A) 0.25\[\overset{0}{\mathop{A}}\,\]
B) 0.5\[\overset{0}{\mathop{A}}\,\]
C) 2.5\[\overset{0}{\mathop{A}}\,\]
D) 1\[\overset{0}{\mathop{A}}\,\]
Correct Answer: A
Solution :
For an accelerating voltage V, the maximum X-ray photon energy is given by \[h{{v}_{\max }}=eV\] Also, \[{{\lambda }_{\min }}=\frac{c}{{{v}_{\max }}}\] \[\therefore \] \[{{\lambda }_{\min }}=\frac{hc}{eV}\] where h is Plancks constant and c is speed of light. For \[h=6.6\times {{10}^{-34}}J-s,\] \[c=3\times {{10}^{8}}\,m/s,\,e=1.6\times {{10}^{-19}}\,C,\] \[V=50\,kV=50\times {{10}^{3}}\text{volt}\] \[\therefore \] \[{{\lambda }_{\min }}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 50\times {{10}^{3}}}=0.25\,\overset{\text{o}}{\mathop{\text{A}}}\,\] Alternative: The minimum wavelength of X-rays is given by \[{{\lambda }_{\min }}=\frac{12375}{V(in\,volts)}\overset{\text{o}}{\mathop{\text{A}}}\,\] \[\therefore \] \[{{\lambda }_{\min }}=\frac{12375}{50000}\overset{\text{o}}{\mathop{\text{A}}}\,=0.25\overset{\text{o}}{\mathop{\text{A}}}\,\]
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