Question 1

Which form of electromagnetic radiation has the shortest wavelengths?

Accepted Answer

Gamma rays have the shortest wavelengths among all forms of electromagnetic radiation, ranging from 10^-12 meters to 10^-16 meters.

Question 2

Which of the following frequencies corresponds to light with the longest wavelength?

Accepted Answer

The wavelength is inversely proportional to frequency. The lowest frequency (4.12 × 10⁵ s^-1) corresponds to the longest wavelength.

Question 3

From the following list of observations, choose the one that most clearly supports the conclusion that the mass of the atom is located mainly in the nucleus.

Accepted Answer

The scattering of alpha particles by metal foil provides evidence that the mass of the atom is located mainly in the nucleus. The fact that some alpha particles are deflected or bounce back indicates that they are encountering a center of mass with significant mass and charge, which is consistent with the presence of a nucleus.

Question 4

Which of the following statements is(are) true?
i. An excited atom can return to its ground state by absorbing electromagnetic radiation.II. The energy of an atom is increased when electromagnetic radiation is emitted from it.III. The energy of electromagnetic radiation increases as its frequency increases.IV. An electron in the n = 4 state in the hydrogen atom can go to the n = 2 state by emitting electromagnetic radiation at the appropriate frequency.V. The frequency and wavelength of electromagnetic radiation are inversely proportional to each other.

Accepted Answer

i. An excited atom can return to its ground state by emitting electromagnetic radiation, not by absorbing it. So, statement i is false.
ii. When electromagnetic radiation is emitted by an atom, its energy decreases, not increases. So, statement ii is false.
iii. This statement is true according to the equation E = hf, where E is the energy of the electromagnetic radiation, h is Planck's constant and f is its frequency. So, statement iii is true.
iv. This statement is also true as it follows the principle of the emission spectrum of hydrogen, where an electron in a higher energy level can emit electromagnetic radiation and go to a lower energy level. So, statement iv is true.
v. This statement is true as the equation c = fλ, where c is the speed of light, f is the frequency, and λ is the wavelength, shows that wavelength and frequency are inversely proportional to each other. So, statement v is true.

Question 5

From the following list of observations, choose the one that most clearly supports the conclusion of de Broglie wavelengths.

Accepted Answer

Diffraction experiments, such as the double-slit experiment, clearly demonstrate the wave-like behavior of particles, supporting the idea that particles also have a wavelength, as proposed by de Broglie.

Question 6

From the following list of observations, choose the one that most clearly supports the conclusion that electrons in atoms have quantized energies.

Accepted Answer

The emission spectrum of hydrogen shows discrete lines of light at specific wavelengths, indicating that electrons in hydrogen atoms have specific energy levels and can only emit or absorb energy in discrete amounts. This is a clear example of quantization. The other options do not directly relate to the quantization of electron energies in atoms.

Question 7

Consider the following portion of the energy-level diagram for hydrogen: 
-For which of the following transitions does the light emitted have the longest wavelength?

Accepted Answer

The wavelength of the light emitted during an electronic transition in an atom is inversely proportional to the energy change associated with the transition. The energy change is smallest when the transition occurs between adjacent energy levels, such as from n = 4 to n = 3, resulting in the longest wavelength of light emitted.

Question 8

A photographic film needs a minimum of 80.0 kJ/mol for exposure.The frequencies of visible light are in the range 7.5 × 10¹⁴ Hz to 4.3 × 10¹⁴ Hz. Is the radiation visible to the human eye?

Accepted Answer

The energy required for exposure is 80.0 kJ/mol, which is equivalent to 80,000 J/mol. To determine if the radiation is visible, we need to calculate the energy associated with the given frequency range using the equation $E = h\nu$, where $E$ is the energy, $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{Js}$), and $\nu$ is the frequency. The highest energy for visible light would correspond to the highest frequency given, which is $4.3 \times 10^{14} \, \text{Hz}$. Calculating the energy for this frequency: $E = (6.626 \times 10^{-34} \, \text{Js}) \times (4.3 \times 10^{14} \, \text{Hz}) = 2.849 \times 10^{-19} \, \text{J}$ per photon. To convert this to J/mol, we multiply by Avogadro's number ($6.022 \times 10^{23} \, \text{mol}^{-1}$): $2.849 \times 10^{-19} \, \text{J/photon} \times 6.022 \times 10^{23} \, \text{mol}^{-1} = 171.5 \, \text{kJ/mol}$. This energy is higher than the minimum required (80.0 kJ/mol), indicating that the highest frequency (and thus the entire range) of visible light can expose the film. However, the question asks if the radiation is visible to the human eye, not if it can expose the film. The provided frequencies are within the visible range, but the question's context about exposure energy and calculation might seem misleading. The correct interpretation based on the question's phrasing should focus on whether the frequencies fall within the visible spectrum, which they do. However, the explanation provided focuses on the energy required for exposure, which might not directly answer the question about visibility. Given the frequencies fall within the visible range, the radiation is visible. The explanation mistakenly concludes the radiation isn't visible based on the energy required for exposure, which was not the question. The correct answer, based on the visibility of the frequencies, should be "yes," but the explanation provided incorrectly justifies a "no" response.

Question 9

A photographic film needs a minimum of 80.0 kJ/mol for exposure.What is the longest wavelength of radiation with sufficient energy to expose the film?

Accepted Answer

The answer of A photographic film needs a minimum of...

Question 10

Consider an atom traveling at 1% of the speed of light. The de Broglie wavelength is found to be 3.31 × 10^-3 pm. Which element is this?

Accepted Answer

The de Broglie wavelength formula is λ = h / (mv), where h is Planck's constant, m is mass, and v is velocity. Given the wavelength and velocity, we can solve for mass and identify the element. The calculated mass corresponds to calcium (Ca).

Question 11

From the following list of observations, choose the one that most clearly supports the conclusion that electromagnetic radiation has wave characteristics.

Accepted Answer

Diffraction is a characteristic property of waves, where waves bend around obstacles and spread out after passing through a small opening. The fact that electromagnetic radiation exhibits diffraction provides evidence that it has wave-like properties.

Question 12

From the following list of observations, choose the one that most clearly supports the conclusion that electrons have wave properties.

Accepted Answer

Diffraction is the bending of waves around obstacles or through openings, which is a characteristic of wave-like behavior. Thus, the observation of diffraction supports the idea that electrons have wave properties.

Question 13

What is the wavelength of light that is emitted when an excited electron in the hydrogen atom falls from the n = 5 level to the n = 2 level?

Accepted Answer

The formula for calculating the wavelength of light emitted from an electron transition in a hydrogen atom is:

λ = (h*c)/(ΔE)

where h is Planck's constant, c is the speed of light, and ΔE is the change in energy between the two levels.

ΔE = E_final - E_initial = (-13.6 eV)*(1/22 - 1/52) = 2.042 eV

Converting to joules:

E = (2.042 eV)*(1.602 × 10^-19 J/eV) = 3.27 × 10^-19 J

Plugging in values:

λ = (6.626 × 10^-34 J*s * 3.00 × 10^8 m/s)/(3.27 × 10^-19 J) = 4.34 × 10^-7 m or 434 nm

Therefore, the correct answer is C.

Question 14

What is the wavelength, in nanometers, of a photon of light whose frequency is 5.40 × 10¹⁴ Hz?

Accepted Answer

The formula relating wavelength ($\lambda$), frequency ($
u$), and speed of light ($c$) is $\lambda = \frac{c}{
u}$. Plugging in the given frequency of $5.40 	imes 10^{14}$ Hz and the speed of light ($3.00 	imes 10^8$ m/s), we get $\lambda = \frac{3.00 	imes 10^8	ext{ m/s}}{5.40 	imes 10^{14}	ext{ Hz}} 	imes 10^9	ext{ nm/m}= 5.56 	imes 10^{-7}	ext{ nm}$. Therefore, the answer is choice (C).

Question 15

In an investigation of the electronic absorption spectrum of a particular element, it is found that a photon having λ = 500 nm provides just enough energy to promote an electron from the second quantum level to the third. From this information, we can deduce

Accepted Answer

The energy difference between the n = 2 and n = 3 levels can be calculated using the formula ΔE = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. Plugging in the given values, we get ΔE = (6.626 x 10^-34 J s)(3.0 x 10^8 m/s)/(500 x 10^-9 m) = 3.98 x 10^-19 J. This is the energy required to promote the electron from the n = 2 to n = 3 level.

Question 16

In general, Z is less than Z_eff.

Accepted Answer

As per the given inequality, Z is greater than or equal to Z_eff. Therefore, the statement "Z is less than Z_eff" is false.

Question 17

In Bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus,

Accepted Answer

According to Bohr's atomic theory, energy is required for an electron to move from a lower energy level to a higher energy level. Therefore, when an electron moves to a more distant energy level, energy is absorbed.

Question 18

From the following list of observations, choose the one that most clearly supports the conclusion that atoms contain electrons.

Accepted Answer

Cathode "rays" were discovered to be negatively charged particles, which were later identified as electrons.

Question 19

Light has a wavelength of 5.6 × 10² nm. What is the energy of a photon of this light?

Accepted Answer

The energy of a photon can be calculated using the equation: 
E = hc/λ 
where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the light.

First, we need to convert the given wavelength from nanometers to meters: 
5.6 x 10^-12 m

Then, we can plug in the values: 
E = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(5.6 x 10^-12 m) 
E = 3.55 x 10^-19 J

Therefore, the answer is choice E.

Question 20

Which of the following is a reasonable criticism of the Bohr model of the atom?

Accepted Answer

The Bohr model only works well for hydrogen and fails to predict the ionization energy of the valence electrons for elements other than hydrogen.

Quiz 12: Quantum Mechanics and Atomic Theory