Answer
$1.97\times10^{-5}\;m$
Work Step by Step
The magnitude of vertical force exerted by the field gradient on the atom due to the magnetic moment of the atom’s electron is $F_z=1.48\times10^{-21}\;N$
Mass of a hydrogen atom is $m_H=1\;amu=1.67\times10^{-24}\;g=1.67\times10^{-27}\;kg$
Therefore, the magnitude of vertical acceleration is given by
$a=\frac{F_z}{m_H}=\frac{1.48\times10^{-21}}{1.67\times10^{-27}}\;m/s^2\approx8.86\times 10^{5}\;m/s^2$
Initial vertical velocity of the hydrogen atom is zero. Therefore, the vertical displacement in time $t$ is given by $\Delta z=\frac{1}{2}at^2$.......................$(1)$
In time $t$ the atoms also travel a horizontal distance $l=80 \;cm=0.8\;m$ with a constant speed of $v=1.2\times 10^{5}\;m/s$. Thus, $t=\frac{l}{v}=\frac{0.8}{1.2\times 10^{5}}\;s=6.67\times10^{-6}\;s$
Substituting the given values in eq. $1$, we get
$\Delta z=\frac{1}{2}\times 8.86\times 10^{7}\times(6.67\times10^{-6})^2\;m\approx 1.97\times10^{-5}\;m$
$\therefore$ The vertical displacement of the atom is $1.97\times10^{-5}\;m$
