A) \[\left( \frac{43}{29} \right)\lambda \]
B) \[\left( \frac{42}{28} \right)\lambda \]
C) \[\left( \frac{9}{4} \right)\lambda \]
D) \[\left( \frac{4}{9} \right)\lambda \]
Correct Answer: C
Solution :
Mosele/s law is given by \[\sqrt{v}=a(Z-b)\] For\[{{K}_{a}}\]line,\[b=1\] \[\sqrt{v}=a(Z-1)\] Squaring, we get \[v={{a}^{2}}{{(Z-1)}^{2}}\] or \[\frac{c}{\lambda }={{a}^{2}}{{(Z-1)}^{2}}\] or \[\lambda =\frac{c}{{{a}^{2}}{{(z-1)}^{2}}}\] For\[Z=43,\]wavelength\[=\lambda \] \[\therefore \] \[\lambda =\frac{c}{{{a}^{2}}{{(43-1)}^{2}}}\] Or \[\lambda =\frac{c}{{{a}^{2}}{{(42)}^{2}}}\] For\[Z=29,\]wavelength \[=\lambda '\] \[\lambda '=\frac{c}{{{a}^{2}}{{(29-1)}^{2}}}\] Or \[\lambda '=\frac{c}{{{a}^{2}}{{(28)}^{2}}}\] ...(ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{\lambda '}{\lambda }={{\left( \frac{42}{28} \right)}^{2}}={{\left( \frac{3}{2} \right)}^{2}}\] \[\therefore \] \[\lambda '=\left( \frac{9}{4} \right)\lambda \]
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