Question

The wavelength of the fourth line in the Lyman series of the hydrogen spectrum is a) \(80.0 \mathrm{nm}\). b) \(85.0 \mathrm{nm}\) c) \(90.2 \mathrm{nm}\). d) \(94.9 \mathrm{nm}\).

Short answer

Answer: The wavelength of the fourth line in the Lyman series of the hydrogen spectrum is approximately 94.97 nm. The closest answer among the given choices is 94.9 nm.

Step-by-step solution

Rydberg formula:

The Rydberg formula for the hydrogen spectrum is given as: $$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ Where \(\lambda\) is the wavelength; \(R_H\) is the Rydberg constant for hydrogen with a value of \(1.097373 \times 10^7 m^{-1}\); \(n_1\) and \(n_2\) are the initial and final energy levels, with \(n_1 \lt n_2\). For the Lyman series, the electron transitions to the ground state (n=1). Therefore, we have \(n_1 = 1\). Since we are looking for the fourth line, this means the electron is jumping from the fifth energy level, so \(n_2 = 5\).

Substitute values into the formula:

We will now plug in the values for n1 and n2 into the Rydberg formula: $$\frac{1}{\lambda} = R_H\left(\frac{1}{1^2} - \frac{1}{5^2}\right)$$

Calculate the wavelength:

Solve the formula to get the wavelength: $$\frac{1}{\lambda} = 1.097373 \times 10^7 \left(1 - \frac{1}{25}\right)$$ $$\frac{1}{\lambda} = 1.097373 \times 10^7 \left(\frac{24}{25}\right)$$ $$\lambda = \frac{1}{(1.097373 \times 10^7) \left(\frac{24}{25}\right)}$$ After calculating, we find: $$\lambda \approx 94.97 \mathrm{nm}$$

Choose the correct option:

Comparing the calculated wavelength to the given options, the closest answer is \(94.9 \mathrm{nm}\). So, the correct option is (d) \(94.9 \mathrm{nm}\).