If kinetic energy of an electron is increased by nine times, the wavelength associated with it would become ...........

Options

A. `1/3` times
B. `1/9` times
C. 9 times
D. 3 times

Answer

A

Explanation

kinetic energy =`1/2 mV^(2)` … and `lambda=(h)/(mV)`
`therefore V=(h)/(lambdav)` and `v^(2) =(h^(2))/(lambda^(2)m^(2))`
`therefore KE alpha (1)/(lambda^(2))`
If KE is increased by nine times
`therefore 9=(lambda_(1)^(2))/(lambda_(2)^(2))`
`therefore 3=(lambda_(1))/(lambda_(2))` and `(lambda_(2))/(lambda_(1))=1/3`