JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2026
- Hard
- Numerical
The ratio of de Broglie wavelength of a deutron with kinetic energy \(E\) to that of an alpha particle with kinetic energy \(2 E\), is \(n: 1\). The value of \(n\) is ___________.
(Assume mass of proton = mass of neutron) :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{ h }{ mv }=\frac{ h }{\sqrt{2 m \cdot KE }}\)
\(\frac{\lambda_{d}}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} \cdot KE_{\alpha}}{m_{d} \cdot KE_{d}}} = \sqrt{\frac{4m \cdot 2E}{2m \cdot E}} = 2:1\)
\(\frac{\lambda_{d}}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} \cdot KE_{\alpha}}{m_{d} \cdot KE_{d}}} = \sqrt{\frac{4m \cdot 2E}{2m \cdot E}} = 2:1\)
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