The de Broglie wavelength associated with an electron accelerated through a potential difference V is \(\lambda_e\) and the de Broglie wavelength associated with a proton accelerated through the same potential difference is \(\lambda_p\). If their corresponding masses are \(m_e\) and \(m_p\), respectively, then the ratio of their de Broglie wavelengths \(\left(\dfrac{\lambda_e}{\lambda_p}\right)\) is ______.

The de Broglie wavelength of a particle accelerated through a potential difference \(V\) is given by \(\lambda = \dfrac{h}{\sqrt{2mqV}}\).

For an electron, the charge is \(e\) and mass is \(m_e\). Its de Broglie wavelength is \(\lambda_e = \dfrac{h}{\sqrt{2m_e eV}}\).

For a proton, the charge is \(e\) and mass is \(m_p\). Its de Broglie wavelength is \(\lambda_p = \dfrac{h}{\sqrt{2m_p eV}}\).

Taking the ratio of the two wavelengths, we get \(\dfrac{\lambda_e}{\lambda_p} = \dfrac{\dfrac{h}{\sqrt{2m_e eV}}}{\dfrac{h}{\sqrt{2m_p eV}}} = \sqrt{\dfrac{m_p}{m_e}}\).

Answer: \(\sqrt{\dfrac{m_p}{m_e}}\)