Energy of an electron is given by E = − 2.178 × 10 − 18 J Z 2 n 2 E=−2.178×10 −18 J n 2 Z 2 . What is the wavelength of light needed to excite an electron in hydrogen atom from n=1 to n=2?

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The energy of an electron in a hydrogen-like atom is given by the formula E equals negative 2.178 times 10 to the negative 18 joules, multiplied by Z squared over n squared. Here, E represents the electron energy, Z is the atomic number, and n is the principal quantum number. For a hydrogen atom, Z equals 1, so the formula simplifies to E equals negative 2.178 times 10 to the negative 18 joules divided by n squared. Now let's calculate the specific energy values for the first two quantum states. For n equals 1, we substitute into our formula: E1 equals negative 2.178 times 10 to the negative 18 joules divided by 1 squared, which gives us negative 2.178 times 10 to the negative 18 joules. For n equals 2, we have E2 equals negative 2.178 times 10 to the negative 18 joules divided by 2 squared, which equals negative 2.178 times 10 to the negative 18 divided by 4, giving us negative 5.445 times 10 to the negative 19 joules. These energy levels are shown in our energy level diagram. To excite an electron from n equals 1 to n equals 2, we need to determine the energy difference between these levels. The energy difference is calculated as delta E equals E2 minus E1. Substituting our values: delta E equals negative 5.445 times 10 to the negative 19 joules minus negative 2.178 times 10 to the negative 18 joules. This simplifies to negative 5.445 times 10 to the negative 19 plus 2.178 times 10 to the negative 18, which equals positive 1.634 times 10 to the negative 18 joules. The positive value indicates that energy must be absorbed for this excitation to occur. To find the required wavelength, we use Planck's equation which relates photon energy to frequency and wavelength. The equation is E equals h f, which can also be written as E equals h c over lambda, where h is Planck's constant equal to 6.626 times 10 to the negative 34 joule seconds, c is the speed of light equal to 3.00 times 10 to the 8 meters per second, and lambda is the wavelength. For excitation to occur, the photon energy must equal our calculated energy difference, so h c over lambda equals delta E. Rearranging this equation to solve for wavelength gives us lambda equals h c over delta E, where delta E is 1.634 times 10 to the negative 18 joules. Now we perform the final calculation by substituting our values into the wavelength equation. Lambda equals h c over delta E, which becomes lambda equals 6.626 times 10 to the negative 34 times 3.00 times 10 to the 8, all divided by 1.634 times 10 to the negative 18. First, we calculate h c equals 1.988 times 10 to the negative 25 joule meters. Then lambda equals 1.988 times 10 to the negative 25 divided by 1.634 times 10 to the negative 18 meters, which gives us 1.22 times 10 to the negative 7 meters. Converting to nanometers, we get 122 nanometers. This wavelength corresponds to ultraviolet light, which makes sense for atomic transitions in hydrogen.