ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter

Electrons are accelerated through a potential difference \(V\) and protons are accelerated through a potential difference \(4\, V\). The de-Broglie wavelengths are \(\lambda_e \) and \(\lambda_p \) for electrons and protons respectively. The ratio of \(\frac{{{\lambda _e}}}{{{\lambda _p}}}\) is given by : (given \(m_e\) is mass of electron and \(m_p\) is mass of proton).

  1. A \(\frac{{{\lambda _e}}}{{{\lambda _p}}} = \sqrt {\frac{{{m_p}}}{{{m_e}}}} \)
  2. B \(\frac{{{\lambda _e}}}{{{\lambda _p}}} = \sqrt {\frac{{{m_e}}}{{{m_p}}}} \)
  3. C \(\frac{{{\lambda _e}}}{{{\lambda _p}}} = \frac{1}{2}\sqrt {\frac{{{m_e}}}{{{m_p}}}} \)
  4. D \(\frac{{{\lambda _e}}}{{{\lambda _p}}} = 2\sqrt {\frac{{{m_p}}}{{{m_e}}}} \)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{{{\lambda _e}}}{{{\lambda _p}}} = 2\sqrt {\frac{{{m_p}}}{{{m_e}}}} \)

Step-by-step Solution

Detailed explanation

Energy in joule \((E)\) \(=\) charge \(\times\) potential diff. in volt \({{\text{E}}_{{\text{electron }}}} = {q_{\text{e}}}{\text{V}}\) and \({{\text{E}}_{{\text{proton }}}} = {q_{\text{p}}}4{\text{V}}\) de-Broglie wavelength…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app