The electron of a Br atom that experiences the lowest effective nuclear charge is:

1. 2p and 3p

2. 4p

3. 2p

4. 3p

The number of electrons that can be present in the subshells having m_s value of \(-\frac{1}{2}\) for n = 4 are:

1. 36

2. 4

3. 16

4. 2

Which among the following are isoelectronic species?

1. ${\mathrm{Na}}^{+}, {\mathrm{K}}^{+}, {\mathrm{Ca}}^{2+}, {\mathrm{Mg}}^{2+}$

2. ${\mathrm{Ca}}^{2+}, {\mathrm{Mg}}^{2+}, {\mathrm{S}}^{2-}, {\mathrm{K}}^{+}$

3. ${\mathrm{Ca}}^{2+}, \mathrm{Ar}, {\mathrm{S}}^{2-}, {\mathrm{K}}^{+}$

4. None of the above

If the position of the electron were measured with an accuracy of +0.002 nm, the uncertainty in the momentum of the electron would be: 

1. 5.637 × 10^–23 kg m s^–1

2. 4.637 × 10^–23 kg m s^–1

3. 2.637 × 10^–23 kg m s^–1

4. 3.637 × 10^–23 kg m s^–1

When a photon with a wavelength of 150 pm strikes an atom, one of its inner bound electrons is ejected at a velocity of 1.5 × 10⁷ m s^–1 . The energy with which it is bound to the nucleus would be:

1. 32.22 × 10^–16 J

2. 12.22 × 10^–16 J

3. 22.27 × 10^–16 J

4. 31.22 × 10^–16 J
 

In Rutherford's experiment, generally, the thin foil of heavy atoms like gold, platinum, etc. have been used to be bombarded by the α-particles.

 If the thin foil of light atoms like aluminum etc. is used in Rutherford’s experiment, the difference that would be observed from the above results is :

1. The same results will be observed.

2. More deflection would be observed.

3. There will not be enough deflection.

4. None of the above.

A certain particle carries 2.5 × 10^–16 C of static electric charge. The number of electrons present in it would be:

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

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 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