Two particles A and B of masses and have de Broglie wavelength and . If , compare their kinetic…
Two particles A and B of masses $m_A$ and $m_B$ have de Broglie wavelength $\lambda_A$ and $\lambda_B$. If $\lambda_A = 2\lambda_B$, compare their kinetic energies $K_A$ and $K_B$.
Answer
Answer
AIWritten by AI (gemini) - it can contain mistakes.
$\frac{(KE)_A}{(KE)_B} = \frac{h^2/(2m_A \lambda_A^2)}{h^2/(2m_B \lambda_B^2)} = \frac{m_B \lambda_B^2}{m_A \lambda_A^2}$. Given $\lambda_A = 2\lambda_B$, $\frac{(KE)_A}{(KE)_B} = \frac{m_B \lambda_B^2}{m_A (2\lambda_B)^2} = \frac{m_B}{4m_A}$. Thus, $(KE)_A : (KE)_B = m_B : 4m_A$.
Final answer: m_B : 4m_A
From ISC 2027 Specimen Physics Paper 1, question 8.