Increasing order (lowest first) for the values of $e/m$ for electron ($e$), proton ($p$), neutron ($n$) and $\alpha$-particles is

Question

Dr. Abubacker · Accepted Answer

(b)
$\frac{e}{m}for\ electron\ (e) = \frac{1.6 	imes 10^{- 19}}{9.1 	imes 10^{- 28}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 1.758 	imes 10^{8}$
$\frac{e}{m}for\ proton\ (p) = \frac{1.6 	imes 10^{- 19}}{1.672 	imes 10^{- 24}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 9.56 	imes 10^{4}$
$\frac{e}{m}for\ neutron\ (n) = \frac{0}{1.675 	imes 10^{- 24}} = 0$
$\frac{e}{m}for\ \alpha - particle = \frac{2}{4} = 0.5$
Hence, the increasing order of $\frac{e}{m}$ is as
$n < \alpha < p < e$

Q. Increasing order (lowest first) for the values of $e/m$ for electron ($e$), proton ($p$), neutron ($n$) and $\alpha$-particles is

Explanation

Reviewed by Dr. Abubacker

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