Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits is mathematically given as

N=1820slits

<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>

Generally, the equation for the wave is mathematically given as

$d\ sin\ (\theta\ m) \ = \ m\ \lambda$

Where the chromatic resolving power (R) is defined by

$R\ =\ \lambda\ / \ d \ \lambda$

R = nN,

Therefore

$\lambda_1 \ = \ (656.45)(1 \ * \ 10^{-9})/1mm$

$\lambda_1= 656.45*10^{-9}$

and

$\lambda_2= (656.27)(1*10^{-9})/1mm$

$\\\\\lambda_2= 656.27*10^{-9}m$

In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

$N\ =\ \dfrac{\lambda}{m\ d\ T\ }\\\\$

Therefore

$N\ =\ \dfrac{656.45 \ * \ 10^{-9}}{2\ * \ (0.18*10^{-9})}$

N=1820slits

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