A) \[\text{25/9}f\]
B) \[\text{16/25}f\]
C) \[\text{9/25}f\]
D) zero
Correct Answer: A
Solution :
Wavelength of \[{{K}_{\alpha }},\] X-ray \[\lambda \propto \frac{1}{{{(Z-1)}^{2}}}\] \[\therefore \] \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{{{({{Z}_{1}}-1)}^{2}}}{{{({{Z}_{2}}-1)}^{2}}}\] But, \[\lambda \propto \frac{1}{frequency\,(f)}\] \[\therefore \] \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{{{({{Z}_{1}}-1)}^{2}}}{{{({{Z}_{2}}-1)}^{2}}}\] or \[\frac{f}{{{f}_{2}}}=\frac{{{(31-1)}^{2}}}{{{(51-1)}^{2}}}=\frac{900}{2500}\] \[\therefore \] \[{{f}_{2}}=\frac{25}{9}f\]
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