Question

The de-broglie wavelength associated with photon changes by 0.25%. If its momentum is changed by `9xx10^(-26)kgms^(-1)`, find the initial momentum of electron.

Answer 1

Correct Answer - `3.6xx10^(23)kgms^(-1)`
`lambda=h/p`......(i)
As p decreases, `lambda` increases.
Let `p_(0)` be the decreases in momentum when wavelength by `0.25%`.
Then, `[lambda+0.25/100lambda]=h/(p-p_(0))`
or `100.25/100 lambda=h/((p-p_(0)))`.......(ii)
From (i) and (ii), we have `100.25/100-p/(p-p_(0))`
On solving, `p=401p_(0)=401xx9xx10^(-26)`
`3.6xx10^(-23)kgms^(-1)`