Answer to Question #237731 in Electric Circuits for hassan ihtisham

Question

What will be the de broglie wavwlength when the kinetic energy of body increases by 5 times

Answer 1

Gives

E'=E

$\lambda'=?$

$\lambda=\frac{h}{\sqrt{2mE}}$

$\frac{\lambda'}{\lambda}=\frac{\sqrt{2mE}}{\sqrt{2mE'}}$

$\frac{\lambda'}{\lambda}=\frac{\sqrt{E}}{\sqrt{E'}}$

$\frac{\lambda'}{\lambda}=\frac{\sqrt{E}}{\sqrt{5E}}$

$\frac{\lambda'}{\lambda}=\frac{\sqrt{1}}{\sqrt{5}}$

${\lambda'}=\frac{{\lambda}}{\sqrt{5}}$