What is the ratio of de Broglie wavelength of an electron to that of proton if the velocity of proton is $\dfrac{1}{6}$ the velocity of electron?

The de Broglie wavelength is given by $\lambda = \dfrac{h}{mv}$.

For an electron, $\lambda_e = \dfrac{h}{m_e v_e}$.

For a proton, $\lambda_p = \dfrac{h}{m_p v_p}$.

The ratio of their wavelengths is:

$\dfrac{\lambda_e}{\lambda_p} = \dfrac{m_p v_p}{m_e v_e}$

Given that the velocity of the proton is $\dfrac{1}{6}$ the velocity of the electron, we have $v_p = \dfrac{v_e}{6}$, which gives $\dfrac{v_p}{v_e} = \dfrac{1}{6}$.

The mass of a proton is approximately $1836$ times the mass of an electron, so $\dfrac{m_p}{m_e} = 1836$.

Substituting these values into the ratio:

$\dfrac{\lambda_e}{\lambda_p} = 1836 \times \dfrac{1}{6} = 306$

The ratio of the de Broglie wavelength of the electron to that of the proton is $306:1$.

Answer: $306:1$