The Lyman series can be expressed in the formula <span><span>1/λ</span>=<span>RH</span><span>(1−<span>1/<span>n2</span></span>) where </span><span><span>RH</span>=1.0968×<span>107</span><span>m<span>−1</span></span>=<span><span>13.6eV</span><span>hc
</span></span></span></span>Where n is a natural number greater than or equal to 2 (i.e. n = 2,3,4,...). Therefore, the lines seen in the image above are the wavelengths corresponding to n=2 on the right, to n=∞on the left (there are infinitely many spectral lines, but they become very dense as they approach to n=∞<span> (Lyman limit), so only some of the first lines and the last one appear).
The wavelengths (nm) in the Lyman series are all ultraviolet
:2 3 4 5 6 7 8 9 10 11
Wavelength (nm) 121.6 102.6 97.3 95 93.8 93.1 92.6 92.3 92.1 91.9 91.18 (Lyman limit)
In your case for the n=5 line you have to replace "n" in the above formula for 5 and you should get a value of 95 x 10^-9 m for the wavelength. then you have to use the other equation that convert wavelength to frequency. </span>
Answer:
Use the Bromotriflouride catalyst, BF₃
Explanation:
The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.
BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:
2 - methylpropene + H-F-BF₃ → H-F + H₃C + benzene
butylbenzene + F-BF₃ → tert-butylbenzene + H-F + BF₃ (regenerated catalyst)
<span>Thomson studied electric discharge in a vacuum and found that the deflection of rays was evidence of atoms containing much smaller particles. He calculated that these particles would have a large charge in relation to their mass. While he did not name electrons, he knew they existed.</span>
The answer is A
Mark Brainliest ☺☻
