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 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

2 ×10⁸ atoms of carbon are arranged side by side. The radius of a carbon atom if the length of this arrangement is 2.4 cm would be 

1. 7.0 x 10^-11 m

2. 5.0 x 10^-11 m

3. 8.0 x 10^-11 m

4. 6.0 x 10^-11 m