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:
 
1.  0.97 eV
2.  97 eV 
3. 4.97 × ${10}^{-19}$ eV
4.  5.84 × 10⁵ eV

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 wavelength of the light emitted when the electron returns to the ground state in the H atom, from n = 5 to n = 1, would be: (The ground-state electron energy is –2.18 × 10^-18 j) 
1. Wavelength = 9.498 x 10^-8 km
2. Wavelength = 12.498 x 10^-8 m
3. Wavelength = 9.498 x 10^-8 m
4. Wavelength = 9.498 x 10^-8 cm

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 

Which of the following sets of quantum numbers is possible-
1. n = 0, l = 0, m_l = 0, m_s = + ½
2. n = 1, l = 0, m_l = 0, m_s = – ½
3. n = 1, l = 1, m_l = 0, m_s = + ½
4. n = 3, l = 3, m_l = –3, m_s = + ½

The total number of electrons in an atom with the following quantum numbers would be
(a) n = 4, m_s = – ½  (b) n = 3, l = 0
1. 16,  2
2. 11, 8
3. 16, 8
4. 12, 7

The transition in the hydrogen spectrum that would have the same wavelength as Balmer transition from n = 4 to n = 2 of He⁺ spectrum is -

1. ${\mathrm{n}}_1$ $=$ $3$ $\mathrm{to}$ ${\mathrm{n}}_2$ $=$ $4$

2. ${\mathrm{n}}_2$  = 3 to n₁ = 2

3. ${\mathrm{n}}_2$  = 3 to n₁ = 1

4. ${\mathrm{n}}_2$  = 2 to n₁ = 1