A) \[0.31\,\overset{o}{\mathop{A}}\,\]
B) \[0.62\,\overset{o}{\mathop{A}}\,\]
C) \[1.62\,\overset{o}{\mathop{A}}\,\]
D) \[1.31\,\overset{o}{\mathop{A}}\,\]
Correct Answer: A
Solution :
Key Idea: At maximum frequency minimum wavelength is obtained. For an accelerating voltage V, the maximum X-ray photon energy is given by \[h{{v}_{\max }}=eV\] The minimum wavelength \[({{\lambda }_{\min }})\]corresponding to this maximum frequency is given by \[{{\lambda }_{\min }}=\frac{hv}{eV}\] where h is Plancks constant and c is speed of light. Given, \[h=6.3\times {{10}^{-34}}J-s,\,c=3\times {{10}^{8}}m/s\], \[V=40\,kV=40\times {{10}^{3}}\] volt, \[e=1.6\times {{10}^{-19}}C\] \[\therefore \] \[{{\lambda }_{\min }}=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{40\times {{10}^{3}}\times 1.6\times {{10}^{-19}}}\] \[{{h}_{\min }}=0.31\,\overset{o}{\mathop{A}}\,\] Note: On keeping the values of h, c, e the standard expression reduces to\[{{h}_{\min }}=\frac{12375}{V}\overset{o}{\mathop{A}}\,\].
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