1.

The de Brogle wavelength of an electron accelrated to a potential of 400 V is approximately

Answer»

0.03 nm
0.04 nm
0.12 nm
0.06 nm

Solution :`lambda=(h)/(SQRT(2meV))`
`lambda=(6.625xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx400))`
`lambda=(6.625xx10^(-34))/(sqrt(11648xx10^(-50)))`
`lambda=(6.625xx10^(-34))/(107.925xx10^(-25))`
`lambda=0.06138xx10^(-34+25)`
`lambda=0.06138xx10^(-9)`
`lambda=0.06nm`


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