The de-Broglie wavelength associated with a material particle when it is accelerated through a potential difference of 150V is 1Å. What will be the de-broglie wavelength associated with the same particle when it is accelerted through a potential difference of 1350V?

Answer

`1//3Å`

Explanation

`lambda=h/(sqrt(2meq)) or lambdaprop 1/(sqrtV)`
`:. (lambda_(2))/(lambda_(1))=sqrt((V_(1))/(V_(2)))=sqrt(150/1350)=1/3`
or `lambda_(2)=lambda_(1)xx1/3=1xx1/3=1/3Å`