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Chapter 1: Problem 12

An atom of silicon in its ground state is subjected to a frequency of light that is high enough to cause electron ejection. An electron from which subshell of silicon would have the highest kinetic energy after ejection? (A) 1 \(\mathrm{s}\) (B) 2\(p\) (C) 3\(p\) (D) 4\(s\)

Short Answer

Expert verified
(D) 4s

Step by step solution

01

Understand electron ejection due to light absorption

When light with high enough frequency is absorbed by an electron, the acquired energy is used by the electron to escape its bonding with the nucleus. This phenomenon is known as the photoelectric effect. If the light's frequency is high enough to overcome the binding energy of the electron, the surplus energy manifests as the electron's kinetic energy.
02

Identify Silicon's electron configuration

Silicon is the 14th element in the periodic table, and its electron configuration in the ground state is \(1s^2 2s^2 2p^6 3s^2 3p^2\). This means that the electrons fill up the 1s, 2s, 2p, 3s, and 3p subshells before populating the 4s subshell.
03

Understand electron ejection from different subshells

An electron from a lower energy level (closer to the nucleus of the atom) requires more energy to be ejected than an electron from a higher energy level. This is because lower energy levels have a stronger attraction to the positively charged nucleus. Hence, an electron from a lower energy level will have lesser surplus energy (kinetic energy) after ejection when exposed to a light of particular frequency.
04

Identify which electron will have the highest kinetic energy upon ejection

Given that the electron from a higher energy level will inherit more kinetic energy after being ejected, the correct choice is (D) the 4s subshell. It’s the highest energy level in the ground state of a silicon atom. Hence, an electron from this subshell would acquire the highest kinetic energy upon ejection.

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Most popular questions from this chapter

Questions 54-56 refer to the following. GRAPH CAN'T COPY Between propane and ethene, which will likely have the higher boiling point and why? (A) Propane, because it has a greater molar mass (B) Propane, because it has a more polarizable electron cloud (C) Ethene, because of the double bond (D) Ethene, because it is smaller in size

A student added 1 liter of a 1.0\(M\) \(\mathrm{KCl}\) solution to 1 liter of a 1.0 \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution. A lead chloride precipitate formed, and nearly all of the lead ions disappeared from the solution. Which of the following lists the ions remaining in the solution in order of decreasing concentration? (A) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]\) (B) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{K}^{+}\right]\) (C) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{NO}_{3}^{-}\right]\) (D) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]\)

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. As \(\Delta T\) increases, what happens to the equilibrium constant and why? (A) The equilibrium constant increases because more products are created. (B) The equilibrium constant increases because the rate of the forward reaction increases. (C) The equilibrium constant decreases because the equilibrium shifts to the left. (D) The value for the equilibrium constant is unaffected by temperature and will not change.

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) The value for \(K_{\mathrm{c}}\) at \(25^{\circ} \mathrm{C}\) is \(8.1 .\) What must happen in order for the reaction to reach equilibrium if the initial concentrations of all three species was 2.0 \(M\) ? (A) The rate of the forward reactions would increase, and \(\left[\mathrm{SO}_{3}\right]\) would decrease. (B) The rate of the reverse reaction would increase, and \(\left[\mathrm{SO}_{2}\right]\) would decrease. (C) Both the rate of the forward and reverse reactions would increase, and the value for the equilibrium constant would also increase. (D) No change would occur in either the rate of reaction or the concentrations of any of the species.

For a reaction involving nitrogen monoxide inside a sealed flask, the value for the reaction quotient \((Q)\) was found to be \(1.1 \times 10^{2}\) at a given point. If, after this point, the amount of NO gas in the flask increased, which reaction is most likely taking place in the flask? (A) \(\operatorname{NOBr}(g) \rightarrow \operatorname{NO}(g)+1 / \operatorname{Br}_{2}(g) \quad K_{\mathrm{C}}=3.4 \times 10^{-2}\) (B) \(2 \mathrm{NOCl}(g) \mapsto 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K_{\mathrm{c}}=1.6 \times 10^{-5}\) (C) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{\mathrm{c}}=4.0 \times 10^{6}\) (D) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g) \quad K_{\mathrm{c}}=4.2 \times 10^{2}\)
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