Question

Two positively charged particles are accelerated by \(200\,\text{keV}\). The masses of particles are \(m_1 = 1\,\text{amu}\) and \(m_2 = 4\,\text{amu}\).

If the de-Broglie wavelength \((\lambda_d)_{m_1}\) is \(x\) times of the second particle \((\lambda_d)_{m_2}\), determine the value of \(x\).

Hint:

For particles accelerated through the {same potential}, de-Broglie wavelength varies as \(\lambda \propto \dfrac{1}{\sqrt{m}}\). Heavier particle $\Rightarrow$ shorter wavelength.

Answer 1

Correct Answer: 2

Concept:
The de-Broglie wavelength of a particle accelerated through a potential difference \(V\) is given by: \[ \lambda = \frac{h}{\sqrt{2mqV}} \] For particles accelerated through the same potential difference
, wavelength depends only on the square root of mass
: \[ \lambda \propto \frac{1}{\sqrt{m}} \]
Step 1: Write Ratio of Wavelengths
\[ \frac{(\lambda_d)_{m_1}}{(\lambda_d)_{m_2}} = \sqrt{\frac{m_2}{m_1}} \]
Step 2: Substitute Given Values
\[ m_1 = 1\,\text{amu}, \quad m_2 = 4\,\text{amu} \] \[ \frac{(\lambda_d)_{m_1}}{(\lambda_d)_{m_2}} = \sqrt{\frac{4}{1}} = 2 \]
Step 3: Determine \(x\)
\[ (\lambda_d)_{m_1} = x(\lambda_d)_{m_2} \Rightarrow x = 2 \] \[ \boxed{x = 2} \]