What is \(\Delta E\) for the transition of an electron from \(n=6\) to \(n=3\) in a Bohr hydrogen atom? What is the frequency of the spectral line produced?

Short Answer

Expert verified
The energy change \(\Delta E\) for the transition of an electron from \(n=6\) to \(n=3\) in a Bohr hydrogen atom is -1.133 eV or -1.815x10^-19 Joules. The frequency of the spectral line produced is 2.74 x 10^{14} Hz.

Step by step solution

01

Calculate Energy at Level 6

First, calculate the energy of the electron at energy level \(n=6\) using the formula \(E_n = -13.6/ n^2\). Here, \(n = 6\), so \(E_{n=6} = -13.6/6^2 = -0.3778\) eV
02

Calculate Energy at Level 3

Next, calculate the energy of the electron at energy level \(n=3\) using the same formula. Here, \(n = 3\). Hence, \(E_{n=3} = -13.6/3^2 = -1.511\) eV
03

Calculate ∆E

Now calculate the change in energy, \(\Delta E \), which is the energy difference between the two levels. \(\Delta E = E_{n=3} - E_{n=6} = -1.511 - (-0.3778) = -1.133\) eV
04

Convert Energy to Joules

We will convert the energy difference from electron Volts (eV) to Joules (J) to utilize it further: 1eV = \(1.602\times10^{-19}\) J. Thus \(\Delta E = -1.133 \times 1.602\times10^{-19} = -1.815\times10^{-19}\) Joules.
05

Calculate Frequency

Now, calculate the frequency of the spectral line produced using the equation \(E = hf\), where \(h\) is Planck's constant. Solving for \(f\), we get \(f = \Delta E / h\). Given \(h = 6.626\times10^{-34}\) Js, \(f = -1.815\times10^{-19} / 6.626\times10^{-34} = 2.74 \times 10^{14}\) Hz.

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