De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to $$(|e| = 1.6 \times 10^{-19}$$ C, $$m_e = 9.1 \times 10^{-31}$$ kg, $$h = 6.6 \times 10^{-34}$$ J s$$)$$

The de-Broglie wavelength $$\lambda$$ of an electron is given by the formula $$\lambda = \frac{h}{p}$$, where $$h$$ is Planck's constant and $$p$$ is the momentum of the electron. When an electron is accelerated by a voltage $$V$$, it gains kinetic energy equal to $$eV$$, where $$e$$ is the charge of the electron. The kinetic energy is also given by $$\frac{1}{2} m v^2$$, so we have:

$$\frac{1}{2} m v^2 = eV$$

Solving for $$v^2$$:

$$v^2 = \frac{2eV}{m}$$

The momentum $$p = m v$$, so $$p^2 = m^2 v^2$$. Substituting $$v^2$$:

$$p^2 = m^2 \cdot \frac{2eV}{m} = 2 m eV$$

Thus, $$p = \sqrt{2 m eV}$$. Now, the wavelength $$\lambda$$ becomes:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m eV}}$$

Given the values: voltage $$V = 50$$ V, charge $$|e| = 1.6 \times 10^{-19}$$ C, electron mass $$m_e = 9.1 \times 10^{-31}$$ kg, and Planck's constant $$h = 6.6 \times 10^{-34}$$ J s.

First, compute the product $$2 m_e e V$$:

$$m_e e = (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19}) = 9.1 \times 1.6 \times 10^{-31-19} = 14.56 \times 10^{-50} = 1.456 \times 10^{-49}$$

Now multiply by $$V = 50$$:

$$m_e e V = 1.456 \times 10^{-49} \times 50 = 1.456 \times 50 \times 10^{-49} = 72.8 \times 10^{-49} = 7.28 \times 10^{-48}$$

Multiply by 2:

$$2 m_e e V = 2 \times 7.28 \times 10^{-48} = 14.56 \times 10^{-48} = 1.456 \times 10^{-47}$$

Now find the square root:

$$\sqrt{2 m_e e V} = \sqrt{1.456 \times 10^{-47}} = \sqrt{1.456} \times \sqrt{10^{-47}} = \sqrt{1.456} \times 10^{-23.5}$$

Calculate $$\sqrt{1.456}$$. Since $$1.206^2 = 1.454436$$ and $$1.207^2 = 1.456849$$, interpolate for $$1.456$$:

$$1.206 + \frac{1.456 - 1.454436}{1.456849 - 1.454436} \times (1.207 - 1.206) = 1.206 + \frac{0.001564}{0.002413} \times 0.001 \approx 1.206 + 0.648 \times 0.001 = 1.206648$$

So $$\sqrt{1.456} \approx 1.2067$$. Now $$10^{-23.5} = 10^{-23} \times 10^{-0.5} = 10^{-23} \times \frac{1}{\sqrt{10}} \approx 10^{-23} \times 0.316227766 \approx 3.16227766 \times 10^{-24}$$. Thus:

$$\sqrt{2 m_e e V} \approx 1.2067 \times 3.16227766 \times 10^{-24} \approx 3.816 \times 10^{-24}$$

Now compute $$\lambda$$:

$$\lambda = \frac{h}{\sqrt{2 m_e e V}} = \frac{6.6 \times 10^{-34}}{3.816 \times 10^{-24}} = \frac{6.6}{3.816} \times 10^{-34+24} = \frac{6.6}{3.816} \times 10^{-10}$$

Calculate $$\frac{6.6}{3.816}$$:

$$3.816 \times 1.7297 \approx 3.816 \times 1.7 = 6.4872, \quad 3.816 \times 0.0297 \approx 0.1133, \quad \text{total} \approx 6.6005$$

So $$\frac{6.6}{3.816} \approx 1.7297$$. Thus:

$$\lambda \approx 1.7297 \times 10^{-10} \text{ m}$$

Convert to Angstroms (1 Å = $$10^{-10}$$ m):

$$\lambda \approx 1.7297 \text{ Å}$$

Comparing with the options:

A. 0.5 Å
B. 1.2 Å
C. 1.7 Å
D. 2.4 Å

The value 1.7297 Å is closest to 1.7 Å.

Hence, the correct answer is Option C.