Question

1. calculate the magnitude of the momentum of a photon whose wavelength is 5.00 × 10² nm.
2. calculate the magnitude of the momentum of a photon whose frequency is 4.5 × 10¹⁵ hz.
3. calculate the magnitude of the momentum of a 1.50 × 10² ev photon.
4. calculate the wavelength of a photon having the same momentum as an electron moving at 1.0 × 10⁶ m/s.

answers

1. 1.33 × 10⁻²⁷ kg·m/s
2. 9.9 × 10⁻²⁷ kg·m/s
3. 8.00 × 10⁻²⁶ kg·m/s
4. 0.73 nm

Explanation:

To solve these problems, we'll use the formulas related to photon momentum and energy, as well as electron momentum. Let's tackle each problem one by one:

Problem 19: Calculate the magnitude of the momentum of a photon whose wavelength is \( 5.00 \times 10^2 \, \text{nm} \).

Step 1: Recall the formula for photon momentum.

The momentum \( p \) of a photon is given by:
\[ p = \frac{h}{\lambda} \]
where \( h = 6.626 \times 10^{-34} \, \text{J·s} \) (Planck's constant) and \( \lambda \) is the wavelength.

Step 2: Convert the wavelength to meters.

The wavelength \( \lambda = 5.00 \times 10^2 \, \text{nm} \). Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \):
\[ \lambda = 5.00 \times 10^2 \times 10^{-9} \, \text{m} = 5.00 \times 10^{-7} \, \text{m} \]

Step 3: Substitute into the momentum formula.

\[ p = \frac{6.626 \times 10^{-34} \, \text{J·s}}{5.00 \times 10^{-7} \, \text{m}} \]
\[ p \approx 1.33 \times 10^{-27} \, \text{kg·m/s} \]

Problem 20: Calculate the magnitude of the momentum of a photon whose frequency is \( 4.5 \times 10^{15} \, \text{Hz} \).

Step 1: Recall the relationship between frequency (\( f \)) and wavelength (\( \lambda \)).

\[ c = f\lambda \implies \lambda = \frac{c}{f} \]
where \( c = 3.00 \times 10^8 \, \text{m/s} \) (speed of light).

Step 2: Substitute \( \lambda = \frac{c}{f} \) into the photon momentum formula.

\[ p = \frac{h}{\lambda} = \frac{hf}{c} \]

Step 3: Plug in the values.

\[ p = \frac{(6.626 \times 10^{-34} \, \text{J·s})(4.5 \times 10^{15} \, \text{Hz})}{3.00 \times 10^8 \, \text{m/s}} \]
\[ p \approx \frac{2.9817 \times 10^{-18}}{3.00 \times 10^8} \]
\[ p \approx 9.9 \times 10^{-27} \, \text{kg·m/s} \]

Problem 21: Calculate the magnitude of the momentum of a \( 1.50 \times 10^2 \, \text{eV} \) photon.

Step 1: Convert energy from eV to joules.

\( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \), so:
\[ E = 1.50 \times 10^2 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} \approx 2.403 \times 10^{-17} \, \text{J} \]

Step 2: Recall the relationship between photon energy (\( E \)) and momentum (\( p \)).

For a photon, \( E = pc \implies p = \frac{E}{c} \)

Step 3: Substitute the values.

\[ p = \frac{2.403 \times 10^{-17} \, \text{J}}{3.00 \times 10^8 \, \text{m/s}} \]
\[ p \approx 8.00 \times 10^{-26} \, \text{kg·m/s} \]

Problem 22: Calculate the wavelength of a photon having the same momentum as an electron moving at \( 1.0 \times 10^6 \, \text{m/s} \).

Step 1: Calculate the electron's momentum.

The momentum of an electron is \( p = m_e v \), where \( m_e = 9.11 \times 10^{-31} \, \text{kg} \) (mass of electron) and \( v = 1.0 \times 10^6 \, \text{m/s} \).
\[ p = (9.11 \times 10^{-31} \, \text{kg})(1.0 \times 10^6 \, \text{m/s}) \approx 9.11 \times 10^{-25} \, \text{kg·m/s} \]

Step 2: Use the photon momentum formula to find wavelength.

For a photon, \( p = \frac{h}{\lambda} \implies \lambda = \frac{h}{p} \)

Step 3: Substitute the values.

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J·s}}{9.11 \times 10^{-25} \, \text{kg·m/s}} \]
\[ \lambda \approx 7.27 \times 10^{-10} \, \text{m} \]
Convert to nanometers (\( 1 \, \text{m} = 10^9 \, \text{nm} \)):
\[ \lambda \approx 0.73 \, \text{nm} \]

Final Answers:

1. \( \boldsymbol{1.33 \times 10^{-27} \, \text{kg·m/s}} \)
2. \( \boldsymbol{9.9 \times 10^{-27} \, \text{kg·m/s}} \)
3. \( \boldsymbol{8.00 \times 10^{-26} \, \text{kg·m/s}} \)
4. \( \boldsymbol{0.73 \, \text{nm}} \)

Answer:

To solve these problems, we'll use the formulas related to photon momentum and energy, as well as electron momentum. Let's tackle each problem one by one:

Problem 19: Calculate the magnitude of the momentum of a photon whose wavelength is \( 5.00 \times 10^2 \, \text{nm} \).

Step 1: Recall the formula for photon momentum.

The momentum \( p \) of a photon is given by:
\[ p = \frac{h}{\lambda} \]
where \( h = 6.626 \times 10^{-34} \, \text{J·s} \) (Planck's constant) and \( \lambda \) is the wavelength.

Step 2: Convert the wavelength to meters.

The wavelength \( \lambda = 5.00 \times 10^2 \, \text{nm} \). Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \):
\[ \lambda = 5.00 \times 10^2 \times 10^{-9} \, \text{m} = 5.00 \times 10^{-7} \, \text{m} \]

Step 3: Substitute into the momentum formula.

\[ p = \frac{6.626 \times 10^{-34} \, \text{J·s}}{5.00 \times 10^{-7} \, \text{m}} \]
\[ p \approx 1.33 \times 10^{-27} \, \text{kg·m/s} \]

Problem 20: Calculate the magnitude of the momentum of a photon whose frequency is \( 4.5 \times 10^{15} \, \text{Hz} \).

Step 1: Recall the relationship between frequency (\( f \)) and wavelength (\( \lambda \)).

\[ c = f\lambda \implies \lambda = \frac{c}{f} \]
where \( c = 3.00 \times 10^8 \, \text{m/s} \) (speed of light).

Step 2: Substitute \( \lambda = \frac{c}{f} \) into the photon momentum formula.

\[ p = \frac{h}{\lambda} = \frac{hf}{c} \]

Step 3: Plug in the values.

\[ p = \frac{(6.626 \times 10^{-34} \, \text{J·s})(4.5 \times 10^{15} \, \text{Hz})}{3.00 \times 10^8 \, \text{m/s}} \]
\[ p \approx \frac{2.9817 \times 10^{-18}}{3.00 \times 10^8} \]
\[ p \approx 9.9 \times 10^{-27} \, \text{kg·m/s} \]

Problem 21: Calculate the magnitude of the momentum of a \( 1.50 \times 10^2 \, \text{eV} \) photon.

Step 1: Convert energy from eV to joules.

\( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \), so:
\[ E = 1.50 \times 10^2 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} \approx 2.403 \times 10^{-17} \, \text{J} \]

Step 2: Recall the relationship between photon energy (\( E \)) and momentum (\( p \)).

For a photon, \( E = pc \implies p = \frac{E}{c} \)

Step 3: Substitute the values.

\[ p = \frac{2.403 \times 10^{-17} \, \text{J}}{3.00 \times 10^8 \, \text{m/s}} \]
\[ p \approx 8.00 \times 10^{-26} \, \text{kg·m/s} \]

Problem 22: Calculate the wavelength of a photon having the same momentum as an electron moving at \( 1.0 \times 10^6 \, \text{m/s} \).

Step 1: Calculate the electron's momentum.

The momentum of an electron is \( p = m_e v \), where \( m_e = 9.11 \times 10^{-31} \, \text{kg} \) (mass of electron) and \( v = 1.0 \times 10^6 \, \text{m/s} \).
\[ p = (9.11 \times 10^{-31} \, \text{kg})(1.0 \times 10^6 \, \text{m/s}) \approx 9.11 \times 10^{-25} \, \text{kg·m/s} \]

Step 2: Use the photon momentum formula to find wavelength.

For a photon, \( p = \frac{h}{\lambda} \implies \lambda = \frac{h}{p} \)

Step 3: Substitute the values.

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J·s}}{9.11 \times 10^{-25} \, \text{kg·m/s}} \]
\[ \lambda \approx 7.27 \times 10^{-10} \, \text{m} \]
Convert to nanometers (\( 1 \, \text{m} = 10^9 \, \text{nm} \)):
\[ \lambda \approx 0.73 \, \text{nm} \]

Final Answers:

1. \( \boldsymbol{1.33 \times 10^{-27} \, \text{kg·m/s}} \)
2. \( \boldsymbol{9.9 \times 10^{-27} \, \text{kg·m/s}} \)
3. \( \boldsymbol{8.00 \times 10^{-26} \, \text{kg·m/s}} \)
4. \( \boldsymbol{0.73 \, \text{nm}} \)