Question

A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glow, sending the light through a diffraction grating, and measuring the positions of first-order spectral lines on a detector $15.0cm$behind the grating. Unfortunately, she has lost the card that gives the specifications of the grating. Fortunately, she has a known compound that she can use to calibrate the grating. She heats the known compound, which emits light at a wavelength of $461nm$, and observes a spectral line $9.95cm$from the center of the diffraction pattern. What are the wavelengths emitted by compounds $A$and $B$that have spectral lines detected at positions $8.55cm$and $12.15cm$, respectively?

Short answer

Compounds that produce wavelengths and have spectral lines recognised at certain places emit wavelengths by

$\begin{array}{r}{\lambda }_A=412.5\mathrm{nm},\\ {\lambda }_B=524.3\mathrm{nm}.\end{array}$

Step-by-step solution

Step: 1 Equating equation:

The concept is to utilise the information supplied for the known compound to determine the spacing of any two subsequent slits in the grating $\left(d\right)$, then we may use localid="1649156823344" $\left(d\right)$to compute the frequency of an unknown compound. The criterion for bight fringes is described as follows:

$\sin \left({\theta }_m\right)=\frac{m\lambda }{d}$

where the first spectral line's angle $\left({\theta }_1\right)$is

${\theta }_1={\sin }^{-1}\left(\frac{\lambda }{d}\right)$

The place of the first spectral line (bright fringe).

$y_1=L\tan \left({\theta }_1\right)=L\tan \left({\sin }^{-1}\left(\frac{\lambda }{d}\right)\right)$

The equation to reorder isolate as

$\begin{array}{l}{\tan }^{-1}\left(\frac{y_1}{L}\right)={\sin }^{-1}\left(\frac{\lambda }{d}\right)\\ \frac{\lambda }{d}=\sin \left({\tan }^{-1}\left(\frac{y_1}{L}\right)\right)\end{array}$

Step: 2 Fining the value:

Apllying the values,

$\begin{array}{l}d=\frac{\lambda }{\sin \left({\tan }^{-1}\left(\frac{y_1}{L}\right)\right)}\\ d=\frac{461\times {10}^{-9}\mathrm{m}}{\sin \left({\tan }^{-1}\left(\frac{9.95\mathrm{cm}}{15\mathrm{cm}}\right)\right)}\\ d=8.33\times {10}^{-7}\mathrm{m}.\end{array}$

Step: 3 Calculating the Wavelength:

Putting the values in equation,

$\lambda =d\sin \left({\tan }^{-1}\left(\frac{y_1}{L}\right)\right)$

Substituting $y_1=8.55cm$in compund A as

role="math" localid="1649127933387" $\begin{array}{l}{\lambda }_A=\left(8.33\times {10}^{-7}\mathrm{m}\right)\times \sin \left({\tan }^{-1}\left(\frac{8.55\mathrm{cm}}{15\mathrm{cm}}\right)\right)\\ {\lambda }_A=412.5\mathrm{nm}.\end{array}$

Substituting $y_1=12.15cm$in compund $B$as

$\begin{array}{l}{\lambda }_B=\left(8.33\times {10}^{-7}\mathrm{m}\right)\times \sin \left({\tan }^{-1}\left(\frac{12.15\mathrm{cm}}{15\mathrm{cm}}\right)\right)\\ {\lambda }_B=524.3\mathrm{nm}.\end{array}$