The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? ($$m_p = 1.00727$$ u, $$m_e = 0.00055$$ u)

The de Broglie wavelength of a particle accelerated through a potential difference $$V$$ is $$\lambda = \frac{h}{\sqrt{2mqV}}$$, where $$m$$ is the mass and $$q$$ is the charge of the particle.

For an electron and a proton accelerated through the same potential $$V = 100$$ V, both having the same charge magnitude $$e$$, the ratio of their wavelengths is: $$\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_p eV}}{\sqrt{2m_e eV}} = \sqrt{\frac{m_p}{m_e}}$$.

Using the given masses: $$m_p = 1.00727$$ u and $$m_e = 0.00055$$ u, we get $$\frac{m_p}{m_e} = \frac{1.00727}{0.00055} = 1831.4$$.

Therefore, $$\frac{\lambda_e}{\lambda_p} = \sqrt{1831.4} \approx 42.8 \approx 43$$.

The ratio of the de Broglie wavelengths is approximately $$43 : 1$$.