For the wavelength associated with a frequency of 50 kHz. Although the question mentions an "electron produced from the surface of a metal," the given frequency is characteristic of an ele

The frequency $f = 50 \, \text{kHz}$.
Since $1 \, \text{kHz} = 10^3 \, \text{Hz}$,

$$f = 50 \times 10^3 \, \text{Hz}$$

The speed of light in a vacuum is approximately $c = 3.00 \times 10^8 \, \text{m/s}$.
The formula is:

$$c = \lambda f$$

To find the wavelength, rearrange the formula:

$$\lambda = \frac{c}{f}$$

$$\lambda = \frac{3.00 \times 10^8 \, \text{m/s}}{50 \times 10^3 \, \text{Hz}}$$ $$\lambda = \frac{3.00 \times 10^8}{5.0 \times 10^4} \, \text{m}$$ $$\lambda = \left(\frac{3.00}{5.0}\right) \times 10^{8-4} \, \text{m}$$ $$\lambda = 0.6 \times 10^4 \, \text{m}$$ $$\lambda = 6 \times 10^3 \, \text{m}$$

(A) $6 \times 10^{-6} \, \text{m}$
(B) $6 \times 10^3 \, \text{m}$
(C) $5 \times 10^{-6} \, \text{m}$
(D) $5 \times 10^3 \, \text{m}$

The calculated wavelength matches option (B).

The final answer is $\boxed{\text{B) } 6 \times 10^3 \, \text{m}}$.