1.

If electron is accelerated under 50 KV in microscope ,find its de-Broglie wavelength.

Answer»

`5.485xx10^(-12)m`
`8.545xx10^(-12)m`
`4.585xx10^(-12)m`
`5.845xx10^(-12)m`

Solution :V=50 KV =`50xx10^(3)V`
`lambda=(h)/(SQRT(2meV))`
`=(6.62xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx50xx10^(3)xx1.6xx10^(-19)))`
`=(6.62xx10^(-34))/(1.207xx10^(-22))`
`5.485xx10^(-12)m`


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