This problem really bugs me and I did figure out the diagram but I cannot figure out the wavelength. Please help me with this problem.

The transitions of
i) n=2 to n=4
ii) n=3 to n=1
iii)n=5 to n=2
iv) n=3 to n=4
which corresponds to emission of radiation with the longest wavelength? Explain.

The levels are in the order

n=5
n=4
n=3
n=2
n=1

without knowing the spacing we know that n=1 to n=5 will have the highest energy and a transition between two adjacent levels will have a lower energy.

So what is the relationship between wavelength and energy? This will then tell you which transition it corresponds to.

To determine which transition corresponds to emission of radiation with the longest wavelength, we can use the Rydberg formula. The Rydberg formula relates the wavelength of light emitted during a transition in the hydrogen atom to the initial and final energy levels of the electron:

1/λ = R * (1/n1^2 - 1/n2^2)

where λ is the wavelength, R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), and n1 and n2 are the initial and final energy levels, respectively.

Now, let's calculate the wavelengths for each transition:

i) n=2 to n=4:
1/λ = R * (1/2^2 - 1/4^2)
Simplifying the equation gives us:
1/λ = R * (1/4 - 1/16)
1/λ = R * (3/16)
Thus, the wavelength for this transition is λ = 16/(3R).

ii) n=3 to n=1:
1/λ = R * (1/3^2 - 1/1^2)
Simplifying the equation gives us:
1/λ = R * (1/9 - 1)
1/λ = R * (-8/9)
Thus, the wavelength for this transition is λ = -9/(8R).

iii) n=5 to n=2:
1/λ = R * (1/5^2 - 1/2^2)
Simplifying the equation gives us:
1/λ = R * (1/25 - 1/4)
1/λ = R * (4/100 - 25/100)
1/λ = R * (-21/100)
Thus, the wavelength for this transition is λ = -100/(21R).

iv) n=3 to n=4:
1/λ = R * (1/3^2 - 1/4^2)
Simplifying the equation gives us:
1/λ = R * (1/9 - 1/16)
1/λ = R * (7/144)
Thus, the wavelength for this transition is λ = 144/(7R).

To determine which transition corresponds to the longest wavelength, we need to compare the magnitude of the values obtained above. It's important to note that the Rydberg constant (R) is positive, so we can disregard the negative signs in the equations.

Comparing the values, we find that:
λ for i) n=2 to n=4: λ = 16/(3R)
λ for ii) n=3 to n=1: λ = 9/(8R)
λ for iii) n=5 to n=2: λ = 100/(21R)
λ for iv) n=3 to n=4: λ = 144/(7R)

Therefore, the transition iii) n=5 to n=2 corresponds to the emission of radiation with the longest wavelength, as the magnitude of λ (100/(21R)) is the smallest among all the calculated values.