1.

Wavelength of an electron having energy 10 keV ……..Å

Answer»

0.12
1.2
12
120

Solution :`LAMBDA=(h)/(sqrt(2ME))`
`=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx10^(4)xx1.6xx10^(-19)))`
`=0.12xx10^(-10)m`
`therefore lambda=0.12 Å`


Discussion

No Comment Found

Related InterviewSolutions