A proton and a $$Li^{3+}$$ nucleus are accelerated by the same potential. If $$\lambda_{Li}$$ and $$\lambda_p$$ denote the de Broglie wavelengths of $$Li^{3+}$$ and proton respectively, then the value of $$\frac{\lambda_{Li}}{\lambda_p}$$ is $$x \times 10^{-1}$$. The value of $$x$$ is ______ (Rounded off to the nearest integer) [Mass of $$Li^{3+}$$ = 8.3 mass of proton]

We are given that a proton and a $$Li^{3+}$$ nucleus are accelerated by the same potential. We need to find $$\frac{\lambda_{Li}}{\lambda_p}$$.

The de Broglie wavelength of a charged particle accelerated through potential $$V$$ is given by:

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

where $$h$$ is Planck's constant, $$m$$ is the mass, $$q$$ is the charge, and $$V$$ is the accelerating potential.

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

For $$Li^{3+}$$: $$\lambda_{Li} = \frac{h}{\sqrt{2 m_{Li} \cdot 3e \cdot V}}$$

Here $$m_{Li} = 8.3 \, m_p$$ and the charge on $$Li^{3+}$$ is $$3e$$.

Taking the ratio:

$$\frac{\lambda_{Li}}{\lambda_p} = \frac{\sqrt{2 m_p \cdot e \cdot V}}{\sqrt{2 m_{Li} \cdot 3e \cdot V}} = \sqrt{\frac{m_p \cdot e}{m_{Li} \cdot 3e}}$$ $$= \sqrt{\frac{m_p}{8.3 \, m_p \times 3}} = \sqrt{\frac{1}{24.9}}$$ $$= \frac{1}{\sqrt{24.9}} = \frac{1}{4.99} \approx 0.2$$

We are told the ratio equals $$x \times 10^{-1}$$. So:

$$0.2 = x \times 10^{-1}$$ $$x = 2$$

So, the answer is $$2$$.