A particle A of mass m and charge q is accelerated by a potential difference of 50 V. Another particle B of mass 4m and charge q is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths $$\frac{\lambda_A}{\lambda_B}$$ is close to:

We start with the de-Broglie hypothesis, which states that a particle of momentum $$p$$ has an associated wavelength

$$\lambda \;=\;\frac{h}{p},$$

where $$h$$ is Planck’s constant.

For a non-relativistic charged particle accelerated from rest through a potential difference $$V$$, the work done by the electric field appears as kinetic energy. The relation between the charge, the potential difference and the gained kinetic energy is

$$\text{K.E.} \;=\; qV,$$

because electric potential energy $$qV$$ is converted entirely into kinetic energy.

Writing the kinetic energy in terms of momentum for a non-relativistic particle, we have the familiar expression

$$\text{K.E.} \;=\; \frac{p^{2}}{2m},$$

where $$m$$ is the mass of the particle. Equating the two expressions for kinetic energy, we get

$$\frac{p^{2}}{2m} \;=\; qV.$$

Solving this equation for the momentum $$p$$ gives

$$p \;=\; \sqrt{2mqV}.$$

Substituting this momentum into the de-Broglie wavelength formula, we obtain

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

We must now apply this expression separately to the two given particles A and B, compare their wavelengths, and then compute the required ratio.

For particle A, the mass is $$m_A = m$$ and the accelerating potential is $$V_A = 50 \text{ V}$$. Its de-Broglie wavelength is therefore

$$\lambda_A \;=\; \frac{h}{\sqrt{2\,m\,q\,\times 50}}.$$

For particle B, the mass is $$m_B = 4m$$ (four times the mass of A) and the accelerating potential is $$V_B = 2500 \text{ V}$$. Its de-Broglie wavelength is

$$\lambda_B \;=\; \frac{h}{\sqrt{2\,(4m)\,q\,\times 2500}}.$$

We now form the required ratio $$\dfrac{\lambda_A}{\lambda_B}$$. Since the constant factors $$h$$ and $$2q$$ appear in both numerators and denominators, they will cancel. Proceeding step by step:

$$\frac{\lambda_A}{\lambda_B} = \frac{\dfrac{h}{\sqrt{2\,m\,q\,\times 50}}}{\dfrac{h}{\sqrt{2\,(4m)\,q\,\times 2500}}} = \frac{\sqrt{2\,(4m)\,q\,\times 2500}}{\sqrt{2\,m\,q\,\times 50}}.$$

Cancelling the common factors $$2q$$ inside both square roots gives

$$\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{(4m)\;2500}}{\sqrt{m\;50}}.$$

We can now deal with the mass factor $$m$$. In the numerator we have $$4m$$, and in the denominator simply $$m$$. Extracting the mass term:

$$\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{4m}\,\sqrt{2500}}{\sqrt{m}\,\sqrt{50}} = \frac{\sqrt{4}\,\sqrt{m}\,\sqrt{2500}}{\sqrt{m}\,\sqrt{50}}.$$

The factors $$\sqrt{m}$$ cancel out, leaving

$$\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{4}\,\sqrt{2500}}{\sqrt{50}}.$$

Next, we simplify each square root:

$$\sqrt{4} = 2,\quad \sqrt{2500} = 50.$$

Substituting these numerical values, we get

$$\frac{\lambda_A}{\lambda_B} = \frac{2 \times 50}{\sqrt{50}} = \frac{100}{\sqrt{50}}.$$

We now simplify the denominator:

$$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}.$$

Putting this into the expression for the ratio, we have

$$\frac{\lambda_A}{\lambda_B} = \frac{100}{5\sqrt{2}} = \frac{20}{\sqrt{2}}.$$

To rationalize (or simply evaluate) the final fraction, multiply numerator and denominator by $$\sqrt{2}$$:

$$\frac{20}{\sqrt{2}} = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2}.$$

Finally, calculate $$10\sqrt{2}$$ numerically:

$$\sqrt{2}\approx 1.414 \quad\Longrightarrow\quad 10\sqrt{2}\approx 10 \times 1.414 = 14.14.$$

Thus,

$$\frac{\lambda_A}{\lambda_B} \approx 14.14.$$

Among the given choices, the value 14.14 corresponds to Option D.

Hence, the correct answer is Option D.