A proton and a Li\(^{3+}\) nucleus are accelerated by the same potential. If \( \lambda_{\text{Li}} \) and \( \lambda_p \) denote the de Broglie wavelengths of Li\(^{3+}\) and proton respectively, then the value of \( \frac{\lambda_{\text{Li}}}{\lambda_p} \) is \( x \times 10^{-1} \). The value of \( x \) is _________. (Rounded off to the nearest integer)
Given Data:
- Mass of Li\(^{3+}\) = \( 8.3 \times \text{mass of proton} \)
Detailed Step-by-Step Solution
To find the ratio of their wavelengths, we will use the relation between the De Broglie wavelength (\(\lambda\)), mass (\(m\)), charge (\(q\)), and accelerating potential (\(V\)).
Step 1: The De Broglie Formula
The kinetic energy of a particle accelerated through a potential \(V\) is \( K = qV \). Thus, the De Broglie wavelength is:
Step 2: Assign Variables for Proton and Li\(^{3+}\)
Let the mass of the proton be \( m_p \) and its charge be \( e \).
- For Proton (\(p\)): Mass = \( m_p \), Charge = \( e \)
- For Li\(^{3+}\): Mass = \( 8.3 m_p \) (given), Charge = \( 3e \) (since it has a +3 charge)
Both are accelerated by the same potential \(V\), so \(V\) is constant.
Step 3: Setup the Ratio
Now we set up the ratio of their wavelengths:
Since \(h\), \(2\), \(m_p\), \(e\), and \(V\) are present in both the numerator and denominator, they cancel out, leaving:
\( \frac{\lambda_{\text{Li}}}{\lambda_p} = \frac{1}{\sqrt{24.9}} \)
Step 4: Approximate and Calculate \(x\)
Notice that \( 24.9 \) is extremely close to \( 25 \). For competitive exams, we can safely approximate:
Therefore, the ratio becomes:
The question requires the answer in the form \( x \times 10^{-1} \):
Conclusion: Comparing our result \( 2 \times 10^{-1} \) with \( x \times 10^{-1} \), we find that the value of \( x \) is exactly 2.
Pro Tips for De Broglie Ratio Problems
In JEE Main, whenever particles are accelerated by the same potential difference, you should immediately use the proportionality shortcut: \( \lambda \propto \frac{1}{\sqrt{mq}} \). This skips the need to write out Planck's constant and the velocity parameters entirely. Also, identifying that \( \sqrt{24.9} \) is essentially \( \sqrt{25} \) saves valuable time.
To master shortcut techniques for the dual nature of matter and other quantum mechanical concepts, dive into our comprehensive notes on the Structure of Atom Class 11 Chemistry.
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