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| 1. |
The magnitude of the de-Broglie wavelength `(lambda)` of an electron `(e)`,proton`(p)`,neutron `(n)` and `alpha` particle `(a)` all having the same energy of `Mev`, in the incresing order will follow the sequence:A. `lambda_(e),lambda_(p),lambda_(n),lambda_(alpha)`B. `lambda_(alpha),lambda_(n),lambda_(p),lambda_(e)`C. `lambda_(e),lambda_(n),lambda_(p),lambda_(alpha)`D. `lambda_(p),lambda_(e),lambda_(alpha),lambda_(n)` |
| Answer» Correct Answer - B | |