A) 1.5 keV
B) 15 keV
C) 150 keV
D) 1.05 keV
Correct Answer: B
Solution :
The de-Broglie wavelength \[\lambda =\frac{h}{p}=\frac{h}{mv}\] Or \[v=\frac{h}{m\lambda }\] \[=\frac{6.6\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times {{10}^{-11}}}\] \[=7.25\times {{10}^{7}}m/s\] Energy of electron\[=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}\times \frac{9.1\times {{10}^{-31}}\times {{(7.25\times {{10}^{7}})}^{2}}}{1.6\times {{10}^{-19}}}\] \[=15\text{ }keV\]
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