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Question: What speed should a galaxy move with respect to us so that the sodium line at \(589.0nm\) is observe...

What speed should a galaxy move with respect to us so that the sodium line at 589.0nm589.0nm is observed at 589.6 nm?589.6{\text{ nm?}}
a.)  3.06×103ms−1 b.)  3.06×104ms−1 c.)  3.06×105ms−1 d.)  3.06×106ms−1  a.){\text{ }}\,{\text{3}}{\text{.06}} \times {\text{1}}{{\text{0}}^3}m{s^{ - 1}} \\\ b.){\text{ }}\,{\text{3}}{\text{.06}} \times {\text{1}}{{\text{0}}^4}m{s^{ - 1}} \\\ c.){\text{ }}\,{\text{3}}{\text{.06}} \times {\text{1}}{{\text{0}}^5}m{s^{ - 1}} \\\ d.){\text{ }}\,{\text{3}}{\text{.06}} \times {\text{1}}{{\text{0}}^6}m{s^{ - 1}} \\\

Explanation

Solution

The measure of the length of the complete wave cycle is known as the wave-length. The distance travelled by the wave from the point is known as the velocity. First convert the given unit of the wavelength in the MKS system. Convert nano-metre in metre. And then substitute the values in the standard formula stating relation between the velocity and wavelength and simplify. Use v=Δλλ×cv = \dfrac{{\Delta \lambda }}{\lambda } \times c

Complete step by step answer:
Given that- the Wavelength of the light is λ=589.0 nm\lambda = 589.0{\text{ nm}}
Observed wavelength is λo=589.6 nm{\lambda _o} = 589.6\,{\text{nm}}
Change in wavelength is Δλ=λo−λ\Delta \lambda = {\lambda _o} - \lambda
Δλ=589.6−589 Δλ=0.6 nm  \Delta \lambda = 589.6 - 589 \\\ \Delta \lambda = 0.6{\text{ nm}} \\\
Convert nano-metres into the metres.
  ⟹  λ=589×10−9m ∴Δλ=0.6×10−9m  \implies \lambda = 589 \times {10^{ - 9}}m \\\ \therefore \Delta \lambda = 0.6 \times {10^{ - 9}}m \\\
Speed of the light is v=3.00×108m/sv = 3.00 \times {10^8}{\text{m/s}}
Now, according to the formula –
Velocity of the light, v=Δλλ×cv = \dfrac{{\Delta \lambda }}{\lambda } \times c
Place the known values in the above equation
v=0.6×10−9589×10−9×3×108v = \dfrac{{0.6 \times {{10}^{ - 9}}}}{{589 \times {{10}^{ - 9}}}} \times 3 \times {10^8}
The terms with the same power and exponents in the division cancels each other.
v=0.6589×3×108v = \dfrac{{0.6}}{{589}} \times 3 \times {10^8}
Simplify the above right hand side of the equation –
v=1.8×108589 v=1800×105589  v = \dfrac{{1.8 \times {{10}^8}}}{{589}} \\\ v = \dfrac{{1800 \times {{10}^5}}}{{589}} \\\
Simplify as per the required answer
v=3.056×105m/sv = 3.056 \times {10^5}m/s
The above value can be re-written as equivalent to
v=3.06×105ms−1v = 3.06 \times {10^5}m{s^{ - 1}}.

Hence, from the given multiple choices – the option C is the correct answer.

Additional Information:
Remember the difference between the frequency wavelength and the wave velocity. The distance travelled by the wave in the medium during the time a particle completes one vibration is the wavelength. Wave velocity is equal to the product of frequency and the wavelength.

Note:
Remember the correct formula to solve these types of solutions. Rest goes well, just substitution and simplification. Take care of all the given units, all the terms should have the same format of the units for further simplification.