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). \includegraphics[width=0.5\linewidth]{75.png}

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

Concept: De Broglie wavelength is given by: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \] where \(E = qV\) is the kinetic energy gained by the particle.
Step 1: Relation of wavelengths. Since both particles are accelerated through the same potential difference, they gain the same kinetic energy \(E\). \[ \lambda \propto \frac{1}{\sqrt{m}} \]
Step 2: Ratio of wavelengths. \[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}} \]
Step 3: Substitute values. \[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{4}{1}} = 2 \] Final Answer: \[ x = 2 \]