A) \[2.7\times {{10}^{-10}}\]
B) \[1.74\times {{10}^{-10}}\]
C) \[3.6\times {{10}^{-9}}\]
D) \[4.9\times {{10}^{-11}}\]
Correct Answer: B
Solution :
[b] The gain of kinetic energy by an electron is eV. \[\frac{1}{2}m{{v}^{2}}=eV\] \[v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2(1.60\times {{10}^{-19}})(50)}{(9.11\times {{10}^{-31}})}}\] \[=4.19\times {{10}^{6}}m{{s}^{-1}}\] Thus, the electron?s de Broglie wavelength is \[\lambda =\frac{h}{mv}=\frac{6.63\times {{10}^{-34}}}{(9.11\times {{10}^{-31}})(4.19\times {{19}^{6}})}\] \[=1.74\times {{10}^{-10}}m\]
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