Understanding De Broglie Wavelength and Particle Mass

The question asks us to find the mass of a particle given its velocity relative to an electron and the ratio of its de Broglie wavelength to that of the electron. This involves applying the concept of de Broglie wavelength, which relates the wave properties of matter to its momentum.

According to de Broglie's hypothesis, any moving particle has a wave associated with it, and the wavelength ($\lambda$) of this matter wave is inversely proportional to its momentum ($p$). The formula for de Broglie wavelength is given by:

\begin{equation} \lambda = \frac{h}{p} \end{equation}

Where:

• $h$ is Planck's constant (a fundamental constant in quantum mechanics).
• $p$ is the momentum of the particle.

Momentum ($p$) is defined as the product of mass ($m$) and velocity ($v$):

\begin{equation} p = mv \end{equation}

Substituting the momentum formula into the de Broglie wavelength formula, we get:

\begin{equation} \lambda = \frac{h}{mv} \end{equation}

Applying De Broglie Wavelength to the Problem

We are given information about two particles: a particle (let's call it 'p') and an electron (let's call it 'e').

For the particle 'p':

\begin{equation} \lambda_p = \frac{h}{m_p v_p} \end{equation}

For the electron 'e':

\begin{equation} \lambda_e = \frac{h}{m_e v_e} \end{equation}

We are given two crucial pieces of information:

1. The particle moves three times as fast as the electron: $v_p = 3v_e$
2. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$: $\frac{\lambda_p}{\lambda_e} = 1.813 \times 10^{-4}$

Calculating the Mass of the Particle

Let's take the ratio of the de Broglie wavelengths using the formulas above:

\begin{equation} \frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{m_p v_p}}{\frac{h}{m_e v_e}} \end{equation}

Simplifying the ratio:

\begin{equation} \frac{\lambda_p}{\lambda_e} = \frac{h}{m_p v_p} \times \frac{m_e v_e}{h} = \frac{m_e v_e}{m_p v_p} \end{equation}

Now, substitute the given relation $v_p = 3v_e$ into this equation:

\begin{equation} \frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p (3v_e)} \end{equation}

The term $v_e$ cancels out:

\begin{equation} \frac{\lambda_p}{\lambda_e} = \frac{m_e}{3 m_p} \end{equation}

We are given that $\frac{\lambda_p}{\lambda_e} = 1.813 \times 10^{-4}$. So, we can write:

\begin{equation} 1.813 \times 10^{-4} = \frac{m_e}{3 m_p} \end{equation}

We need to find the mass of the particle, $m_p$. Let's rearrange the equation to solve for $m_p$:

\begin{equation} 3 m_p \times (1.813 \times 10^{-4}) = m_e \end{equation}

\begin{equation} m_p = \frac{m_e}{3 \times (1.813 \times 10^{-4})} \end{equation}

The mass of an electron ($m_e$) is approximately $9.109 \times 10^{-31}$ kg.

Now, substitute the value of $m_e$ and calculate $m_p$:

\begin{equation} m_p = \frac{9.109 \times 10^{-31} \text{ kg}}{3 \times 1.813 \times 10^{-4}} \end{equation}

\begin{equation} m_p = \frac{9.109 \times 10^{-31}}{5.439 \times 10^{-4}} \text{ kg} \end{equation}

Performing the division:

\begin{equation} m_p \approx 1.6748 \times 10^{-31 - (-4)} \text{ kg} \end{equation}

\begin{equation} m_p \approx 1.6748 \times 10^{-27} \text{ kg} \end{equation}

Comparing this value with the given options, we find that it is very close to $1.67 \times 10^{-27}$ kg.

The calculated mass of the particle is approximately $1.675 \times 10^{-27}$ kg, which corresponds to the mass of a proton or neutron (nucleons).

Let's check the options:

• Option 1: $1.67 \times 10^{-27}$ kg
• Option 2: $1.675 \times 10^{-31}$ kg (This is close to electron mass)
• Option 3: $1.675 \times 10^{-29}$ kg
• Option 4: $1.675 \times 10^{-30}$ kg

Our calculated value $1.6748 \times 10^{-27}$ kg is closest to Option 1, $1.67 \times 10^{-27}$ kg. The slight difference is likely due to rounding of the values used (like $m_e$ or the given ratio) or the expected precision in the options.

Summary of Steps

1. Write down the de Broglie wavelength formula $\lambda = h/mv$.
2. Write the formulas for the particle ($\lambda_p$) and the electron ($\lambda_e$).
3. Use the given information relating their velocities ($v_p = 3v_e$) and wavelength ratio ($\lambda_p / \lambda_e = 1.813 \times 10^{-4}$).
4. Set up the ratio of wavelengths and substitute the velocity relation.
5. Simplify the equation to relate the masses ($m_p$ and $m_e$).
6. Rearrange the equation to solve for the unknown mass $m_p$.
7. Substitute the known value of the electron mass ($m_e$) and perform the calculation.
8. Compare the result with the provided options.

Revision Table: De Broglie Concepts

Additional Information on De Broglie Waves and Quantum Physics

The de Broglie hypothesis is a cornerstone of quantum mechanics, proposing that all matter exhibits wave-like properties. This idea, put forward by Louis de Broglie in 1924, extended the concept of wave-particle duality, which had previously only been applied to light, to matter.

• Wave-Particle Duality: This principle states that every elementary particle or quantum entity exhibits the properties of both particles and waves.
• Planck's Constant (h): This fundamental constant links the energy of a photon to its frequency ($E=h\nu$) and, in the case of de Broglie's hypothesis, momentum to wavelength. Its small value ($6.626 \times 10^{-34}$ Js) is why wave properties are only noticeable for very small particles like electrons and at very low momenta for macroscopic objects.
• Experimental Verification: The wave nature of electrons was experimentally confirmed by the Davisson-Germer experiment in 1927, where electrons were shown to diffract, just like waves.
• Significance: De Broglie waves are crucial for understanding the behavior of particles in quantum systems, such as the structure of atoms (electron orbitals) and the behavior of particles in confined spaces.

In this problem, by comparing the de Broglie wavelengths and velocities of the particle and the electron, we were able to determine the particle's mass relative to the electron's mass. This highlights the inverse relationship between de Broglie wavelength and mass (when momentum or velocity is considered in relation).