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JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    The wavelength of \[{{K}_{\alpha }}\] X-ray from an element of atomic number 41 is \[\lambda ,\] then the wavelength of \[{{K}_{\alpha }}\] X-ray from an element of atomic number 21 is

    A)  \[4\lambda \]                                  

    B)  \[\lambda /4\]

    C)  \[\frac{\lambda }{2}\]                                  

    D)  \[2\lambda \]

    Correct Answer: A

    Solution :

    From Moseley's law\[\frac{1}{\lambda }=\frac{3R}{4}{{(z-1)}^{2}}\]for\[{{K}_{\alpha }}\]X-ray Let \[\lambda '\] is the required wavelength then \[\frac{\lambda '}{\lambda }=\frac{{{(41-1)}^{2}}}{{{(21-1)}^{2}}}\Rightarrow \lambda '=4\lambda \]


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