A) 0.012 Å
B) 0.024 Å
C) 0.012 Å to ¥
D) 0.024 Å to ¥
Correct Answer: A
Solution :
Since electron and positron annihilate \[\lambda =\frac{hc}{{{E}_{Total}}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{(0.51+0.51)\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}\] \[=1.21\times {{10}^{-12}}m=0.012{\AA}\].You need to login to perform this action.
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