1.

An electron is moving with a kinetic energy of `4.55 xx 10^-25 J`. What will be Broglie wavelength for this electron ?

Answer» `KE=(1)/(2)mv^(2)=4.55xx10^(-25)`
or `(1)/(2)xx9.1xx10^(-31)xxv^(2)=4.55xx10^(-25)`
or `v^(2)=(2xx4.55xx10^(-25))/(9.1xx10^(-31))`
Applying de Broglie equation,
`lamda=(h)/(mv)=(6.6xx10^(34))/(9.1xx10^(-31)xx10^(3))=0.72xx10^(-6)m`.


Discussion

No Comment Found

Related InterviewSolutions