A) 500 ke V
B) 100 ke V
C) 1 ke V
D) 25 ke V
Correct Answer: D
Solution :
\[k=\frac{{{P}^{2}}}{2m}=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] |
\[=\frac{{{(6.62\times {{10}^{-34}})}^{2}}}{2\times 9.1\times {{10}^{-31}}\times (7.5\times {{10}^{-12}})}\times \frac{1}{1.6\times {{10}^{-19}}}eV\] |
Alternate method\[\lambda =\sqrt{\frac{150}{V}}\] \[\Rightarrow \] \[7.5\times {{10}^{-12}}=\sqrt{\frac{150}{V}}\] |
\[V=\frac{80}{3}kV\] |
Energy \[=\frac{80}{3}keV\simeq 25keV\] |
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