91Ó°ÊÓ

1. The magnitude of magnetic field,$\left|\stackrel{→}{\mathrm{B}}\right|=0.200\mathrm{T}$
2. It is assumed that l = 0.

Bohr magneton is the magnitude of the magnetic dipole of an electron orbiting an atom with such an angular force.

In atomic physics, Bohr magneton is a static unit that is a natural unit of expression for the electron moment magnetic field caused by its orbital or spin angular force.

Using the energy equations from the Stern-Gerlach experiment and Planck's relation, we can get the required value of the wavelength of the photon by substituting the given data.

Formulas:

The energy equation from the Stern-Gerlach experiment,

$∆E=2{\mathrm{μ}}_{B}B$ ….. (1)

Here, B is the magnetic field and ${\mathrm{μ}}_B$is the Bohr magneton.

The energy due to Planck-Einstein relation,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{λ}}$ ….. (2)

Here, c is the speed of light,$\mathrm{λ}$ is the wavelength, and h is the Plank’s constant.

Consider the known data as below.

The magnetic field,$\mathrm{B}=0.200T$

Plank’s constant,$\mathrm{h}=6.63\mathrm{x}{10}^{-34}\mathrm{J}.\mathrm{s}$

Speed of light,$\mathrm{c}=3\mathrm{x}{10}^8\mathrm{m}/\mathrm{s}$

Bohr magneton,${\mathrm{μ}}_{\mathrm{B}}=9.27×{10}^{-24}\mathrm{J}/\mathrm{T}$

Comparing both the equations (1) and (2) and substituting the given data in the derived equation.

The wavelength of the photon that will induce a transition of an electron spin is as follow.

$$\begin{gathered} 2{\mathrm{μ}}_{\mathrm{B}}\mathrm{B}=\frac{\mathrm{hc}}{\mathrm{λ}} \\ \mathrm{λ}=\frac{\mathrm{hc}}{2{\mathrm{μ}}_{\mathrm{B}}\mathrm{B}} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{λ}=\frac{\left(6.63×{10}^{-34}J.s\right)\left(9.27×{10}^{-24}J/T\right)\left(3×{10}^{8}m/s\right)}{2\left(9.27×{10}^{-24}J/T\right)\left(0.200T\right)} \\ =5.35×{10}^{-2}m \end{gathered}$$

Hence, the value of the wavelength is $5.35×{10}^{-2}m$.