Hydrogen atoms are excited by a laser to the =4 state and then allowed to emit. What is the maximum number of distinct emission spectral lines (lines of different wavelengths) that can be observed from this system? Calculate the wavelength of the 4⟶3 transition.

1875 nm

Explanation:

Given the Rydberg formula for hydrogen: 1/λ = R(1/n₁² - 1/n₂²)

where R ≈ 1.097* 10^7 /m

Hence;

1/λ = 1.097* 107/m * (1/3² - 1/4²) = 5.3 * 10^5/m

λ = 1.875 * 10-6 m = 1875 nm

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answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.