Answer
a) $6.02\times10^{-3 }\;\rm m/s$
b) $1.235\times10^{-7 }\;\rm K$
Work Step by Step
a)
Since we need the nonrelativistic recoil speed of the atom, so we can use the conservation of momentum.
$$p_i=p_f=0$$
Thus,
$$m_{atom}v_{atom}=p_{proton}$$
The momentum of the proton is given by $p=\dfrac{h}{\lambda}$
Hence,
$$m_{atom}v_{atom}=\dfrac{h}{\lambda}$$
So, the recoil speed of the atom is given by
$$v_{atom}=\dfrac{h}{\lambda m_{atom}}$$
Plugging the known;
$$v_{atom}=\dfrac{6.626\times10^{-34}}{780\times10^{-9}\times 85\times 1.66\times10^{-27}}$$
$$v_{atom}=\color{red}{\bf 6.02\times10^{-3}}\;\rm m/s$$
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b) The kinetic energy of the gas is given by
$$\overline{KE}=\frac{1}{2}mv^2=\dfrac{3kT }{2}$$
Thus, the temperature is given by
$$ T= \dfrac{mv^2}{3k} $$
Plugging the known;
$$ T= \dfrac{85\times 1.66\times10^{-27} (6.02\times10^{-3})^2}{3 \times1.38\times10^{-23}} $$
$$ T=\color{red}{\bf1.235\times10^{-7}}\;\rm K$$
