1.

de- Broglie wavelength associated with an electron, accelerating through a potential

Answer»

GAMMA rays
X-rays
ultraviolet
visible REGION

Solution :DE Broglie wavelength
`lamda=h/p=1.227/sqrtV nm`
`lamda=1.227/sqrt100=1.227/10`
`THEREFORE lamda=0.1227 nm`
This wavelength FALLS in the region of X-rays.


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