Question: In the Stern-Gerlach experiment how much would a hydrogen atom emanating from a 500 K oven$\left(KE=\frac{3}{2}{k}_{B}T\right)$be deflected in traveling 1 m through a magnetic field whose rate of change is 10 T/m?

The kinetic energy is the measure of the work that an object does by virtue of its motion.

As we know the kinetic energy (KE) is given by

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ =\frac{3}{2}{k}_{B}T \end{gathered}$$

Here mass of the proton, $k_B=1.38\times {10}^{-23}J/K$, and T= 500K

Therefore,

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{1}{2}\left(1.67\times {10}^{-27}kg\right)v^2 \\ \frac{3}{2}{k}_{B}T= \frac{3}{2} \times \left(1.38\times {10}^{-23}\right)\times 500K \end{gathered}$$

Comparing the above two expressions,we get

$$\begin{gathered} \frac{1}{2}\left(1.67\times {10}^{-27}kg\right)v^2 = \frac{3}{2} \times \left(1.38\times {10}^{-23}J/K\right)\times 500K \\ v=\sqrt{\left(\frac{3\times 1.38\times {10}^{-23}J/K500K}{1.67\times {10}^{-27}kg}\right)} \\ v=3.52\times {10}^{3}m/s \end{gathered}$$

Which gives the speed of travel 3.52 x 10³m/s

Now the time taken by the H-atom to travel 1 m distance at this speed is

_{$$\begin{gathered} t=\frac{dis\tan ce}{speed} \\ =\frac{1m}{3.52\times {10}^{3}m/s} \\ =2.84\times {10}^{-4}s \end{gathered}$$}

Now according to the equation, the z-component of force on the H-atom by the field is

$F_z=-\left(\frac{e}{m_e}{S}_z\right)\frac{\partial B_z}{\partial z}$

Here,$S_z=m_s\hslash ;and{m}_s=-s,...,+s$

$$\begin{gathered} e=1.6\times {10}^{-19}C \\ \hslash =\frac{h}{2\pi }=1.05\times {10}^{-34} J·s \\ {m}_e=9.1\times {10}^{-31}kg \\ \frac{\partial B_z}{\partial z}=10T/m \end{gathered}$$

Therefore, by substituting the aboveexpression, we get the force is

$$\begin{gathered} F_z=\left(\frac{1.6\times {10}^{-19}C}{9.1\times {10}^{-31}kg}\times \frac{1}{2}\times 1.05\times {10}^{-34} J·s\right)\times 10 \\ =9.23 \times {10}^{-23} N \end{gathered}$$

As we know the acceleration is

$$\begin{gathered} a=\frac{F_z}{m} \\ =\frac{9.23 \times {10}^{-23} N}{6.67\times {10}^{-27}kg} \\ =5.55\times {10}^4 \end{gathered}$$

Now the displacement is

$$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ =\frac{1}{2}\left(5.55\times {10}^4\right){\left(2.84\times {10}^{-4}s\right)}^2 \\ =2.2\times {10}^{-3}m \end{gathered}$$

Here the initial velocity u = 0

Therefore the displacement is $2.2\times {10}^{-3}m$