KCET · Chemistry · Structure of Atom
Two particles \(A\) and \(B\) are in motion. If the wavelength associated with particle \(A\) is \(5 \times 10^{-8} \mathrm{~m}\). Calculated the wavelength (in Ã…) associated with particle B if its momentum is half of \(\mathrm{A}\).
- A \( 1.25 \times 10^{-7} \mathrm{~m} \)
- B \( 1.0 \times 10^{-7} \mathrm{~m} \)
- C \( 1.0 \times 10^{-8} \mathrm{~m} \)
- D \( 2.5 \times 10^{-8} \mathrm{~m} \)
Answer & Solution
Correct Answer
(B) \( 1.0 \times 10^{-7} \mathrm{~m} \)
Step-by-step Solution
Detailed explanation
By de Broglie equation, \(\lambda_{A}=h / P_{A}\)
And \(\lambda_{B}=h / P_{B}\)
\(\lambda_{A} / \lambda_{B}=P_{A} / P_{B}\)
\(P_{B}=1 / 2 P_{A}(\) Given \()\)
\(\lambda_{A} / \lambda_{B}=1 / 2 P_{A} / P_{A}=P_{A}\)
\(\lambda_{B}=2 \lambda_{A}\)
\(\lambda_{B}=2 \times 5 \times 10^{-8} \mathrm{~m}\)
\(=10^{-7} \mathrm{~m}\)
And \(\lambda_{B}=h / P_{B}\)
\(\lambda_{A} / \lambda_{B}=P_{A} / P_{B}\)
\(P_{B}=1 / 2 P_{A}(\) Given \()\)
\(\lambda_{A} / \lambda_{B}=1 / 2 P_{A} / P_{A}=P_{A}\)
\(\lambda_{B}=2 \lambda_{A}\)
\(\lambda_{B}=2 \times 5 \times 10^{-8} \mathrm{~m}\)
\(=10^{-7} \mathrm{~m}\)
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