A) \[1:2:2\]
B) \[2:2:1\]
C) \[1:2:1\]
D) \[1:1:2\]
Correct Answer: D
Solution :
K.E. gained by charged particle of charge q when accelerated under a pot. diff. V will be \[{{E}_{k}}=qV;\]; For a given V, \[E\propto q\]. For proton, deutron and a-particle, the ratio of charges is \[1:1:2\].You need to login to perform this action.
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