If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series emission spectrum. Calculate the wavelength λ associated with the fifth line.

Question

Express your answer with the appropriate units.

Answer 1

To calculate the wavelength associated with the fifth line of the Balmer series, we will use the Balmer formula:

1/λ = R_H * (1/n1² - 1/n2²)

where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m⁻¹), n1 is the principal quantum number of the lower energy level, and n2 is the principal quantum number of the higher energy level.

For the Balmer series, the lower energy level (n1) is always 2 since the electron transitions occur to the second energy level. We're looking for the fifth line, which means that the electron is transitioning from the sixth energy level (n2 = 6) to the second energy level (n1 = 2).

1/λ = R_H * (1/2² - 1/6²)
1/λ = 1.097 × 10^7 * (1/4 - 1/36)

Now, we can calculate the numerical value:

1/λ = 1.097 × 10^7 * (0.25 - 0.0278)
1/λ = 1.097 × 10^7 * 0.2222

1/λ ≈ 2.436 × 10^6

Now, to find the wavelength, we can take the reciprocal of this value:

λ ≈ 4.106 × 10^-7 m

We can convert this to nanometers (nm) by multiplying by 10^9:

λ ≈ 410.6 nm

Therefore, the wavelength associated with the fifth line in the Balmer series emission spectrum is approximately 410.6 nm.