A galaxy is moving away from the earth at a speed of 286 km s$$^{-1}$$. The shift in the wavelength of a red line at 630 nm is $$x \times 10^{-10}$$ m. The value of $$x$$, to the nearest integer, is ___.
[Take the value of the speed of the light $$c$$, as $$3 \times 10^8$$ m s$$^{-1}$$]

When a galaxy moves away from the earth, the observed wavelength is red-shifted. The Doppler shift in wavelength for light is given by $$\Delta\lambda = \frac{v}{c}\lambda$$, where $$v$$ is the recessional speed, $$c$$ is the speed of light, and $$\lambda$$ is the original wavelength.

Substituting the given values: $$\Delta\lambda = \frac{286 \times 10^3}{3 \times 10^8} \times 630 \times 10^{-9}$$.

Computing step by step: $$\frac{286 \times 10^3}{3 \times 10^8} = \frac{286}{3 \times 10^5} = 9.533 \times 10^{-4}$$.

Then $$\Delta\lambda = 9.533 \times 10^{-4} \times 630 \times 10^{-9} = 6.006 \times 10^{-10}$$ m.

Since $$\Delta\lambda = x \times 10^{-10}$$ m, the value of $$x$$ to the nearest integer is $$6$$.