A) 0.31 \[\overset{o}{\mathop{A}}\,\]
B) 0.62 \[\overset{o}{\mathop{A}}\,\]
C) 0.155 \[\overset{o}{\mathop{A}}\,\]
D) 0.62 \[\overset{o}{\mathop{A}}\,\]
Correct Answer: A
Solution :
For the most favourable collision in which the electron loses the whole of its energy in a single collision with the target atom, an Z-ray photon of maximum energy \[{{h}_{\max }}\] emitted. Thus, for an accelerating voltage V, the maximum X-ray photon energy is given by \[h{{v}_{\max }}=eV\] Also, \[\gamma =\frac{C}{\lambda }=\frac{velocity\text{ }of\text{ }light}{wavelength}\] \[\therefore \] \[{{\lambda }_{\min }}=\frac{ch}{eV}=\frac{3\times {{10}^{8}}\times 6.63\times {{10}^{-34}}}{1.6\times {{10}^{-19}}\times 40\times {{10}^{3}}}\] \[=0.31\times {{10}^{-10}}m\] \[\simeq 0.31\,\overset{o}{\mathop{A}}\,\] Note: From the formula \[{{\lambda }_{\min }}=\frac{hc}{eV}\] Substituting \[h=6.6\times {{10}^{-34}}J\,s,c=3\times {{10}^{8}}m/s\] \[e=1.6\times {{10}^{-19}}C\] \[{{\lambda }_{\min }}=\frac{1.2375\times {{10}^{-6}}}{V}m=\frac{12375}{V}\overset{o}{\mathop{A}}\,\] In this expression V is in volts. For solving further such numerical problems, this expression is very useful.
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