Structure of Atom Part 3

Concept:The de Broglie wavelength of a charged particle depends inversely on the square root of its mass when the accelerating potential is the same.Explanation:The de Broglie wavelength formula is $\lambda = \frac{h}{\sqrt{2mK}}$.Here, $h$ is Planck's constant and $K$ is the kinetic energy.Both particles $m_1$ and $m_2$ are accelerated through the same potential difference of 200 keV.Therefore, they acquire the same kinetic energy $K$.Since $h$ and $K$ are identical for both particles, $\lambda \propto \frac{1}{\sqrt{m}}$.Mass of $m_1$ is 1 amu and mass of $m_2$ is 4 amu.The ratio of their wavelengths is:$\frac{\lambda_{m_1}}{\lambda_{m_2}} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{4}{1}} = 2$.Thus, $\lambda_{m_1} = 2 \times \lambda_{m_2}$.This means the de Broglie wavelength of $m_1$ is 2 times that of $m_2$.Shortcut:For constant kinetic energy, $\lambda \propto \frac{1}{\sqrt{m}}$.Given the mass ratio $m_2 : m_1 = 4 : 1$.The wavelength ratio is $\lambda_{m_1} : \lambda_{m_2} = \sqrt{m_2} : \sqrt{m_1} = 2 : 1$.Therefore, $\lambda_{m_1} = 2 \cdot \lambda_{m_2}$.Answer:$x = 2$ (Option A)