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The increasing order (lowest first) for ...

The increasing order (lowest first) for the values of `e//m` (charge//mass) for electron `(e)`, proton `(p)`, neutron `(n)`, and alpha particle `(alpha)` is

A

`e, p, n, alpha`

B

`n, p, e, alpha`

C

`n, p, alpha, e`

D

`n, alpha, p, e`

Text Solution

Verified by Experts

The correct Answer is:
D
Neutron has no charge, hence e/m is zero for neutron. Next, `alpha-` particle `(He^(2+))` has very high mass compared to proton and electron, therefore very small e/m ratio. Proton and electron have same charge (magnitude) but former is heavier, hence has smaller value of e/m. `(e)/(m), n lt alpha lt p lt e`
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