and wave number of this yellow light would be, respectively:

1. 517 × 10¹⁴ s^–1 , 172 × 10⁶ m^–1

2. 6.17 × 10¹⁴ s^–1 , 1.72 × 10⁶ m^–1

3. 4.17 × 10¹⁴ s^–1 , 2.72 × 10⁶ m^–1

4. 5.17 × 10¹⁴ s^–1 , 1.72 × 10⁶ m^–1

The energy of the photon with a frequency of 3 × 10¹⁵ Hz is:

1. 1.988 × ${10}^{-18}$ J
2. 1.588 × ${10}^{-18}$ J
3. 1.288 × ${10}^{-18}$ J
4. 2.988 × ${10}^{-18}$ J

A photon of wavelength 4 × 10^–7 m strikes a metal surface, the work function of the metal being 2.13 eV.  The kinetic energy of emission would be:

The correct comparison between energy required to ionize an H atom if the electron occupies n = 5 orbits and the ionization enthalpy of H atom ( energy required to remove the electron from n =1 orbit) would be -

1) More energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state

2) More energy is required to ionize an electron in the 8th orbital of hydrogen atom as compared to that in the ground state

3) Less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state

4) Less energy is required to ionize a neutron in the 6th orbital of hydrogen atom as compared to that in the ground state

The possible values of n, l, and m for the electron present in 3d would be respectively:

1. n = 3, l = 1, m = – 2, – 1, 3, 1, 2

2. n = 3, l = 3, m = – 2, – 1, 0, 1, 2

3. n = 3, l = 2, m = – 2, – 1, 0, 1, 2

4. n = 5, l = 2, m = – 2, – 1, 0, 1, 2

The number of electrons in the species ${\mathrm{H}}_2^{+}$ $,$ ${\mathrm{H}}_2$ $,$ ${\mathrm{O}}_2^{+}$,  are respectively:

1. 15, 2, 1

2. 15, 1, 2

3. 1, 2, 15

4. 2, 1, 15

Orbital that does not exist:

The set of quantum numbers which represent 3p is :
1. n = 1, l = 0;
2, n = 3; l = 1
3. n = 4; l = 2;
4. n = 4; l = 3