Question 1

Light shines through atomic hydrogen gas.It is seen that the gas absorbs light readily at a wavelength of 91.63 nm.What is the value of n of the level to which the hydrogen is being excited by the absorption of light of this wavelength? Assume that the most of the atoms in the gas are in the lowest level.(h = 6.626 × 10^-34 J ∙ s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19 J,the Rydberg constant is R = 1.097 × 10⁷ m^-1.)

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy of a photon is also equal to the difference in energy between two energy levels, ΔE = E2 - E1. In this case, the photon is absorbed by the hydrogen atom, which means that the atom is excited to a higher energy level. The energy of the photon is equal to the difference in energy between the ground state and the excited state, ΔE = E2 - E1. The energy of the ground state is given by E1 = -13.6 eV, and the energy of the excited state is given by E2 = -3.4 eV. Therefore, the difference in energy between the two levels is ΔE = E2 - E1 = -10.2 eV. The wavelength of the photon is given by λ = hc/ΔE, where h is Planck's constant, c is the speed of light, and ΔE is the difference in energy between the two levels. Substituting the values of h, c, and ΔE into the equation, we get λ = hc/ΔE = (6.626 × 10^-34 J ∙ s)(3.00 × 10^8 m/s)/(-10.2 eV) = 91.63 nm. Therefore, the correct answer is A.

Question 2

Calculate the kinetic energy (in eV)of a nonrelativistic neutron that has a de Broglie wavelength of (h = 6.626 × 10^-34 J ∙ s,m_neutron = 1.675 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

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Question 3

In a double slit experiment,a beam of electrons strikes a pair of slits.The slits are 15 μm apart,and the first interference maximum lies at an angle of 0.50 µrad from the center of the interference pattern.What is the momentum of the incoming electrons? (h = 6.626 × 10^-34 J ∙ s,m_el = 9.11 × 10^-31 kg)

Accepted Answer

The answer of In a double slit experiment,a beam of...

Question 4

Suppose that in a parallel universe,the proton and electron were identical to their counterparts in our own universe EXCEPT that the electron had twice as much charge as our electron.In our present universe,the radius of the first Bohr orbit for hydrogen is a₀ and the speed of an electron in that orbit is v₀.In the parallel universe, (a)what would be the radius (in terms of a₀)of the first Bohr orbit for hydrogen? (b)what would be the speed (in terms of v₀)of an electron in the first Bohr orbit for hydrogen?

Accepted Answer

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Question 5

A hydrogen atom makes a downward transition from the state to the n = 5 state.Find the wavelength of the emitted photon.The lowest level energy state of hydrogen is -13.6 eV. (h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s)

Accepted Answer

The answer of A hydrogen atom makes a downward transition...

Question 6

The energy of the ground state in the Bohr model of the hydrogen atom is -13.6 eV.The energy of the n = 2 state of hydrogen in this model is closest to

Accepted Answer

The energy levels in the Bohr model of the hydrogen atom are given by the formula $E_n = \frac{-13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. For the $n = 2$ state, the energy is $E_2 = \frac{-13.6 \text{ eV}}{2^2} = \frac{-13.6 \text{ eV}}{4} = -3.4 \text{ eV}$.

Question 7

Electrons emerge from an electron gun with a speed of 2.0 × 10⁶ m/s and then pass through a pair of thin parallel slits.Interference fringes with a spacing of 2.7 mm are detected on a screen far from the double slit and fairly close to the center of the pattern.What would the fringe spacing be if the electrons were replaced by neutrons with the same speed? (m_el = 9.11 × 10^-31 kg,m_neutron = 1.67 × 10^-27 kg)

Accepted Answer

The fringe spacing is given by:$$d = \frac{\lambda L}{a}$$where:* d is the fringe spacing* λ is the wavelength of the electron* L is the distance between the slits and the screen* a is the distance between the slitsThe wavelength of the electron is given by:$$\lambda = \frac{h}{p}$$where:* h is Planck's constant* p is the momentum of the electronThe momentum of the electron is given by:$$p = mv$$where:* m is the mass of the electron* v is the speed of the electronSubstituting these equations into the equation for the fringe spacing, we get:$$d = \frac{hL}{mv}$$Since the speed of the electron is the same for both electrons and neutrons, the fringe spacing will be inversely proportional to the mass of the particle. Therefore, the fringe spacing for neutrons will be:$$d_{neutron} = d_{electron} \frac{m_{electron}}{m_{neutron}}$$Substituting the given values, we get:$$d_{neutron} = 2.7 	ext{ mm} \frac{9.11 	imes 10^{-31} 	ext{ kg}}{1.67 	imes 10^{-27} 	ext{ kg}} = 1.5 \mu 	ext{m}$$Therefore, the correct answer is A.

Question 8

What is the energy of a photon that has a wavelength equal to the de Broglie wavelength of a proton having a speed of 7.1 × m/s? (m_proton = 1.67 × 10^-27 kg,c = 3.00 × 10⁸ m/s)

Accepted Answer

We can use the de Broglie wavelength formula to find the wavelength of the proton:
λ = h/p
where h is Planck's constant and p is the momentum of the proton.
p = mv = (1.67 × 10^-27 kg)(7.1 × 10^5 m/s) = 1.185 × 10^-19 kg*m/s
λ = (6.626 × 10^-34 J*s)/(1.185 × 10^-19 kg*m/s) = 5.59 × 10^-15 m
Now, we can use the formula for the energy of a photon:
E = hc/λ
where c is the speed of light.
E = (6.626 × 10^-34 J*s)(3.00 × 10^8 m/s)/(5.59 × 10^-15 m) = 3.56 × 10^-11 J
Convert to keV:
E = 3.56 × 10^-11 J / (1.602 × 10^-16 J/keV) = 222 keV
The closest answer choice is A, 220 keV.

Question 9

A photon of wavelength 18.0 pm is scattered through an angle of 120° by a stationary electron.What is the wavelength of the scattered photon? (m_el = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s,c = 3.00 × 10⁸ m/s)

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The answer of A photon of wavelength 18.0 pm is...

Question 10

A photon of wavelength 29 pm is scattered by a stationary electron.What is the maximum possible energy loss of the photon? (m_el = 9.11 × 10^-31 kg,h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s)

Accepted Answer

The maximum possible energy loss of a photon in the Compton scattering process can be calculated using the equation:

ΔE = (h*c/λ) * (1 - cosθ)

where h is Planck's constant, c is the speed of light, λ is the wavelength of the incident photon, and θ is the angle between the incident photon and the scattered photon.

In this case, the incident photon has a wavelength of 29 pm (picometers), which is equivalent to 2.9 × 10^-11 m. The electron is stationary, so the scattered photon will be deflected at an angle of 180 degrees (i.e. backscatter).

Using these values, we can calculate the maximum possible energy loss:

ΔE = (6.626 × 10^-34 J s * 3.00 × 10^8 m/s / (2.9 × 10^-11 m)) * (1 - cos 180°)
    = 6.1 keV

Therefore, the correct answer is D (6.1 keV).

Question 11

A hydrogen atom is excited to the n = 10 stated.It then decays to the n = 4 state by emitting a photon which is detected in a photographic plate.What is the frequency of the detected photon? The lowest level energy state of hydrogen is -13.6 eV.(h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The answer of A hydrogen atom is excited to the...

Question 12

A hydrogen atom is in its n = 2 excited state when its electron absorbs a photon of energy .What is the energy of the resulting free electron? The lowest level energy state of hydrogen is -13.6 eV.(h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

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The answer of A hydrogen atom is in its n...

Question 13

How fast must a nonrelativistic electron move so its de Broglie wavelength is the same as the wavelength of a 3.4-eV photon? (m_el = 9.11 × 10^-31 kg,c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The de Broglie wavelength of an electron is given by:$$\lambda = \frac{h}{p}$$where h is Planck's constant and p is the momentum of the electron. The momentum of a photon is given by:$$p = \frac{E}{c}$$where E is the energy of the photon and c is the speed of light.We can set the de Broglie wavelength of the electron equal to the wavelength of the photon and solve for the momentum of the electron:$$\frac{h}{p} = \frac{hc}{E}$$$$p = \frac{E}{c}$$We can then use the momentum of the electron to find its velocity:$$v = \frac{p}{m}$$where m is the mass of the electron.Plugging in the given values, we get:$$v = \frac{3.4 	ext{ eV} 	imes 1.60 	imes 10^{-19} 	ext{ J/eV}}{9.11 	imes 10^{-31} 	ext{ kg} 	imes 3.00 	imes 10^8 	ext{ m/s}} = 2000 	ext{ m/s}$$Therefore, the correct answer is A.

Question 14

The energy of the ground state in the Bohr model of the hydrogen atom is -13.6 eV.In a transition from the n = 2 state to the n = 4 state,a photon of energy

Accepted Answer

The answer of The energy of the ground state in...

Question 15

Light excites atomic hydrogen from its lowest level to the n = 4 level.What is the energy of the light? The energy of the lowest level is -13.6 eV.

Accepted Answer

The energy difference between the n = 1 level and the n = 4 level in a hydrogen atom can be calculated using the formula for the energy levels of hydrogen: $E_n = -13.6 \, \text{eV} \times \frac{1}{n^2}$. For n = 1, $E_1 = -13.6 \, \text{eV}$, and for n = 4, $E_4 = -13.6 \, \text{eV} \times \frac{1}{4^2} = -13.6 \, \text{eV} \times \frac{1}{16} = -0.85 \, \text{eV}$. The energy of the light required to excite an electron from n = 1 to n = 4 is the difference in energy between these two levels: $E = E_4 - E_1 = -0.85 \, \text{eV} - (-13.6 \, \text{eV}) = 12.75 \, \text{eV}$, which rounds to 12.8 eV.

Question 16

A hydrogen atom initially in the n = 6 state decays to the n = 2 state.The emitted photon is detected in a photographic plate.What is the wavelength of the detected photon? The lowest level energy state of hydrogen is -13.6 eV.(h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J, c = 3.00 × 10⁸ m/s)

Accepted Answer

The energy difference between the initial and final states is calculated using the formula for the energy levels of a hydrogen atom: $E_n = \frac{-13.6 \text{ eV}}{n^2}$. The energy of the photon emitted is $E = E_{initial} - E_{final} = \frac{-13.6 \text{ eV}}{6^2} - \frac{-13.6 \text{ eV}}{2^2}$. Converting this energy difference to joules and using $E = \frac{hc}{\lambda}$ allows solving for $\lambda$, the wavelength of the photon, which is approximately 410 nm.

Question 17

X-rays of energy 2.9 × 10⁴ eV are scattered by a free stationary electron through an angle of 135°.What is the energy of the scattered X-rays,in electron volts? (m_el = 9.11 × 10^-31 kg, e = - 1.60 × 10^-19 C,h = 6.626 × 10^-34 J ∙ s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19 J)

Accepted Answer

The answer of X-rays of energy 2.9 × 10⁴ eV...

Question 18

What is the frequency of the light emitted by atomic hydrogen with m = 8 and n = 12? (The Rydberg constant is R = 1.097 × 10⁷ m^-1,^c= 3.00 × 10⁸ m/s.)

Accepted Answer

The frequency of the light emitted can be calculated using the formula derived from the Rydberg formula for hydrogen emission: $f = R \cdot c \cdot \left( \frac{1}{m^2} - \frac{1}{n^2} \right)$, where $R$ is the Rydberg constant, $c$ is the speed of light, $m$ is the lower energy level, and $n$ is the higher energy level. Substituting $R = 1.097 \times 10^{7} \, \text{m}^{-1}$, $c = 3.00 \times 10^{8} \, \text{m/s}$, $m = 8$, and $n = 12$, we get the frequency $f$. Calculating this gives a value close to $2.86 \times 10^{13} \, \text{Hz}$.

Question 19

The Bohr radius of the hydrogen atom is 0.529 × 10^-10 m.What is the radius of the n = 2 state?

Accepted Answer

The radius of the state n can be calculated by multiplying the Bohr radius by n². Therefore, the radius of the n=2 state can be calculated as follows:
r₂ = 0.529 × 10⁻¹⁰ m × 2²
r₂ = 2.12 × 10⁻¹⁰ m
Therefore, the best choice is B, 2.12 × 10⁻¹⁰ m.

Question 20

What is the orbital radius of the excited state in the Bohr model of the hydrogen atom? The ground-state radius of the hydrogen atom is 0.529 × 10^-10 m.

Accepted Answer

The answer of What is the orbital radius of the...

Quiz 38: Quantization

Calculate the kinetic energy (in eV)of a nonrelativistic neutron that has a de Broglie wavelength of (h = 6.626 × 10-34 J ∙ s,mneutron = 1.675 × 10-27 kg, 1 eV = 1.60 × 10-19 J)

A hydrogen atom makes a downward transition from the state to the n = 5 state.Find the wavelength of the emitted photon.The lowest level energy state of hydrogen is -13.6 eV. (h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J,c = 3.00 × 108 m/s)

The energy of the ground state in the Bohr model of the hydrogen atom is -13.6 eV.The energy of the n = 2 state of hydrogen in this model is closest to

What is the energy of a photon that has a wavelength equal to the de Broglie wavelength of a proton having a speed of 7.1 × m/s? (mproton = 1.67 × 10-27 kg,c = 3.00 × 108 m/s)

A photon of wavelength 18.0 pm is scattered through an angle of 120° by a stationary electron.What is the wavelength of the scattered photon? (mel = 9.11 × 10-31 kg, h = 6.626 × 10-34 J ∙ s,c = 3.00 × 108 m/s)

A photon of wavelength 29 pm is scattered by a stationary electron.What is the maximum possible energy loss of the photon? (mel = 9.11 × 10-31 kg,h = 6.626 × 10-34 J ∙ s, c = 3.00 × 108 m/s)

A hydrogen atom is excited to the n = 10 stated.It then decays to the n = 4 state by emitting a photon which is detected in a photographic plate.What is the frequency of the detected photon? The lowest level energy state of hydrogen is -13.6 eV.(h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J)

A hydrogen atom is in its n = 2 excited state when its electron absorbs a photon of energy .What is the energy of the resulting free electron? The lowest level energy state of hydrogen is -13.6 eV.(h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

How fast must a nonrelativistic electron move so its de Broglie wavelength is the same as the wavelength of a 3.4-eV photon? (mel = 9.11 × 10-31 kg,c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J)