Home / AP Chemistry: 3.12 Photoelectric Effect- Exam Style questions with Answer- FRQ

AP Chemistry: 3.12 Photoelectric Effect- Exam Style questions with Answer- FRQ

Question.

The first ionization energies for Na and K are given in the table above. Na metal reacts vigorously with water to form hydrogen gas and a metal hydroxide. K metal reacts vigorously as well, but it bursts into a violet-colored flame.
(a) Write the electron configuration for a\( K^{+}\) ion.
(b) Based on principles of atomic structure, explain why the first ionization energy of K is lower than the first ionization energy of Na.
(c) A student hypothesizes that the flame is violet colored because violet light consists of photons that have the energy needed to ionize K atoms. The wavelength of the violet light is measured to be 423 nm.
(i) Calculate the energy, in J, of one photon of violet light with a wavelength of 423 nm.
(ii) Is the energy of one photon of the violet light sufficient to cause the ionization of a K atom? Justify your answer.

▶️Answer/Explanation

(a) \(1s^22s^22p^63s^23p^6\)       OR [Ne] \(3s^23p^6\)

(b) The valence electron in K is in a higher principal energy level and farther from the nucleus than the valence electron in Na, therefore the valence electron in K is easier to remove.

(c) \(c=\lambda v\)
\(v=\frac{c}{\lambda }=\frac{2.998\times 10^{8}ms^{-1}}{4.23\times 10^{-7}m}=7.09\times 10^4s^{-1}\)
\(E=hv=(6.626\times10^{-34}js)(7.09\times 10^14s^{-1})\)
\(=4.70\times 10^{-19}j\) J required to ionize one atom  This is greater than the energy of one photon of violet light, so it is not sufficient.

Question

Answer the following questions regarding light and its interactions with molecules, atoms, and ions.

(a) The longest wavelength of light with enough energy to break the Cl-Cl bond in \(Cl_2(g)\) is 495 nm.

(i) Calculate the frequency, in \(s^{-1}\), of the light.

(ii) Calculate the energy, in J, of a photon of the light.

(iii) Calculate the minimum energy, in kJ \(mol^{-1}\), of the Cl-Cl bond. 

(b) A certain line in the spectrum of atomic hydrogen is associated with the electronic transition in the H from the sixth energy level (n = 6) to the second energy level (n = 2). atom

(i) Indicate whether the H atom emits energy or whether it absorbs energy during the transition. Justify your answer.

(ii) Calculate the wavelength, in nm, of the radiation associated with the spectral line.

(iii) Account for the observation that the amount of energy associated with the same electronic transition (n = 6 to rz = 2) in the \(He^+\) ion is greater than that associated with the corresponding transition in the H atom.

▶️Answer/Explanation

Answer:

(a) (i) \(v=\frac{c}{\lambda}=\frac{3.00 \times 10^{17}nm/sec}{495nm}(or,=\frac{3.00 \times 10^8m/sec}{495 \times 10^{-9}m})=6.06 \times 10^{14} sec^{-1}\)
(ii) \(E=hv=(6.626 \times 10^{-34}J sec)(6.06 \times 10^{14} sec^{-1}) = 4.02 \times 10^{-19}J\)
(iii) \((4.02 \times 10^{-19}J)(6.022 \times 10^{-23}mol^{-1})(0.00100kJ/J)=242 kJ/mol\)

(b) (i) Energy is emitted.
The n = 6 state is at a higher energy than the n=2 state. Going from a high energy state to low energy state means that energy must be emitted.
(ii) \(E_2=\frac{-2.178 \times 10^{-18} J}{2^2}=-5.45 \times 10^{-19}J, E_6 = \frac{-2.178 \times 10^{-18}J}{6^2}=-6.05 \times 10^{-20}J\)
\(\Delta E = E_6 – E_2 = -6.05 \times 10^{-20}J- (-5.45 \times 10^{-19}J) = 4.84 \times 10^{-19}J)= 4.84 \times 10^{-19}J\)
OR,
\(\Delta E = 2.178 \times 10^{-18}(\frac{1}{2^2}-\frac{1}{6^2})J = 2.178 \times 10^{-18}(0.2222)J = 4.84 \times 10^{-19}J\)
\(E=\frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E}\)
OR,
\(v=\frac{E}{h}=\frac{4.84 \times 10^{-19}J}{6.626 \times 10^{-34}J sec}=7.30 \times 10^{14} sec^{-1}\)
\(\lambda = \frac{(6.626 \times 10^{-34}J sec)(3.00 \times 10^{17} nm sec^{-1}}{4.84 \times 10^{-19}J}=411 nm\)
OR,
\(\lambda = \frac{(6.626 \times 10^{-34}J sec)(3.00 \times 10^{17}nm sec^{-1})}{4.84 \times 10^{-19}J}=411 nm\)
(iii) The positive charge holding the electron is greater for \(He^+\), which has a 2+ nucleus, than for H with its 1+ nucleus. The stringer attraction means that it requires more energy for the electron to move to higher energy levels. Therefore, transitions form high energy state to lower states will be more energetic for \(He^+\) than for H.

Question

Answer the following questions about the element Si and some of its compounds.
(a) The mass spectrum of a pure sample of Si is shown below.

(i) How many protons and how many neutrons are in the nucleus of an atom of the most abundant isotope of Si ?
(ii) Write the ground-state electron configuration of Si.
Two compounds that contain Si are SiO2 and SiH4.
(b) At 161 K, SiH4 boils but SiO2 remains as a solid. Using principles of interparticle forces, explain the difference in boiling points.

At high temperatures, SiH4 decomposes to form solid silicon and hydrogen gas.
(c) Write a balanced equation for the reaction.
A table of absolute entropies of some substances is given below.

(d) Explain why the absolute molar entropy of Si(s) is less than that of H2(g).
(e) Calculate the value, in J/(mol. K), of ∆S° for the reaction.
(f) The reaction is thermodynamically favorable at all temperatures. Explain why the reaction occurs only at high temperatures.

(g) A partial photoelectron spectrum of pure Si is shown below. On the spectrum, draw the missing peak that corresponds to the electrons in the 3p sublevel.

(h) Using principles of atomic structure, explain why the first ionization energy of Ge is lower than that of Si.

(i) A single photon with a wavelength of 4.00 ×  10−7 m is absorbed by the Si sample. Calculate the energy of the photon in joules.

▶️Answer/Explanation

Ans:

(a)  (i) For the correct answer:
14 protons and 14 neutrons

(ii) For the correct answer:
Accept one of the following:
• 1s2 2s2 2p6 3s2 3p2
• [Ne] 3s2 3p2

(b) For a correct explanation:
SiH4 is composed of molecules, for which the only intermolecular forces are London dispersion forces. SiO2 is a network covalent compound with covalent bonds between
silicon and oxygen atoms. London dispersion forces are much weaker than covalent bonds, so SiH4 boils at a much lower temperature than SiO2.

(c) For the correct balanced equation (state symbols not required):
SiH4 ( g)  → Si(s)  + 2 H2(g)

(d) For a correct explanation:
The H2(g) molecules are more highly dispersed than the Si(s) atoms and, therefore, have a higher absolute molar entropy. Silicon is a solid; therefore, its atoms are in fixed positions,
are less dispersed, and have a lower absolute molar entropy.

(e) For the correct calculated value:
ΔS0rxn = (18 + 2(131)) – 205 = +75 J/(molrxn • K)

(f) For a correct explanation:
High temperature is required for the reactant particles to have sufficient thermal energy to overcome the activation energy of the reaction.

(g) For the correct peak height and location:
The peak should be drawn to the right of the other peaks, and it should reach the second line above the horizontal axis.

(h) For a correct explanation:
The valence electrons of a Ge atom occupy a higher shell (n=4) than those of a Si atom (n=3), so the average distance between the nucleus and the valence electrons is greater in
Ge than in Si. This greater separation results in weaker Coulombic attractions between the Ge nucleus and its valence electrons, making them less tightly bound and, therefore, easier
to remove compared to those in Si.

(i) For the correct calculated value:

\(E = hv = h\left ( \frac{c}{\lambda } \right )=6.626\times 10^{-34}J\cdot s\left ( \frac{2.998\times 10^{8}ms^{-1}}{4.00\times 10^{-7}m} \right )=4.97 \times 10^{-19} J\)

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