Answer to Question #280553 in Atomic and Nuclear Physics for BIGIRMANA Elie

Question

3. (i) In the presence of an external uniform magnetic field, the normal Zeeman effect is observed

and the original transition generates into three transitions of frequencies:

ν = ν0 −



eB

4πm

; ν = ν0; ν = ν0 +



eB

4πm

where m is the mass of the electron. Using a diagram, indicate what are allowed normal

Zeeman transitions between d(l = 2) state and p(l = 1) state. [6 marks]

(ii) Find the minimum magnetic field needed for the Zeeman effect to be observed in a

spectral line of 400 nm wavelength when a spectrometer whose resolution is 0.010 nm is

used.

Answer 1

the presence of an external uniform magnetic field, the normal Zeeman effect is observed

Photon energy $E=h\upsilon$

In presence of magnetic field 3 spectrum line correspondingPhoton energy

We can written as

$E_1=E_0-\mu_B B$

$E_2=E_0$

$E_3=E_0+\mu_B B$

Now frequency

$\upsilon_1=\upsilon_0-\frac{\mu_B B}{h}$

$\upsilon_2=\upsilon_0$

$\upsilon_2=\upsilon_0+\frac{\mu_B B}{h}$

$\frac{\mu_B}{h}=\frac{e}{4\pi m}$

$\upsilon_1=\upsilon_0-\frac{eB}{4\pi m} (when ∆m=-1)$

$\upsilon_2=\upsilon_0(when∆m=0)$

$\upsilon_3=\upsilon_0+\frac{eB}{4\pi m}(∆m=+1)$

We that

$\nu=\frac{c}{\lambda}$

$\nu=\frac{3\times10^8}{400\times10^{-9}}=7.5\times10^{-14}Hz$

$\nu_1=\frac{c}{\lambda_1}$

$\nu_1=\frac{3\times10^8}{400\times10^{-9}+0.010\times10^{-9}}$

$\nu_1=\frac{3\times10^8}{400\times10^{-9}}\times\frac{1}{1+\frac{0.1}{400}}$

$∆\nu=7.5\times10^{14}(1-2.5\times10^{-5})$

$∆\nu=1.875\times10^{10}Hz$

$B=\frac{4\pi m∆\nu}{e}$

$$B=\frac{4\times3.14\times9.1\times10^{-31}\times1.875\times10^{10}}{1.06\times10^{-19}}=1.34T$$