1.

de-Broglie wavelength of electron in ground state is 2.116Å then its velocity will be………. ms^(-1)

Answer»

`0.034xx10^(8)`
`3.4xx10^(8)`
`34xx10^(-8)`
`0.034xx10^(-8)`

Solution :`lamda=(H)/(P)=(lamda)/(MV)`
`:.V=(h)/(mlamda)=(6.62xx10^(-34))/(9.1xx10^(-31)xx2.116xx10^(-10))`
`:.v=0.034xx10^(8)ms^(-1)`


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