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 $$\left(\frac{\lambda_e}{\lambda_p}\right)$$ is ______.

The de Broglie wavelength of any particle is given by $$\lambda=\frac{h}{p}$$, where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle.

When a charged particle of charge magnitude $$e$$ is accelerated from rest through a potential difference $$V$$, the gain in kinetic energy is $$eV$$. Hence for a particle of mass $$m$$:

$$\text{K.E.}=eV=\frac{p^{2}}{2m}\; \Longrightarrow\; p=\sqrt{2meV}$$

Substituting this in the de Broglie formula:

$$\lambda=\frac{h}{\sqrt{2meV}}$$ $$-(1)$$

Both the electron (mass $$m_e$$) and the proton (mass $$m_p$$) are accelerated through the same potential difference $$V$$, so the factor $$\sqrt{2eV}$$ is common. Writing their wavelengths from $$(1)$$:

$$\lambda_e=\frac{h}{\sqrt{2m_e eV}}, \qquad \lambda_p=\frac{h}{\sqrt{2m_p eV}}$$

Taking the ratio:

$$\frac{\lambda_e}{\lambda_p} =\frac{h}{\sqrt{2m_e eV}}\;\Big/\;\frac{h}{\sqrt{2m_p eV}} =\frac{1}{\sqrt{m_e}}\times\sqrt{m_p} =\sqrt{\frac{m_p}{m_e}}$$

Therefore $$\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}}$$.

Option A which is: $$\sqrt{\frac{m_p}{m_e}}$$