Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    The \[{{K}_{\alpha }}\] line for molybdenum (atomic no. = 42) has a wavelength of \[0.7078\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]. The wavelength of\[{{K}_{\alpha }}\] line of zinc (atomic no. = 30) will be

    A) \[1\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) \[1.388\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[0.3541\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D) \[0.5\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: B

    Solution :

    Energy in an orbit \[E=-{{Z}^{2}}\frac{Rhc}{{{n}^{2}}}\]                 or            \[E\propto {{Z}^{2}}\]                    ??(i)                 But         \[E=\frac{hc}{\lambda }\] or \[E\propto \frac{1}{\lambda }\]       ?(ii) From Eqs. (i) and (ii)                 \[\frac{{{Z}_{1}}^{2}}{{{Z}_{2}}^{2}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}\] or            \[{{\left( \frac{42}{30} \right)}^{2}}=\frac{{{\lambda }_{2}}}{0.7078}\] or            \[{{\lambda }_{2}}=1.388\overset{\text{o}}{\mathop{\text{A}}}\,\]


You need to login to perform this action.
You will be redirected in 3 sec spinner