QuestionA proton is roughly 1800 times more massive than an electron. If a proton and an electron are traveling at the same speed, the wavelength of the photon will be about $(1800)^{1 / 2}$ times longer than the wavelength of the electron. the wavelength of the photon will be about 1800 times longer than the wavelength of the electron. the wavelength of the photon will be roughly equal to the wavelength of the electron. the wavelength of the electron will be about $(1800)^{1 / 2}$ times longer than the wavelength of the photon. the wavelength of the electron will be about 1800 times longer than the wavelength of the photon.

Studdy Solution

STEP 1

1. We are using the de Broglie wavelength formula, which relates the wavelength of a particle to its momentum.
2. The mass of the proton is approximately 1800 times the mass of the electron.
3. The proton and electron are traveling at the same speed.

STEP 2

1. Recall the de Broglie wavelength formula.
2. Apply the formula to both the proton and the electron.
3. Compare the wavelengths of the proton and the electron.

STEP 3

Recall the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$ where $\lambda$ is the wavelength, $h$ is Planck's constant, $m$ is the mass of the particle, and $v$ is the velocity of the particle.

STEP 4

Apply the de Broglie formula to the proton: $\lambda_p = \frac{h}{m_p v}$ Apply the de Broglie formula to the electron: $\lambda_e = \frac{h}{m_e v}$

STEP 5

Since the speed $v$ is the same for both particles, compare the wavelengths: $\frac{\lambda_p}{\lambda_e} = \frac{m_e}{m_p}$
Given $m_p \approx 1800 \times m_e$ , we have: $\frac{\lambda_p}{\lambda_e} = \frac{m_e}{1800 \times m_e} = \frac{1}{1800}$
Therefore, the wavelength of the electron is about 1800 times longer than the wavelength of the proton.
The correct statement is:
"The wavelength of the electron will be about 1800 times longer than the wavelength of the photon."

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