91Ó°ÊÓ

the detected wavelengths of light from a distant galaxy is$0.50\%$longer than the actual wavelength. So, the measured wavelength is,role="math" localid="1649946954794" $\mathrm{λ}=1.005{\mathrm{λ}}_o$

The Doppler effect for receding light waves, the detected wavelength is longer than the emitted wavelength by the source.

Here according to the question, the detected wavelengths of light from a distant galaxy is$0.5\%$longer than the actual wavelength. So, the galaxy is receding away.

Consider the galaxy has a speed of$v_c$

Now, according to the equation of detected wavelength of receding light wave,

role="math" localid="1649947051967" $$\begin{gathered} \mathrm{λ}=\sqrt{\frac{1+{\displaystyle \frac{v_s}{c}}}{1-\frac{v_s}{c}}}{\mathrm{λ}}_o \\ 1.005{\mathrm{λ}}_o=\sqrt{\frac{1+{\displaystyle \frac{v_s}{c}}}{1-\frac{v_s}{c}}}{\mathrm{λ}}_o\left[âˆ\mu \mathrm{λ}=1.005{\mathrm{λ}}_o\right] \\ {\left(1.005\right)}^2=\frac{1+{\displaystyle \frac{v_s}{c}}}{1-\frac{v_s}{c}} \\ {\left(1.005\right)}^2=\frac{c+v_c}{c-v_c} \\ {\left(1.005\right)}^{2}c-{\left(1.005\right)}^2{v}_c=c+v_c \\ 0.01c=2.01v_c \\ {v}_c=0.005c \\ {v}_c=1.5×{10}^{6}m/s \end{gathered}$$