Question 1

The first Bohr radius, a₀, is 0.0529 nm and the corresponding energy, E₀, is 13.6 eV. The wavelength of the light emitted as a hydrogen atom undergoes a transition from state n = 3 to n = 2 is about

Accepted Answer

The energy of a hydrogen atom in the nth state is given by:$$E_n = -\frac{13.6}{n^2} eV$$The wavelength of light emitted when an electron transitions from state n = 3 to n = 2 is given by:$$\frac{1}{\lambda} = R_H (\frac{1}{n_f^2} - \frac{1}{n_i^2})$$where $R_H$ is the Rydberg constant.Plugging in the values for $n_i$, $n_f$, and $R_H$, we get:$$\frac{1}{\lambda} = (1.097 	imes 10^7 m^{-1}) (\frac{1}{2^2} - \frac{1}{3^2})$$$$\lambda = 656 nm$$Therefore, the correct answer is A) 656 nm.

Question 2

Which of the following statements is true?

Accepted Answer

All the statements provided are true regarding atomic structure. The nucleus is significantly smaller than the atom itself, the size of an atom is indeed determined by its electron cloud, most of an atom's volume is empty space, and the mass of an atom is primarily due to its nucleus, which contains protons and neutrons.

Question 3

If the potential energy of an electron in the Bohr model is -U₀, then the kinetic energy and the total energy of the electron are, respectively,

Accepted Answer

In the Bohr model, the total energy of the electron is the sum of its kinetic and potential energy. Therefore, the kinetic energy of the electron can be calculated by subtracting the potential energy from the total energy. The total energy of the electron is -U₀, as given in the problem. So, total energy = potential energy + kinetic energy => kinetic energy = total energy − potential energy = (-U₀) - (-U₀) = -U₀ + U₀ = 0 Therefore, the kinetic energy of the electron is 0. Using the formula for total energy in the Bohr model, Total energy = -hcR/n^2, where h is Planck's constant, c is the speed of light, R is the Rydberg constant, and n is the principal quantum number. From the given potential energy, we can write -hcR/n^2 = -U₀ => hcR/n^2 = U₀ Solving for n, we get n=2. Now, kinetic energy = (hcR/2n^2) − (hcR/n^2) = (hcR/4) − (hcR/2) = -0.5hcR Substituting the values of h, c, and R, we get kinetic energy = -0.5U₀_. Therefore, the kinetic energy and total energy of the electron are 0.5U₀ and -0.5U₀_., respectively, which is option B.

Question 4

The energy of the n = 1 level of hydrogen is -13.6 eV. The energy of the n = 4 level is

Accepted Answer

The energy levels of a hydrogen atom can be calculated using the formula $E_n = \frac{-13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. For the $n = 4$ level, the energy is $E_4 = \frac{-13.6 \text{ eV}}{4^2} = \frac{-13.6 \text{ eV}}{16} = -0.85 \text{ eV}$.

Question 5

The kinetic energy of an electron moving in a circular orbit of radius r about a positive charge Ze varies

Accepted Answer

The kinetic energy of an electron moving in a circular orbit is given by KE = (1/2)mv^2, where m is the mass of the electron, and v is its velocity. For a circular orbit, v is proportional to 1/r, where r is the radius of the orbit. Therefore, KE is proportional to (1/2)m(1/r)^2, or inversely proportional to r.

Question 6

An electron in the n = 3 state of a hydrogen atom falls into the n = 2 state. The Rydberg constant R is 1.097 × 10^-2/nm. The wavelength of the photon emitted is approximately

Accepted Answer

The answer of An electron in the n = 3...

Question 7

The order-of-magnitude of the diameter of an atom is closest to

Accepted Answer

The diameter of an atom is typically on the order of 1 angstrom (10^-10 meters). Therefore, the closest option is C, which is 10^-10 meters written in scientific notation as 10^-10S1S1P0 meters. Choices A, B, D, and E are either too small or too large to be on the same order of magnitude as the diameter of an atom.

Question 8

The ground-state energy of hydrogen is -13.6 eV. The difference in energy between the n = 3 and n = 4 levels magnitude only) is

Accepted Answer

The answer of The ground-state energy of hydrogen is -13.6...

Question 9

The quantum theory suggests that the stable orbits of electrons around a nucleus correspond to standing waves for the orbit. For a hydrogen atom, the radius of the ground state or the first orbit is the Bohr's radius, 0.0529 nm. What is the wavelength of the ground state of the hydrogen atom?

Accepted Answer

The answer of The quantum theory suggests that the stable...

Question 10

According to the Bohr theory, a hydrogen atom

Accepted Answer

Bohr's theory posits that a hydrogen atom does not emit radiation while the electron is in a stationary state. It only emits or absorbs energy when the electron transitions between these stationary states.

Question 11

The limit n = ∞) of a particular spectral series is the line of

Accepted Answer

As n approaches infinity, the energy difference between levels becomes the smallest, leading to the highest frequency possible in the series according to the Rydberg formula for spectral lines.

Question 12

An electron in the hydrogen atom ground-state energy = -13.6 eV) makes a transition from the n = 2 to the n = 4 energy level. Calculate the magnitude of the energy of the photon involved in this process and state whether the photon was absorbed or emitted.

Accepted Answer

The energy levels of a hydrogen atom are given by the formula $E_n = -\frac{13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. For a transition from $n = 2$ to $n = 4$, the energy difference is calculated as follows: $E_{initial} = -\frac{13.6}{2^2} = -3.4 \text{ eV}$ and $E_{final} = -\frac{13.6}{4^2} = -0.85 \text{ eV}$. The energy of the photon involved is $E_{photon} = E_{final} - E_{initial} = -0.85 - (-3.4) = 2.55 \text{ eV}$. Since the energy of the final state is higher than the initial state, the photon is absorbed. To find the wavelength, we use the relation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light. Rearranging for $\lambda$ gives $\lambda = \frac{hc}{E}$. Substituting $h = 4.135667696 \times 10^{-15} \text{ eV·s}$, $c = 3 \times 10^8 \text{ m/s}$, and $E = 2.55 \text{ eV}$, we find $\lambda \approx 486 \text{ nm}$, indicating that the photon was absorbed.

Question 13

The order-of-magnitude of the diameter of the nucleus is closest to

Accepted Answer

The answer of The order-of-magnitude of the diameter of the...

Question 14

An electron in the hydrogen atom ground-state energy = -13.6 eV) makes a transition from the n = 3 to the n = 1 energy level. Calculate the magnitude of the energy of the photon involved in this process and state whether the photon was absorbed or emitted.

Accepted Answer

The energy difference between the n = 3 and n = 1 levels in a hydrogen atom is 12.1 eV. Since the electron is transitioning to a lower energy level, the photon is emitted.

Question 15

The wavelength of the photon emitted when a hydrogen atom undergoes a transition from the n = 10 state to the n = 1 state is approximately

Accepted Answer

The answer of The wavelength of the photon emitted when...

Question 16

An electron in the hydrogen atom makes a transition from an energy level of -1.51 eV to one of energy -3.40 eV. In this process a photon is emitted with what wavelength?

Accepted Answer

To find the wavelength of the emitted photon, we can use the equation:
$\Delta E = E_2 - E_1 = h
u = \frac{hc}{\lambda}$

where $\Delta E$ is the energy difference between the initial and final energy levels, $E_1$ and $E_2$ respectively, $h$ is Planck's constant, $
u$ is the frequency of the emitted photon, $c$ is the speed of light, and $\lambda$ is the wavelength of the emitted photon.

Substituting the given values, we get:

$\Delta E = (-3.40 	ext{ eV}) - (-1.51 	ext{ eV}) = -1.89 	ext{ eV}$

$\Delta E$ must be converted to joules:
$-1.89	ext{ eV}	imes\frac{1.602	imes10^{-19}	ext{ J}}{	ext{eV}}=-3.03	imes10^{-19}	ext{ J}$

$\lambda = \frac{hc}{\Delta E} = \frac{(6.626	imes10^{-34}	ext{ J s})(2.998	imes10^{8}	ext{ m/s})}{-3.03	imes10^{-19}	ext{ J}}\approx656	ext{ nm}$

Therefore, the wavelength of the emitted photon is approximately 656 nm, which is best described by choice B.

Question 17

The first Bohr radius, a₀, is 0.0529 nm and the corresponding energy, E₀, is 13.6 eV. The wavelength of the light emitted as a hydrogen atom undergoes a transition from state n = 4 to n = 2 is

Accepted Answer

The answer of The first Bohr radius, a₀, is 0.0529...

Question 18

An electron in a hydrogen atom jumps from n = 5 to n = 2. The wavelength of the emitted photon is

Accepted Answer

The transition from n=5 to n=2 in a hydrogen atom corresponds to the Balmer series, which emits visible light. The wavelength for this transition is approximately 435 nm.

Question 19

In the Bohr model of the atom

Accepted Answer

In the Bohr model, electrons in stable orbits do not radiate energy, which prevents them from spiraling into the nucleus.

Question 20

The wavelength of the photon emitted when a hydrogen atom undergoes a transition from the n = 5 state to the n = 1 state is approximately

Accepted Answer