What is the least amount of energy that can be emitted by an excited electron in a hydrogen atom falling from an excited state directly to the n = 3 state? what is the quantum number n for the excited state? humans cannot visually observe the photons emitted in this process. why not?

1.

2.

3. The associated wavelength belongs to the infrared spectrum which is invisible for humans.

Explanation:

Hello,

1. At first, according to the equation:

Whereas is the emitted energy, the first level energy, the atomic number, the first level and the second level. In such a way, the closest the 's are, the least the amount of emitted energy, therefore, (based on the statement) and , thus, the least amount of energy turns out being:

2. Secondly, and based on the first question, the quantum number for the excited state is mandatorily

3. Finally, to substantiate why we cannot observe the emitted photons we apply the following equation:

Whereas the obtained wavelength corresponds to the infrared spectrum which is not observable by humans.

Best regards.

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms

the answer is 1 i believe