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Correct decreasing order of specific charge for the following particles is

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$e > p > α > n$

$e > p > n > α$

$e > n > α > p$

$α > n > e > p$

answer is A.

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Detailed Solution

$\begin{array}{l} \bar{e} p α n \\ \frac{e}{m} = \frac{1}{(\frac{1}{1836})} \frac{1}{1836} \frac{2}{4 (1836)} \frac{o}{1839} \\ 1836 \frac{1}{1836} \frac{1}{2 (1836)} zero \end{array}$

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