1.

The kinetic energy of a proton and that of an alpha - particle are 4 eV and 1 eV , respectively. The ratio of the de - Broglie wavelengths associated with them , will be

Answer»

<P>`2:1`
`1:1`
`1:2`
`4:1`

Solution :[Hint : As per RELATION `lamda_(DE)=h/(sqrt(2mK)),lamda_p/lamda_(ALPHA)=sqrt((m_alphaK_(alpha))/(m_pK_(p)))=sqrt((4m_pxx1eV)/(m_pxx4eV))=1`]


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