Express the de Broglie wavelength in terms of the kinetic energy of a relativistic particle. What is the kinetic energy for which the nonrelativistic formula leads to an error of less than 1% ?

Answer

`lamda=(hc)/(sqrt(K(2epsilon_(0)+K)));Klt0.04epsilon_(0)`.

Explanation

The kinetic energy of a particle is `K=epsilon-epsilon_(0)=sqrt(epsilon_(0)^(2)+p^(2)c^(2))-epsilon_(0)` from which we obtain for the momentum `p=1/csqrt(K(2epsilon_(0)+K))`, and for the de Broglie wave
`lamda=h/p=(hc)/(sqrt(K(2epsilon_(0)+K)))`
For `Kltlt epsilon _(0)`, we obtain the nonrelativistic approximation:
`lamda_("nonrel")=(hc)/(sqrt(2epsilon_(0)K))=h/(sqrt(2mK))`
The error due to the substitution of the nonrelativistic formula for the relativistic one is
`delta=(lamda_("nonrel")-lamda)/lamda=sqrt((2epsilon _(0)+K)/(2 epsilon_(0)))-1`, which gives
`K/(2epsilon_(0))=(1+delta)^(2)-1~~1delta`
Since `deltaltlt1`. Hence, the error introduced by the substitution of the nonrelativistic formula for the relativistic one will be less than `delt , if Kltlt4delta epsilon_(0)`.