Question

Hydrogen atoms are bombarded with \(13.1-\mathrm{eV}\) electrons. Determine the shortest wavelength of light the atoms will emit.

Short answer

Answer: The shortest wavelength of light emitted by hydrogen atoms when bombarded with 13.1 eV electrons is approximately \(1.215 \times 10^{-7} \, m\).

Step-by-step solution

Determine the greatest energy level transition

Since the electrons have an energy of 13.1 eV, the hydrogen atoms can undergo a transition from the ground state, where the principal quantum number n is equal to 1, up to a higher energy level, where n is greater than 1. To find the greatest transition, we'll determine the highest energy level n that can be reached by the hydrogen atoms using the energy level formula for hydrogen: \(E_n=-13.6 \,eV \left(\frac{1}{n^2}\right)\) We need to find the maximum n such that the energy difference between the ground state (n=1) and the higher energy level is less than or equal to 13.1 eV. Let's denote the energy difference as \(ΔE\): \(ΔE = E_{n=1} - E_n\)

Calculate the maximum value of n

Now we will solve for the maximum value of n using the energy difference formula: \(13.1 \,eV = -13.6 \,eV \left(\frac{1}{1^2}\right) - (-13.6 \, eV \left(\frac{1}{n^2}\right))\) Solving for n, we get: \(n^2 \approx \frac{13.6}{13.1+13.6}\) \(n^2 \approx 3.79\) Since n must be an integer, the maximum value of n that can be reached is 2.

Use the Rydberg formula to find the shortest wavelength

We will use the Rydberg formula to determine the shortest wavelength of light emitted. The Rydberg formula for the emission spectrum of hydrogen is as follows: \(\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\) In our case, we are looking for the shortest wavelength, which corresponds to the largest energy transition between energy levels. We know that the initial energy level (n1) is 1 and the final energy level (n2) is 2. The Rydberg constant (R) for hydrogen is approximately \(1.097 \times 10^7 \, m^{-1}\). Substituting the values, we obtain: \(\frac{1}{\lambda} = 1.097 \times 10^7 \, m^{-1} \left(\frac{1}{1^2} - \frac{1}{2^2}\right)\) \(\frac{1}{\lambda} \approx 8.232 \times 10^6 \, m^{-1}\)

Calculate the shortest wavelength

Finally, we will find the shortest wavelength by taking the reciprocal of the expression obtained in the previous step: \(\lambda = \frac{1}{8.232 \times 10^6 \, m^{-1}}\) \(\lambda \approx 1.215 \times 10^{-7} \, m\) So, the shortest wavelength of light emitted by the hydrogen atoms when bombarded with 13.1 eV electrons is approximately \(1.215 \times 10^{-7} \, m\).