The wavelength of an electron of kinetic energy $$4.50 \times 10^{-29}$$ J is ______ $$\times 10{-5}$$ m. (Nearest integer)
Given: mass of electron is $$9 \times 10{-31}$$ kg, h = $$6.6 \times 10{-34}$$ Js

The de Broglie wavelength is given by:

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

where $$h = 6.6 \times 10^{-34}$$ Js, $$m = 9 \times 10^{-31}$$ kg, and $$K = 4.50 \times 10^{-29}$$ J.

We have $$2mK = 2 \times 9 \times 10^{-31} \times 4.50 \times 10^{-29} = 81 \times 10^{-60} = 8.1 \times 10^{-59}$$.

So $$\sqrt{2mK} = \sqrt{8.1 \times 10^{-59}} = 9 \times 10^{-30}$$ kg m/s.

Now the wavelength is:

$$\lambda = \frac{6.6 \times 10^{-34}}{9 \times 10^{-30}} = 0.733 \times 10^{-4} = 7.33 \times 10^{-5}\;\text{m}$$

Rounding to the nearest integer, $$\lambda \approx 7 \times 10^{-5}$$ m. So, the answer is $$7$$.