KCET · Physics · Dual Nature of Matter
The kinetic energy of an electron gets tripled, then the de-Broglie wavelength associated with it changes by a factor
- A \(\frac{1}{3}\)
- B \(\sqrt{3}\)
- C \(\frac{1}{\sqrt{3}}\)
- D 3
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
de-Broglie wavelength of an electron is given by
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}\)
or \(\quad \lambda \propto \frac{1}{\sqrt{\mathrm{K}}}\)
\(\therefore \quad \frac{\lambda^{\prime}}{\lambda}=\frac{1}{\sqrt{3 \mathrm{~K}}} \frac{\sqrt{\mathrm{K}}}{1}=\frac{1}{\sqrt{3}}\)
or \(\quad \lambda^{\prime}=\frac{\lambda}{\sqrt{3}}\)
Hence, de-Broglie wavelength will change by factor \(\frac{1}{\sqrt{3}}\).
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}\)
or \(\quad \lambda \propto \frac{1}{\sqrt{\mathrm{K}}}\)
\(\therefore \quad \frac{\lambda^{\prime}}{\lambda}=\frac{1}{\sqrt{3 \mathrm{~K}}} \frac{\sqrt{\mathrm{K}}}{1}=\frac{1}{\sqrt{3}}\)
or \(\quad \lambda^{\prime}=\frac{\lambda}{\sqrt{3}}\)
Hence, de-Broglie wavelength will change by factor \(\frac{1}{\sqrt{3}}\).
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