Question

A proton and an \(\propto\) -particle are passed through same potential difference. If their initial velocity is zero, the ratio of their de Broglie's wavelength after getting accelerated is \(\ldots \ldots\) (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(2 \sqrt{2}: 1\)

Short answer

The ratio of the de Broglie wavelengths of a proton and an alpha particle after being accelerated through the same potential difference is: (D) \(2 \sqrt{2}: 1\)

Step-by-step solution

Calculate the final velocities of the proton and alpha particle

We can calculate the final velocity of the proton and alpha particle after passing through the potential difference using the formula: \(v = \sqrt{\dfrac{2eV}{m}}\) where \(v\) is the final velocity, \(e\) is the charge of the particle, \(V\) is the potential difference, and \(m\) is the mass of the particle. For the proton, the mass is \(1.67 \times 10^{-27}\) kg and the charge is \(1.6 \times 10^{-19}\) C. For the alpha particle, the mass is \(6.68 \times 10^{-27}\) kg and the charge is \(2 \times 1.6 \times 10^{-19}\) C. Note that the potential difference is the same for both particles, which means that their final velocities can be expressed as ratios.

Calculate the momenta of the proton and alpha particle

Once we have the final velocities, we can find the momenta of the proton and alpha particle using the formula: \(p = mv\) where \(p\) is the momentum, \(m\) is the mass, and \(v\) is the final velocity. As we found the final velocities in Step 1 as ratios, we can keep the momenta of the particles as ratios.

Calculate the de Broglie wavelengths

Finally, using the formula for the de Broglie wavelength (\(\lambda = \dfrac{h}{p}\)), we can calculate the wavelengths for both particles, expressed as ratios by replacing the momenta found in Step 2.

Compute the ratio of the de Broglie wavelengths

Now that we have the de Broglie wavelengths of both particles, we can compute the ratio of their wavelengths. For the proton, we can denote its wavelength as \(λ₁\) and for the alpha particle, we can denote its wavelength as \(λ₂\). We are looking for the ratio \(\dfrac{λ₁}{λ₂}\). By using the ratios found in Step 1 and Step 2 we can determine the ratio of the de Broglie wavelengths: \(\dfrac{λ₁}{λ₂} = \dfrac{p₂}{p₁}\) Plugging the ratios for velocities and momenta into the expression for the de Broglie wavelengths (since the Planck constant is the same for both particles, it cancels out), we get: \(\dfrac{λ₁}{λ₂} = \dfrac{m_2v_2\sqrt{2eV}}{(2eV)m_1v_1} = \dfrac{m_2v_2}{m_1v_1}\) Now plug in the values and solve for the ratio to find that the ratio of their de Broglie wavelengths is: \(\dfrac{λ₁}{λ₂}=\dfrac{6.68 \times 10^{-27} \cdot \sqrt{\dfrac{2eV}{6.68 \times 10^{-27}}}}{1.67 \times 10^{-27} \cdot \sqrt{\dfrac{2 \cdot 1.6 \times 10^{-19}V}{1.67 \times 10^{-27}}}}\) Which after canceling out terms simplifies to: \(\dfrac{λ₁}{λ₂}=\dfrac{2}{\sqrt{2}}\) So the ratio of their de Broglie wavelengths is: \(\dfrac{λ₁}{λ₂} = 2 \sqrt{2}: 1\) The correct answer is: (D) \(2 \sqrt{2}: 1\)