A) \[{{\lambda }_{0}}=\frac{2mc{{\lambda }^{2}}}{h}\]
B) \[{{\lambda }_{0}}=\frac{2h}{m}\]
C) \[{{\lambda }_{0}}=\frac{2{{m}^{2}}-{{c}^{2}}{{\lambda }^{3}}}{{{h}^{2}}}\]
D) \[{{\lambda }_{0}}=\lambda \]
Correct Answer: A
Solution :
\[P=h/\lambda \] And \[K=\frac{{{P}^{2}}}{2m}\,=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] For X-ray photons, it is also maximum energy So, \[\frac{hc}{{{\lambda }_{0}}}\,=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\]Or, \[{{\lambda }_{0}}\,=\frac{2m{{\lambda }^{2}}c}{h}\,\]
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