A proton, an electron, and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:
- \(\lambda_e>\lambda_\alpha>\lambda_p\)
- \(\lambda_\alpha<\lambda_p<\lambda_e\)
- \(\lambda_p<\lambda_e<\lambda_\alpha\)
- \(\lambda_p>\lambda_e>\lambda_\alpha\)
The Correct Option is B
Approach Solution - 1
To compare the de-Broglie wavelengths of particles with the same energies, we need to understand the de-Broglie wavelength formula and how it is affected by the mass and energy of the particles.
The de-Broglie wavelength \(\lambda\) is given by the formula:
\(\lambda = \dfrac{h}{p}\)
where:
- \(h\) is the Planck's constant.
- \(p\) is the momentum of the particle.
For a particle with kinetic energy \(E\), the momentum can be expressed as:
\(p = \sqrt{2mE}\)
Substituting this in the de-Broglie equation, we get:
\(\lambda = \dfrac{h}{\sqrt{2mE}}\)
From this equation, we see that for particles with the same energy \(E\), the wavelength \(\lambda\) is inversely proportional to the square root of their masses:
\(\lambda \propto \dfrac{1}{\sqrt{m}}\)
Now, let's compare the masses of the given particles:
- Proton (\(m_p\)): Approximately \(1.67 \times 10^{-27}\) kg
- Electron (\(m_e\)): Approximately \(9.11 \times 10^{-31}\) kg
- Alpha particle (\(m_\alpha\)): Consists of 2 protons and 2 neutrons, approximately \(4 \times 1.67 \times 10^{-27}\) kg = \(6.68 \times 10^{-27}\) kg
Based on these masses:
- The electron is the lightest, hence it has the longest wavelength.
- The proton is heavier than the electron, but lighter than the alpha particle, so it has an intermediate wavelength.
- The alpha particle is the heaviest, hence it has the shortest wavelength.
Therefore, the correct order of the de-Broglie wavelengths is:
\(\lambda_\alpha < \lambda_p < \lambda_e\)
Thus, the correct answer is:
\(\lambda_\alpha<\lambda_p<\lambda_e\)
Approach Solution -2
The de-Broglie wavelength is given by:
\[ \lambda_{DB} = \frac{h}{p} = \frac{h}{\sqrt{2mK}}, \]
where \(m\) is the mass, \(K\) is the kinetic energy, and \(p\) is the momentum. Since the particles have the same energy \(K\), the de-Broglie wavelength becomes:
\[ \lambda_{DB} \propto \frac{1}{\sqrt{m}}. \]
Step 1: Compare masses
- Electron (\(m_e\)): Lightest particle.
- Proton (\(m_p\)): Heavier than an electron.
- Alpha particle (\(m_\alpha\)): Heaviest, \(m_\alpha = 4m_p\).
Step 2: Compare wavelengths
Since \(\lambda_{DB} \propto \frac{1}{\sqrt{m}}\), the lighter the mass, the longer the wavelength:
\[ \lambda_e > \lambda_p > \lambda_\alpha. \]
Final Answer:
\[ \lambda_\alpha < \lambda_p < \lambda_e. \]
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