KCET · Physics · Dual Nature of Matter
The de-Broglie wavelength of a proton (charge \(=16 . \times 10^{-19} \mathrm{C}\), mass \(\left.=16 . \times 10^{-27} \mathrm{~kg}\right)\) accelerated through a potential difference of \(1 \mathrm{kV}\) is
- A \(600 Å\)
- B \(09 . \times 10^{-12} \mathrm{~m}\)
- C \(7 Å\)
- D \(0.9 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(B) \(09 . \times 10^{-12} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
According to de-Broglie hypothesis
\[
\begin{aligned}
\lambda &=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m q V}} \\
\therefore \quad \lambda &=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(1.6 \times 10^{-27}\right)\left(1.6 \times 10^{-19}\right) \times 1000}} \\
&=\frac{6.6 \times 10^{-34}}{7.16 \times 10^{-22}} \\
&=0.9 \times 10^{-12} \mathrm{~m}
\end{aligned}
\]
\[
\begin{aligned}
\lambda &=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m q V}} \\
\therefore \quad \lambda &=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(1.6 \times 10^{-27}\right)\left(1.6 \times 10^{-19}\right) \times 1000}} \\
&=\frac{6.6 \times 10^{-34}}{7.16 \times 10^{-22}} \\
&=0.9 \times 10^{-12} \mathrm{~m}
\end{aligned}
\]
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