Q155 / 160 · TS EAMCET 2016 · Physics

A charged particle is accelerated from rest through a certain potential difference. The de-Broglie wavelength is $\lambda_1$ when it is accelerated through $V_1$ and is $\lambda_2$ when accelerated through $V_2$. The ratio $\lambda_1 / \lambda_2$ is

  • A
    $V_1^{3 / 2}: V_2^{3 / 2}$
  • Correct answer
    B
    $V_2^{1 / 2}: V_1^{1 / 2}$
  • C
    $V_1^{1 / 2}: V_2^{1 / 2}$
  • D
    $V_1^2: V_2^2$

Solution

The de-Broglie wavelength associated with a charged particle is given by $$ \lambda=\frac{h}{\sqrt{m q V}} $$ where, $h=$ planck's constant $m=$ mass of charged particle $q=$ charge on the particle $V=$ potential at which the charged particle is accelerated. $\therefore$ $$ \lambda_1=\frac{h}{\sqrt{m_1 q V_1}}, \lambda_2=\frac{h}{\sqrt{m_2 q V_2}} $$ Here, $$ \begin{aligned} m_1 & =m_2 \\ \frac{\lambda_1}{\lambda_2} & =\sqrt{\frac{m_2 q V_2}{m_1 q V_1}}=\sqrt{\frac{V_2}{V_1}} \\ & =V_2^{\frac{1}{2}} \cdot V_1^{\frac{1}{2}} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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