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

Question #280553

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.

1
Expert's answer
2021-12-19T18:17:27-0500

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

Photon energy E=hυE=h\upsilon

In presence of magnetic field 3 spectrum line correspondingPhoton energy

We can written as

E1=E0μBBE_1=E_0-\mu_B B

E2=E0E_2=E_0

E3=E0+μBBE_3=E_0+\mu_B B



Now frequency

υ1=υ0μBBh\upsilon_1=\upsilon_0-\frac{\mu_B B}{h}

υ2=υ0\upsilon_2=\upsilon_0

υ2=υ0+μBBh\upsilon_2=\upsilon_0+\frac{\mu_B B}{h}

μBh=e4πm\frac{\mu_B}{h}=\frac{e}{4\pi m}

υ1=υ0eB4πm(whenm=1)\upsilon_1=\upsilon_0-\frac{eB}{4\pi m} (when ∆m=-1)

υ2=υ0(whenm=0)\upsilon_2=\upsilon_0(when∆m=0)

υ3=υ0+eB4πm(m=+1)\upsilon_3=\upsilon_0+\frac{eB}{4\pi m}(∆m=+1)

We that

ν=cλ\nu=\frac{c}{\lambda}

ν=3×108400×109=7.5×1014Hz\nu=\frac{3\times10^8}{400\times10^{-9}}=7.5\times10^{-14}Hz

ν1=cλ1\nu_1=\frac{c}{\lambda_1}

ν1=3×108400×109+0.010×109\nu_1=\frac{3\times10^8}{400\times10^{-9}+0.010\times10^{-9}}

ν1=3×108400×109×11+0.1400\nu_1=\frac{3\times10^8}{400\times10^{-9}}\times\frac{1}{1+\frac{0.1}{400}}

ν=7.5×1014(12.5×105)∆\nu=7.5\times10^{14}(1-2.5\times10^{-5})

ν=1.875×1010Hz∆\nu=1.875\times10^{10}Hz

B=4πmνeB=\frac{4\pi m∆\nu}{e}


B=4×3.14×9.1×1031×1.875×10101.06×1019=1.34TB=\frac{4\times3.14\times9.1\times10^{-31}\times1.875\times10^{10}}{1.06\times10^{-19}}=1.34T


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