Two positively charged particles m1 and m2 have been accelerated across the same potential difference of 200 keV as shown below.

[Given mass of $$m_{1}$$ = 1amu and $$m_{2}$$ = 4amu]
The deBroglie wavelength of $$m_{1}$$ will be x times of $$m_{2}$$. The value of x is_______(nearest integer)

The two particles are released from rest and are accelerated through the same potential difference of $$V = 200 \,\text{keV}$$.

For a particle with charge $$q$$ accelerated through a potential difference $$V$$, the kinetic energy gained is
$$K = qV$$.

The (non-relativistic) de Broglie wavelength of a particle is related to its momentum by
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}.$$(Here $$h$$ is Planck’s constant.)

Substituting $$K = qV$$ gives
$$\lambda = \frac{h}{\sqrt{2mqV}}.$$

The potential difference $$V$$ and (for singly-charged ions) the charge magnitude $$q = e$$ are the same for both particles, so the wavelength varies only with the inverse square root of the mass:

$$\lambda \propto \frac{1}{\sqrt{m}}.$$

Therefore the ratio of the wavelengths is
$$\frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{m_{2}}{m_{1}}}.$$

Given $$m_{1}=1\,\text{amu}$$ and $$m_{2}=4\,\text{amu}$$, we obtain
$$\frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{4}{1}} = 2.$$

Hence, the de Broglie wavelength of particle $$m_{1}$$ is 2 times that of particle $$m_{2}$$.

Nearest integer value: 2.