Understanding De Broglie Wavelength and Electron Energy

This problem asks us to find the kinetic energy of an electron that has the same de Broglie wavelength as light from a spectral emission line of sodium. We are given the wavelength of the light, Planck's constant, and the mass of the electron. The key concept connecting wavelength and momentum for a particle is the de Broglie hypothesis.

De Broglie Hypothesis and Kinetic Energy

According to the de Broglie hypothesis, a particle with momentum \(p\) has a corresponding wavelength \(\lambda\) given by the formula:

\(\lambda = \frac{h}{p}\)

where \(h\) is Planck's constant.

For a non-relativistic particle like an electron moving with velocity \(v\), the momentum \(p\) is given by:

\(p = m_e v\)

where \(m_e\) is the mass of the electron.

The kinetic energy (\(KE\)) of the electron is given by:

\(KE = \frac{1}{2}m_e v^2\)

We need to relate the kinetic energy (\(KE\)) to the momentum (\(p\)). From the momentum formula, \(v = \frac{p}{m_e}\). Substituting this into the kinetic energy formula:

\(KE = \frac{1}{2}m_e \left(\frac{p}{m_e}\right)^2 = \frac{1}{2}m_e \frac{p^2}{m_e^2} = \frac{p^2}{2m_e}\)

So, the momentum squared is \(p^2 = 2m_e \cdot KE\), which means \(p = \sqrt{2m_e \cdot KE}\).

Now, we can substitute this expression for momentum into the de Broglie wavelength formula:

\(\lambda = \frac{h}{\sqrt{2m_e \cdot KE}}\)

We are given \(\lambda\), \(h\), and \(m_e\), and we need to find \(KE\). Let's rearrange the formula to solve for \(KE\). Squaring both sides:

\(\lambda^2 = \frac{h^2}{2m_e \cdot KE}\)

Now, solve for \(KE\):

\(KE = \frac{h^2}{2m_e \lambda^2}\)

Calculating the Electron's Kinetic Energy

We are given the following values:

• Wavelength of light, \(\lambda = 662 \text{ nm}\). We need to convert this to meters: \(662 \text{ nm} = 662 \times 10^{-9} \text{ m}\).
• Planck's constant, \(h = 6.62 \times 10^{-34} \text{ J}\cdot\text{s}\).
• Mass of the electron, \(m_e = 9 \times 10^{-31} \text{ kg}\).

Now, substitute these values into the formula for \(KE\):

\(KE = \frac{(6.62 \times 10^{-34} \text{ J}\cdot\text{s})^2}{2 \times (9 \times 10^{-31} \text{ kg}) \times (662 \times 10^{-9} \text{ m})^2}\)

\(KE = \frac{(6.62)^2 \times (10^{-34})^2 \text{ J}^2\cdot\text{s}^2}{18 \times 10^{-31} \text{ kg} \times (662)^2 \times (10^{-9})^2 \text{ m}^2}\)

\(KE = \frac{(6.62)^2 \times 10^{-68}}{18 \times 10^{-31} \times (662)^2 \times 10^{-18}} \frac{\text{J}^2\cdot\text{s}^2}{\text{kg}\cdot\text{m}^2}\)

Recall that \(1 \text{ J} = 1 \text{ kg}\cdot\text{m}^2/\text{s}^2\), so \(1 \text{ J}^2\cdot\text{s}^2 = 1 \text{ J} \cdot (1 \text{ kg}\cdot\text{m}^2/\text{s}^2) \cdot \text{s}^2 = 1 \text{ J}\cdot\text{kg}\cdot\text{m}^2\). Therefore, the units cancel out correctly to give Joules:

\(\frac{\text{J}^2\cdot\text{s}^2}{\text{kg}\cdot\text{m}^2} = \frac{\text{J}\cdot\text{kg}\cdot\text{m}^2}{\text{kg}\cdot\text{m}^2} = \text{J}\)

Continuing the numerical calculation:

\(KE = \frac{(6.62)^2}{18 \times (662)^2} \times \frac{10^{-68}}{10^{-31} \times 10^{-18}}\)

\(KE = \frac{(6.62)^2}{18 \times (100 \times 6.62)^2} \times \frac{10^{-68}}{10^{-49}}\)

\(KE = \frac{(6.62)^2}{18 \times (100)^2 \times (6.62)^2} \times 10^{-68 - (-49)}\)

\(KE = \frac{1}{18 \times (10^2)^2} \times 10^{-19}\)

\(KE = \frac{1}{18 \times 10^4} \times 10^{-19}\)

\(KE = \frac{1}{180000} \times 10^{-19}\)

Now, calculate the numerical value of \(\frac{1}{180000}\):

\(\frac{1}{180000} = \frac{1}{1.8 \times 10^5} \approx 0.555 \times 10^{-5}\)

Alternatively, using \(\frac{1}{18} \times 10^{-4}\):

\(\frac{1}{18} \approx 0.0555\)

\(0.0555 \times 10^{-4} = 5.55 \times 10^{-2} \times 10^{-4} = 5.55 \times 10^{-6}\)

So, \(KE \approx 5.55 \times 10^{-6} \times 10^{-19}\)

\(KE \approx 5.55 \times 10^{-25} \text{ J}\)

Result

The kinetic energy at which an electron would have the same de Broglie wavelength as 662 nm light is approximately \(5.55 \times 10^{-25} \text{ J}\).

Revision Table: Key Concepts in Wave-Particle Duality

Additional Information on De Broglie Wavelength and Electron Energy

The de Broglie hypothesis, proposed by Louis de Broglie in 1924, was a groundbreaking idea in quantum mechanics. It suggested that particles, not just light, exhibit wave-like properties. This concept of wave-particle duality is fundamental to understanding the behavior of matter at the atomic and subatomic levels.

• Wave Nature of Matter: The de Broglie wavelength shows that even particles like electrons can behave like waves. This wave nature is typically only observable for particles with very small mass and high momentum, which is why we don't observe everyday objects exhibiting significant wavelengths.
• Experimental Verification: The wave nature of electrons was experimentally confirmed by the Davisson-Germer experiment and G. P. Thomson's experiment, which showed electron diffraction patterns similar to those of X-rays.
• Electron Microscopy: The wave property of electrons is utilized in electron microscopes. Electrons with very short de Broglie wavelengths (achieved by accelerating them to high energies) can be used to image structures much smaller than what is possible with visible light.
• Relation to Energy Levels: In atoms, the de Broglie waves of electrons are related to the quantized energy levels. Electrons can only exist in orbits where their de Broglie wavelength fits an integer number of times around the circumference, leading to standing waves and discrete energy states.
• Sodium Emission Line: The question refers to a spectral emission line of sodium. Sodium lamps produce a characteristic yellow light, which is actually a doublet (two closely spaced lines) at wavelengths around 589 nm and 589.6 nm. The given wavelength of 662 nm is different from this typical sodium doublet, but the principle of calculating the equivalent electron wavelength remains the same.