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Chapter 18: Problem 2488

which of the following atom has the lowest ionization potential? (A) \({ }^{14}{ }_{7} \mathrm{~N}\) (B) \({ }^{40}{ }_{18} \mathrm{Ar}\) (C) \({ }^{133} 55 \mathrm{Cs}\) (D) \({ }^{16}{ }_{8} \mathrm{O}\)

Short Answer

Expert verified
The atom with the lowest ionization potential is (C) \({ }^{133} 55 \mathrm{Cs}\) (Cesium), as it is farther down and to the left in the periodic table compared to the other options.

Step by step solution

01

Identify each atom's position on the periodic table

To analyze the ionization potential, first, we need to determine the position of each atom in the periodic table. The atomic number (number of protons) will help in identifying the element, and the period and group they belong to. (A) \({ }^{14}{ }_{7} \mathrm{~N}\) : Nitrogen, Period 2, Group 15 (5A) (B) \({ }^{40}{ }_{18} \mathrm{Ar}\) : Argon, Period 3, Group 18 (8A) (C) \({ }^{133} 55 \mathrm{Cs}\) : Cesium, Period 6, Group 1 (1A) (D) \({ }^{16}{ }_{8} \mathrm{O}\) : Oxygen, Period 2, Group 16 (6A)
02

Compare the atoms' ionization potentials based on their position in the periodic table

As mentioned before, ionization potential decreases down a group and increases across a period. We can now compare the atoms according to their position: - Nitrogen and Oxygen are in the same period (Period 2), but Oxygen has a higher ionization potential because it is farther to the right. - Argon is in a higher period (Period 3) than Nitrogen and Oxygen and is in Group 18, which is the farthest right group on the periodic table, thus having a very high ionization potential. - Cesium is in Period 6 and Group 1. It is farther down and to the left on the periodic table, indicating lower ionization potential compared to the other atoms.
03

Determine which atom has the lowest ionization potential

Based on the positions of each atom on the periodic table, we can conclude that Cesium (\({ }^{133} 55 \mathrm{Cs}\)) has the lowest ionization potential among the given options. Thus, the correct answer is (C) \({ }^{133} 55 \mathrm{Cs}\).

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Most popular questions from this chapter

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy \(\mathrm{K}_{1}\) is ro. The distance of the closest approach when the \(\alpha\) - particle is fired at the same nucleus with kinetic energy \(2 \mathrm{k}_{1}\) will be. (A) \(\left(\mathrm{r}_{0} / 2\right)\) (B) \(4 r_{0}\) (C) \(\left(\mathrm{r}_{0} / 4\right)\) (D) \(2 \mathrm{r}_{0}\)

It the radius of \({ }^{27}{ }_{13} \mathrm{~A} \ell\) nucleus is $3.6 \mathrm{fm}\( the radius of \){ }^{125}{ }_{52} \mathrm{Te}$ nucleus is nearly equal to (A) \(8 \mathrm{fm}\) (B) \(6 \mathrm{fm}\) (C) \(4 \mathrm{fm}\) (D) \(5 \mathrm{fm}\)

The binding energy Per nucleon of deuteron $\left({ }^{2}{ }_{1} \mathrm{H}\right)\( and Lielium nucleus \){ }_{2}{ }^{4}{ }_{2} \mathrm{He}$ ) is \(1.1 \mathrm{MeV}\) and \(7.0 \mathrm{MeV}\). respectively. If two deuteron react to form a single helium nucleus, the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)

The half time of a radioactive substance is \(20 \mathrm{~min}\), difference between Points of time when it is \(33 \%\) disintegrated and \(67 \%\) disintegrated is approximately (A) \(10 \mathrm{~min}\) (B) \(20 \mathrm{~min}\) (C) \(40 \mathrm{~min}\) (D) \(30 \mathrm{~min}\)

The half life time of a radioactive elements of \(\mathrm{x}\) is the same as the mean life of another radioactive element \(\mathrm{y}\). Initially they have same number of atoms, then (A) \(\mathrm{y}\) will decay faster then \(\mathrm{x}\) (B) \(\mathrm{x}\) will decay faster then \(\mathrm{y}\) (C) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate at all time (D) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate initially.
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