MHT CET · Physics · Dual Nature of Matter
Graph shows the variation of de-Broglie wavelength \((\lambda)\) versus \(\frac{1}{\sqrt{V}}\) where ' \(V\) ' is the accelerating potential for four particles A, B, C, \(\mathrm{D}\) carrying same charge but of masses \(\mathrm{m}_1, \mathrm{~m}_2\), \(\mathrm{m}_3, \mathrm{~m}_4\). Which on represents a particle of largest mass?

- A \(\mathrm{m}_1\)
- B \(\mathrm{m}_2\)
- C \(\mathrm{m}_3\)
- D \(\mathrm{m}_4\)
Answer & Solution
Correct Answer
(A) \(\mathrm{m}_1\)
Step-by-step Solution
Detailed explanation
de Broglie wayelength \(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)
\(\therefore \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}\)
\(\therefore \quad \lambda \sqrt{\mathrm{v}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq}}} \Rightarrow \frac{\lambda}{\left(\frac{1}{\sqrt{\mathrm{v}}}\right)}=\frac{1}{\sqrt{2 \mathrm{mq}}}\)
\(\therefore \quad\) Slope of the graph \(=\frac{1}{\sqrt{2 \mathrm{mq}}}\)
The slope will be maximum for minimum mass.
\(\therefore \quad \mathrm{m}_4\) is minimum and \(\mathrm{m}_1\) will be minimum.
\(\therefore \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}\)
\(\therefore \quad \lambda \sqrt{\mathrm{v}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq}}} \Rightarrow \frac{\lambda}{\left(\frac{1}{\sqrt{\mathrm{v}}}\right)}=\frac{1}{\sqrt{2 \mathrm{mq}}}\)
\(\therefore \quad\) Slope of the graph \(=\frac{1}{\sqrt{2 \mathrm{mq}}}\)
The slope will be maximum for minimum mass.
\(\therefore \quad \mathrm{m}_4\) is minimum and \(\mathrm{m}_1\) will be minimum.
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