An $$\alpha$$ particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are $$\lambda_\alpha$$ and $$\lambda_p$$ respectively. The ratio $$\frac{\lambda_p}{\lambda_\alpha}$$ is:

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

For the proton: $$\lambda_p = \frac{h}{\sqrt{2m_p \cdot e \cdot V}}$$

For the $$\alpha$$ particle (mass $$= 4m_p$$, charge $$= 2e$$): $$\lambda_\alpha = \frac{h}{\sqrt{2 \cdot 4m_p \cdot 2e \cdot V}} = \frac{h}{\sqrt{16m_p eV}}$$

Taking the ratio:

$$\frac{\lambda_p}{\lambda_\alpha} = \frac{\sqrt{16m_p eV}}{\sqrt{2m_p eV}} = \sqrt{\frac{16}{2}} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$

Rounding to one decimal place, this gives approximately $$2.8$$.

The correct answer is Option (3): 2.8.