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

To determine the value of \(x\) such that \((\lambda_d)_{m_1} = x \times (\lambda_d)_{m_2}\), we will use the de-Broglie wavelength formula:

\(\lambda_d = \frac{h}{\sqrt{2mE}}\)

where \(h\) is the Planck’s constant, \(m\) is the mass of the particle, and \(E\) is the energy.

The energy \(E\) provided is \(200\,\text{keV}\), which is equal for both particles.

For the first particle with mass \(m_1 = 1\,\text{amu}\):

\((\lambda_d)_{m_1} = \frac{h}{\sqrt{2 \cdot 1 \cdot \text{amu} \cdot 200\,\text{keV}}}\)

For the second particle with mass \(m_2 = 4\,\text{amu}\):

\((\lambda_d)_{m_2} = \frac{h}{\sqrt{2 \cdot 4 \cdot \text{amu} \cdot 200\,\text{keV}}}\)

Dividing the wavelengths:

\(\frac{(\lambda_d)_{m_1}}{(\lambda_d)_{m_2}} = \frac{\sqrt{2 \cdot 4 \cdot 200\,\text{keV}}}{\sqrt{2 \cdot 1 \cdot 200\,\text{keV}}}\)

This simplifies to:

\(\frac{(\lambda_d)_{m_1}}{(\lambda_d)_{m_2}} = \sqrt{4} = 2\)

Thus, \(x = 2\).

The computed value \(x=2\) is confirmed to be within the given range of \(2,2\).