A) \[0.1\overset{o}{\mathop{\text{A}}}\,\]
B) \[0.5\overset{o}{\mathop{\text{A}}}\,\]
C) \[1\overset{o}{\mathop{\text{A}}}\,\]
D) \[5\overset{o}{\mathop{\text{A}}}\,\]
Correct Answer: B
Solution :
When electrons accelerated through a potential difference V strike a target, the maximum frequency of the emitted X-rays is given by \[eV=h{{v}_{\max }}\] where, e is charge of the electron and h the Plancks constant. But\[{{v}_{\max }}=\frac{c}{{{\lambda }_{\min }}},\]where c is the speed of light and\[{{\lambda }_{\min }}\]the minimum wavelength. \[\therefore \] \[eV=\frac{hc}{{{\lambda }_{\min }}}\] \[\Rightarrow \] \[{{\lambda }_{\min }}=\frac{hc}{eV}\] Given, \[V=24.75\text{ }kV=24.75\times {{10}^{3}}V,\] \[h=6.6\times {{10}^{-34}}Js,\text{ }c=3\times {{10}^{8}}m{{s}^{-1}},\] \[e=1.6\times {{10}^{-19}}C\] \[\therefore \] \[{{\lambda }_{\min }}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 24.75\times {{10}^{3}}}\] \[=0.5\times {{10}^{-10}}\] \[=0.5\overset{o}{\mathop{\text{A}}}\,\]
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