1.

Electron starting from rest is acclerated between two point having p.d. of 20 V and 40 V. de-Broglie wavelength of electron……

Answer»

0.75 Å
7.5 Å
2.75 Å
0.75 nm

Solution :Here `V=V_(2)-V_(1)=40-20=20V`
`lambda=(H)/(sqrt(2meV))`
`=(6.62xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx20))`
`0.274xx10^(-9)=2.75Å`


Discussion

No Comment Found

Related InterviewSolutions