MCQ
$A$ proton and an $\alpha$-particle are accelerated through a potential difference of $100 V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha$-particle is
  • A
    $\sqrt 2 :1$
  • B
    $2:1$
  • $2\sqrt 2 :1$
  • D
    $\frac{1}{{2\sqrt 2 }}:1$

Answer

Correct option: C.
$2\sqrt 2 :1$
c
(c) $\lambda = \frac{h}{{\sqrt {2mQV} }} $

$\Rightarrow \lambda \propto \frac{1}{{\sqrt {mQ} }}$

$ \Rightarrow \frac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \sqrt {\frac{{{m_\alpha }{Q_\alpha }}}{{{m_p}{Q_p}}}} $

$ = \sqrt {\frac{{4{m_p} \times 2{Q_p}}}{{{m_p} \times {Q_p}}}} = 2\sqrt 2 $

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