Answer
$2.43\times10^{-12}m$
Work Step by Step
Maximum wavelength $\Delta\theta_{max}$, where " $\theta$ "represents "lambda", shift is given by
$\Delta\theta_{max}$ = 2 $\theta_{c}$
So, Half of the maximum wavelength can be defined as
$\theta_{c}$= $\frac{\Delta\theta_{max}}{2}$
So basically we have to find $\theta_{c}$ in order to find the answer. We already know that,
= $\theta_{c}$=$\frac{h}{mc}$
Putting in the values of h,m,c according to the conventions,
= $h=6.625\times10^{-34}m^{2}kg/s$, $m=9.11\times10^{-31}kg$ and $c=2.99\times10^{8}m/s$
$\theta_{c}$=$\frac{6.625\times10^{-34}}{9.11\times10^{-31}\times2.99\times10^{8}}m$
adjusting the powers in the equation, we get,
= $\frac{6.625\times10^{-11}}{9.11\times2.99}m$
= $\frac{6.625\times10^{-11}}{27.2}m$
=$0.243\times10^{-11}m$
=$2.43\times10^{-12}m$
Hence, maximum wavelength shift's half is $2.43\times10^{-12}m$
