Answer to Question #279270 in Atomic and Nuclear Physics for BIGIRIMANA Elie

Question

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 0.01nm is used.

Answer 1

We know that

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

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

Again

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

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

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

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

$∆\nu=1.875\times10^{10}Hz$

Magnetic field

$B=\frac{4\pi m∆\nu}{e}$

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