Answer:
n ≈ √87
Step-by-step explanation:
To determine the value of the principal quantum number (n) for the higher energy level, we can use the formula: ΔE = hf where ΔE represents the change in energy, h is Planck's constant, and f is the frequency of the emitted photon. In this case, we are given the frequency of the emitted photon as 2.98×10^14 Hz. The change in energy (ΔE) can be calculated using the equation: ΔE = E_final - E_initial where E_final represents the energy of the final state and E_initial represents the energy of the initial state. Since the electron is transitioning from an excited state to the n = 3 level, the initial state has a higher energy than the final state. Therefore, we can set the value of the final state to be zero for simplicity. Now, we can rewrite the equation for ΔE as: ΔE = 0 - E_initial = -E_initial Substituting the values into the formula ΔE = hf: -E_initial = hf Solving for E_initial, we divide both sides of the equation by -1: E_initial = -hf Plugging in the given values: E_initial = -(6.63×10^-34 J·s)(2.98×10^14 Hz) Calculating this, we find: E_initial ≈ -1.98×10^-19 J To find the value of the principal quantum number (n) for the higher energy level, we can use the formula for the energy of an electron in the hydrogen atom: E = -13.6 eV / n^2 where E is the energy and n is the principal quantum number. We can rearrange the formula to solve for n: n^2 = -13.6 eV / E Plugging in the values for E and converting electronvolts to joules (1 eV = 1.6×10^-19 J): n^2 = -13.6 eV / -1.98×10^-19 J Calculating this, we find: n^2 ≈ 87 Taking the square root of both sides of the equation: n ≈ √87 Therefore, the value of the principal quantum number (n) for the higher energy level is approximately the square root of 87.