BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    The \[{{K}_{\alpha }}\] line of singly ionized calcium has a wavelength of 393.3 nm as measured on earth. In the spectrum of one of the observed galaxies, the spectral line is located at 401.8 nm. The speed with which this galaxy is moving away from us, will be

    A)  7400 m/s                            

    B)  \[32.\,4\times {{10}^{2}}m/s\]

    C)  6480 km/s         

    D)         none of these

    Correct Answer: C

    Solution :

    From Hubbles law, the red shift (Z) is directly proportional to the distance (r) of the galaxy  from us that is \[Z\propto r\]                    Also, \[Z=\frac{d\lambda }{\lambda }=\frac{v}{c}=\frac{\text{speed}\,\text{of}\,\text{galaxy}}{\text{speed}\,\text{of}\,\text{light}}\] Given, \[d\lambda =401.8-393.3=8.5\,nm,\] \[\lambda =393.3\,nm.\] \[\therefore \]  \[Z=\frac{8.5}{393.3}=0.0216\] Also       \[v=cZ\]                 \[=3\times {{10}^{8}}\times 0.0216\] \[=64.8\times {{10}^{8}}\,m/s\] Since, \[1\,km={{10}^{3}}m,\]therefore \[v=6480\,km/s\]


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