Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The initial velocity of the rubidium atoms is given as $ v = 500 \, \text{m/s} $, and the mass of each rubidium atom is $ m = 1.4 \times 10^{-25} \, \text{kg} $.
The momentum of the atoms is given by
$$
p_{\text{atom}} = m v$$
$$ p_{\text{atom}} = (1.4 \times 10^{-25} )(500 ) =\color{red}{\bf 7.0 \times 10^{-23}} \, \text{kg} \cdot \text{m/s}
$$
The momentum of a photon is given by
$$
p_{\text{photon}} =\dfrac{E_{\rm photon}}{c}$$
where $E_{\rm photon}=hc/\lambda$, so
$$p_{\text{photon}} = \frac{h}{\lambda}
$$
Plug the known;
$$p_{\text{photon}} = \frac{(6.63 \times 10^{-34} )}{(780\times 10^{-9})}
$$
$$
p_{\text{photon}} = 8.5\times 10^{-28}\, \text{kg} \cdot \text{m/s}
$$
And since the photon is traveling in the opposite direction to the atoms, the momentum of the photon is negative.
$$
p_{\text{photon}} = \color{red}{\bf -8.5\times 10^{-28}}\; \text{kg} \cdot \text{m/s}
$$
$$\color{blue}{\bf [b]}$$
When the atom absorbs a photon, the total momentum of the system (atom + photon) is conserved. Before absorption, the atom has a momentum $ (p_{\text{atom}})_i $, and after absorption, it has a momentum
$$(p_{\text{atom}})_f = (p_{\text{atom}} )_i+ p_{\text{photon}} $$
Thus,
$$(p_{\text{atom}})_f - (p_{\text{atom}} )_i= p_{\text{photon}} $$
$$\Delta p_{\text{atom}} = p_{\text{photon}} $$
Hence, the momentum change per absorption is equal to the momentum of the photon.
$$
\Delta p_{\text{atom}} = p_{\text{photon}} = {\bf-8.50 \times 10^{-28}} \, \text{kg} \cdot \text{m/s}
$$
The change in the atom's momentum is negative since it loses momentum.
To stop the atoms, the momentum must go to zero. Therefore, the total number of photons that must be absorbed to bring the atom to rest is:
$$
N_{\text{absorbed}} = \dfrac{(p_{\text{atom}})_i}{|\Delta p_{\text{atom}}| }= \dfrac{(p_{\text{atom}})_i}{|p_{\text{photon}}|} $$
$$N_{\text{absorbed}} = \frac{7.0 \times 10^{-23} }{8.50 \times 10^{-28} } = \color{red}{\bf 82\;353}\, \text{photons}
$$
$$\color{blue}{\bf [c]}$$
The atom spends $ 15 \, \text{ns} $ in the excited state before emitting a photon and returning to the ground state. This means that the atom can absorb another photon every 15 ns. Therefore, the total time required to absorb 82,353 photons is:
$$
\Delta t_{tot}= N_{\text{absorbed}} \;t= (82\;353 )(15 \times 10^{-9} ) $$
$$\Delta t_{tot}= 1.235\times 10^{-3}\text{s} =\color{red}{\bf 1.235}\, \text{ms}
$$
$$\color{blue}{\bf [d]}$$
The force exerted on the atom by the photons can be found using Newton’s second law
$$ F = \dfrac{\Delta p_{atom}}{\Delta t} = \dfrac{ p_{\rm photon}}{\Delta t} $$
Plug the known;
$$
F = \frac{-8.50 \times 10^{-28} }{15 \times 10^{-9} } =\color{red}{\bf -5.67 \times 10^{-20}} \, \text{N}
$$
The acceleration is given by
$$a=\dfrac{F}{m}=\dfrac{-5.67 \times 10^{-20}}{1.4\times 10^{-25}}$$
$$a=\color{red}{\bf -4.05\times 10^5}\;\rm m/s^2$$
$$\color{blue}{\bf [e]}$$
The distance over which the atoms are brought to rest can be calculated using the kinematic equation
$$ v_f^2 = v_i^2 + 2a \Delta x $$
where the final velocity $ v_f = 0 $, the initial velocity $ v_i = 500 \, \text{m/s} $, and $ a = -4.05 \times 10^5 \, \text{m/s}^2 $:
$$
0 = (500 )^2 + 2(-4.05 \times 10^5 ) \Delta x
$$
Solving for $ \Delta x $:
$$
\Delta x = \frac{(500 )^2}{2(4.05 \times 10^5 )} = \color{red}{\bf 0.309}\, \text{m}
$$
