The wavelength of electrons accelerated from rest through a potential difference of 40 kV is X $$\times 10^{-12}$$ m. The value of X is. (Nearest integer)
Given: Mass of electron = 9.1 $$\times 10^{-31}$$ kg
Charge on an electron = 1.6 $$\times 10^{-19}$$ C
Planck's constant = 6.63 $$\times 10^{-34}$$ Js

The de Broglie wavelength of an electron accelerated through a potential difference $$V$$ is given by:

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

Substituting the given values: $$h = 6.63 \times 10^{-34}$$ Js, $$m = 9.1 \times 10^{-31}$$ kg, $$e = 1.6 \times 10^{-19}$$ C, and $$V = 40{,}000$$ V:

$$2meV = 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 40000$$

$$= 2 \times 9.1 \times 1.6 \times 4 \times 10^{-31-19+4}$$

$$= 2 \times 58.24 \times 10^{-46} = 116.48 \times 10^{-46} = 1.1648 \times 10^{-44} \text{ kg}^2\text{m}^2\text{s}^{-2}$$

$$\sqrt{2meV} = \sqrt{1.1648 \times 10^{-44}} = 1.0793 \times 10^{-22} \text{ kg m s}^{-1}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{1.0793 \times 10^{-22}} = 6.14 \times 10^{-12} \text{ m}$$

So $$\lambda = X \times 10^{-12}$$ m, where $$X \approx 6$$.

The value of X is $$\mathbf{6}$$.