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  • A proton, a neutron, an electron and an α-particle have the same energy. Then their de Broglie wavelengths compare as (a) lambda p = lambda n > lambda e > lambda a (b) lambda a < lambda p = lambda n < lambda e

 A proton, a neutron, an electron and an α-particle have the same energy. Then their de Broglie wavelengths compare as
(a) \lambda _{p} = \lambda _{n} > \lambda _{e} > \lambda _{\alpha} \\ (b) \lambda _{\alpha} < \lambda _{p} = \lambda _{n} < \lambda _{e}\\ (c) \lambda _{e} < \lambda _{p} = \lambda _{n} > \lambda _{\alpha} \\ (d) \lambda _{e} = \lambda _{p} = \lambda _{n} = \lambda _{\alpha}

Answers (1)

The answer is the option (b)
de-Broglie wavelength \lambda is given by 

  \lambda=\frac{h}{p}=\frac{h}{mv}

2K=mv^{2}\\ 2Km=m^{2}v^{2}\\ 2mK=p^{2}\\ p=\sqrt{2mK}\\ y_{d}=\frac{h}{p}\\ y_{d}=\frac{h}{\sqrt{2mK}}\\ Since\: \lambda\alpha \frac{1}{\sqrt{m}}\\ M_{a}>m_{p}=m_{n}>m_{e}\\ \lambda_{a}>\lambda_{p}=\lambda_{n}>\lambda_{e}

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