A) \[30\,\,kV\]
B) \[50\,\,kV\]
C) \[80\,\,kV\]
D) \[60\,\,kV\]
Correct Answer: A
Solution :
In the most favorable collision in which the electron loses the whole of its energy in a single collision with target atom, an x-ray photon of maximum energy \[h{{v}_{\max }}\] is emitted. Thus, for an accelerating voltage V, the maximum X-ray photon energy is \[h{{v}_{\max }}=ev\] Minimum wavelength corresponding to this maximum frequency is given by \[{{\lambda }_{\min }}=\frac{hc}{eV}\] \[V=\frac{hc}{e{{\lambda }_{\min }}}\] \[\Rightarrow \] \[V=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 0.4125\times {{10}^{-10}}}\] \[=30\times {{10}^{3}}V=30\,\,kV\] Alternative: In the expression \[h=6.6\times {{10}^{-34}}\,Js\] \[c=3\times {{10}^{8}}\,\,m/s\,\,and\,\,e=1.6\times {{10}^{-19}}\,\,C\] \[\therefore \] \[V=\frac{1.2375\times {{10}^{-6}}}{{{\lambda }_{\min }}}=\frac{12375}{{{\lambda }_{\min }}\left( in\,\,A \right)}V\] \[\therefore \] \[\therefore V=\frac{12375}{0.4125}=30\,\,kV\] Note: This expression can be memorized for solving further such numerical
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