Question 1

An $\mathrm{x}$-ray photon of energy $3.20 \times 10$-15 J has what wavelength?&#10;A) $200 \mathrm{~nm}$&#10;B) $1.07 \mathrm{~nm}$&#10;C) $32.7 \mathrm{~nm}$&#10;D) $5.73 \mathrm{pm}$&#10;E) $62.1 \mathrm{pm}$

Accepted Answer

$62.1 \mathrm{pm}$&#10;

Question 2

A  photon of frequency $6.40 \times 1014 \mathrm{~Hz}$ has what energy?&#10;A) $2.65 \mathrm{eV}$&#10;B) $3.17 \mathrm{eV}$&#10;C) $1.93 \mathrm{eV}$&#10;D) $1.26 \mathrm{eV}$&#10;E) $2.11 \mathrm{eV}$

Accepted Answer

The energy of a photon is given by $E = hf$, where $h$ is Planck's constant and $f$ is the frequency of the photon. Substituting the given values into this equation, we get:$$E = (6.626 \times 10^{-34} \mathrm{~J\cdot s})(6.40 \times 10^{14} \mathrm{~Hz}) = 4.22 \times 10^{-19} \mathrm{~J}$$Converting this to electron volts, we get:$$E = \frac{4.22 \times 10^{-19} \mathrm{~J}}{1.602 \times 10^{-19} \mathrm{~J/eV}} = 2.65 \mathrm{~eV}$$Therefore, the correct answer is A.

Question 3

A $1.00 \mathrm{~mW}$ laser produces photons of wavelength $633 \mathrm{~nm}$. How many photons per second does the laser emit?&#10;A) $1.58 \times 10^{3}$&#10;B) $6.24 \times 1015$&#10;C) $3.19 \times 10^{15}$&#10;D) $1.60 \times 1019$&#10;E) $1.06 \times 1010$

Accepted Answer

The energy of one photon is calculated using $E = \frac{hc}{\lambda}$. For $633 \mathrm{~nm}$, $E \approx 3.14 \times 10^{-19} \mathrm{~J}$. The number of photons per second is $\frac{1.00 \times 10^{-3} \mathrm{~W}}{3.14 \times 10^{-19} \mathrm{~J}}$, which is approximately $3.19 \times 10^{15}$.

Question 4

The work function of a surface is $6.53 \times 10^{-19} \mathrm{~J}$. What is this quantity in $\mathrm{eV}$ ?&#10;A) 3.13&#10;B) 4.70&#10;C) 2.46&#10;D) 4.08&#10;E) 3.93

Accepted Answer

The answer of The work function of a surface is...

Question 5

Photons of energy $4.20 \mathrm{eV}$ bombard a surface, which emits photoelectrons with kinetic energies from 0 to $1.80 \mathrm{eV}$. What is the threshold frequency for this surface?&#10;A) $1.35 \times 1014 \mathrm{~Hz}$&#10;B) $5.80 \times 10^{14} \mathrm{~Hz}$&#10;C) $7.74 \times 1015 \mathrm{~Hz}$&#10;D) $7.74 \times 1014 \mathrm{~Hz}$&#10;E) $1.35 \times 10^{15} \mathrm{~Hz}$

Accepted Answer

The threshold frequency ($f_0$) can be found using the photoelectric equation $E_{\text{photon}} = \phi + K_{\text{max}}$, where $E_{\text{photon}}$ is the energy of the incoming photon, $\phi$ is the work function (or threshold energy), and $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons. Given $E_{\text{photon}} = 4.20 \, \text{eV}$ and $K_{\text{max}} = 1.80 \, \text{eV}$, we find $\phi = 4.20 \, \text{eV} - 1.80 \, \text{eV} = 2.40 \, \text{eV}$. To find the threshold frequency, convert the work function from eV to Joules ($1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$), then use $\phi = h f_0$, where $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$ is Planck's constant. Solving for $f_0$ gives $f_0 = \frac{\phi}{h}$ in Hz. Substituting $\phi = 2.40 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}$ yields $f_0 \approx 5.80 \times 10^{14} \, \text{Hz}$.

Question 6

An x-ray tube produces photons with wavelengths down to a minimum of $0.0827 \mathrm{~nm}$. What is the accelerating potential for this tube?&#10;A) $9.22 \mathrm{kV}$&#10;B) $15.0 \mathrm{kV}$&#10;C) $10.0 \mathrm{kV}$&#10;D) $123 \mathrm{kV}$&#10;E) $8.00 \mathrm{kV}$

Accepted Answer

The answer of An x-ray tube produces photons with wavelengths...

Question 7

An $x$-ray photon with wavelength $15.0 \mathrm{pm}$ is scattered at $84.0^{\circ}$ by an electron. What is the wavelength of the scattered photon?&#10;A) $31.0 \mathrm{pm}$&#10;B) $12.8 \mathrm{pm}$&#10;C) $14.9 \mathrm{pm}$&#10;D) $1.57 \mathrm{pm}$&#10;E) $17.2 \mathrm{pm}$

Accepted Answer

The wavelength of the scattered photon can be calculated using the Compton scattering formula: $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta)$, where $\lambda'$ is the wavelength of the scattered photon, $\lambda$ is the initial wavelength (15.0 pm), $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg} / \text{s}$), $m_e$ is the electron mass ($9.109 \times 10^{-31} \, \text{kg}$), $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$), and $\theta$ is the scattering angle ($84.0^\circ$). Plugging in the values and solving for $\lambda'$ gives a result that matches the option indicating a wavelength of $17.2 \, \text{pm}$.

Question 8

An $x$-ray photon with wavelength $15.0 \mathrm{pm}$ is scattered through $84.0^{\circ}$ by an electron. What is the resulting kinetic energy of the electron?&#10;A) $10.5 \mathrm{keV}$&#10;B) $59.5 \mathrm{keV}$&#10;C) $15.0 \mathrm{keV}$&#10;D) $568 \mathrm{keV}$&#10;E) $30.1 \mathrm{keV}$

Accepted Answer

The kinetic energy of the electron can be found using the Compton scattering formula and the conservation of energy principle. The change in wavelength ($\Delta \lambda$) due to scattering can be calculated using the Compton formula: $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta)$, where $h$ is Planck's constant, $m_e$ is the electron mass, $c$ is the speed of light, and $\theta$ is the scattering angle. The energy of the photon before and after scattering can be calculated using $E = \frac{hc}{\lambda}$, and the difference in energy gives the kinetic energy transferred to the electron. For a scattering angle of $84.0^\circ$ and an initial wavelength of $15.0 \, \text{pm}$, the calculation yields a kinetic energy of approximately $10.5 \, \text{keV}$ for the electron.

Question 9

In a Compton scattering experiment, the scattered photon has a wavelength of $6.55 \mathrm{pm}$. If the scattering angle is $50^{\circ}$, what was the wavelength of the incident photon?&#10;A) $0.868 \mathrm{pm}$&#10;B) $7.42 \mathrm{pm}$&#10;C) $3.28 \mathrm{pm}$&#10;D) $5.68 \mathrm{pm}$&#10;E) $2.43 \mathrm{pm}$

Accepted Answer

The Compton wavelength shift formula is $\Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)$. Given $\lambda' = 6.55 \mathrm{pm}$ and $\theta = 50^{\circ}$, solve for $\lambda$. The shift $\Delta \lambda$ is approximately $0.87 \mathrm{pm}$, so $\lambda = 6.55 \mathrm{pm} - 0.87 \mathrm{pm} = 5.68 \mathrm{pm}$.

Question 10

The mathematician Johann Jakob Balmer found that the formula $1 / \lambda=\mathrm{R}\left(1 / \mathrm{n}_{\mathrm{f}}^{2}-1 / \mathrm{n}_{\mathrm{i}}^{2}\right)$ works for the hydrogen emission lines in the visible range, with what value of $n_{\mathrm{f}}$ ?

A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

Balmer's formula specifically applies to the hydrogen emission lines in the visible spectrum, and it uses $n_{\mathrm{f}} = 2$. This is because the Balmer series describes the transitions of electrons from higher energy levels (n > 2) down to the n = 2 energy level.

Question 11

According to the Bohr model, what is the potential energy of an electron in the ground state of the hydrogen atom?&#10;A) $-27.2 \mathrm{eV}$&#10;B) $27.2 \mathrm{eV}$&#10;C) $-13.6 \mathrm{eV}$&#10;D) 0&#10;E) $13.6 \mathrm{eV}$

Accepted Answer

The answer of According to the Bohr model, what is...

Question 12

What is the highest energy of a photon in the Balmer series?&#10;A) $0.94 \mathrm{eV}$&#10;B) $3.40 \mathrm{eV}$&#10;C) $1.89 \mathrm{eV}$&#10;D) $1.51 \mathrm{eV}$&#10;E) $0.66 \mathrm{eV}$

Accepted Answer

The answer of What is the highest energy of a...

Question 13

For the ion $\mathrm{He}^{+}$, what is the energy of the ground state?&#10;A) $-13.6 \mathrm{eV}$&#10;B) $-27.2 \mathrm{eV}$&#10;C) $-3.40 \mathrm{eV}$&#10;D) $-6.80 \mathrm{eV}$&#10;E) $-54.4 \mathrm{eV}$

Accepted Answer

The answer of For the ion $\mathrm{He}^{+}$, what is the...

Question 14

What is the radius of the ground state of $\mathrm{He}^{+}$?&#10;A) $0.0529 \mathrm{~nm}$&#10;B) $0.212 \mathrm{~nm}$&#10;C) $0.193 \mathrm{~nm}$&#10;D) $0.0265 \mathrm{~nm}$&#10;E) $0.106 \mathrm{~nm}$

Accepted Answer

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

Question 15

According to the Bohr model, what is the kinetic energy of an electron in the ground state of hydrogen?&#10;A) $6.80 \mathrm{eV}$&#10;B) $27.2 \mathrm{eV}$&#10;C) 0&#10;D) $1.88 \mathrm{eV}$&#10;E) $13.6 \mathrm{eV}$

Accepted Answer

The answer of According to the Bohr model, what is...

Question 16

What is the smallest energy photon that a hydrogen atom in the ground state can absorb?&#10;A) $3.40 \mathrm{eV}$&#10;B) any energy&#10;C) $1.90 \mathrm{eV}$&#10;D) $13.6 \mathrm{eV}$&#10;E) $10.2 \mathrm{eV}$

Accepted Answer

The answer of  What is the smallest energy photon...

Question 17

If a hydrogen atom in the ground state absorbs a $20.0 \mathrm{eV}$ photon, what kinetic energy of the electron results?&#10;A) $33.6 \mathrm{eV}$&#10;B) this cannot happen&#10;C) $13.6 \mathrm{eV}$&#10;D) $6.4 \mathrm{eV}$&#10;E) $20.0 \mathrm{eV}$

Accepted Answer

The answer of If a hydrogen atom in the ground...

Question 18

What is the maximum wavelength photon that can produce a pair of tau particles (mass $1.777 \mathrm{GeV} / \mathrm{c}^{2}$ ) by pair production?&#10;A) $6.98 \times 10^{-7} \mathrm{~nm}$&#10;B) $1.74 \times 10^{-7} \mathrm{~nm}$&#10;C) $8.72 \times 10^{-8} \mathrm{~nm}$&#10;D) $3.49 \times 10^{-7} \mathrm{~nm}$

Accepted Answer

The answer of  What is the maximum wavelength photon...

Deck 27: Early Quantum Physics and the Photon