Question

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?

Short answer

Answer

The emanating displacement of a hydrogen atom is$S=2.2 x {10}^{-3}m$

Step-by-step solution

Definition of kinetic energy

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

Determine the speed of the travel

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

Calculate the time taken by the hydrogen atom

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

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

Determine the force and acceleration

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

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

Find the emerging displacement of the hydrogen atom

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 $2.2\times {10}^{-3}m$