Question

Two positively charged particles \(m_1\) and \(m_2\) have been accelerated across the same potential difference of 200 keV. Given mass of \(m_1 = 1 \,\text{amu}\) and \(m_2 = 4 \,\text{amu}\). The de Broglie wavelength of \(m_1\) will be \(x\) times that of \(m_2\). The value of \(x\) is _______ (nearest integer). 

Hint:

For particles accelerated through the same potential difference: - Kinetic energy is the same. - De Broglie wavelength varies inversely with the square root of mass.

Answer 1

Correct Answer: 2

To solve the problem, we need to calculate the de Broglie wavelengths of particles \(m_1\) and \(m_2\) and determine how many times the wavelength of \(m_1\) is that of \(m_2\).

First, recall the de Broglie wavelength formula: \[\lambda = \frac{h}{p}\] where \(h\) is Planck’s constant and \(p\) is the momentum. The momentum \(p\) for a charged particle accelerated through a potential difference \(V\) is given by:

\[p = \sqrt{2m e V}\]

where \(e\) is the charge of the electron and \(m\) is the mass of the particle.

For particle \(m_1\):

\[\lambda_1 = \frac{h}{\sqrt{2 \times m_1 \times e \times V}}\]

For particle \(m_2\):

\[\lambda_2 = \frac{h}{\sqrt{2 \times m_2 \times e \times V}}\]

The ratio of the wavelengths is:

\[\frac{\lambda_1}{\lambda_2} = \frac{\sqrt{m_2}}{\sqrt{m_1}}\]

Given \(m_1 = 1 \, \text{amu}\) and \(m_2 = 4 \, \text{amu}\), we have:

\[\frac{\lambda_1}{\lambda_2} = \frac{\sqrt{4}}{\sqrt{1}} = \frac{2}{1} = 2\]

Thus, the de Broglie wavelength of \(m_1\) is 2 times that of \(m_2\). The value of \(x\) is 2, which falls within the expected range of [2, 2].