Calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level. Recall that for hydrogen E_n = -2.18 × 10^-18 J(1/n² ). (h = 6.63 × 10^-34 J • s) A) 3.08 × 10¹⁵s^-1 B) 1.03 × 10⁸s^-1 C) 2.06 × 10¹⁴s^-1 D) 1.35 × 10^-51s^-1 E) 8.22 × 10¹⁴s^-1

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The frequency of the light emitted during the transition can be calculated using the energy difference between the two levels and Planck's constant. The energy of a level in a hydrogen atom is given by $E_n = -2.18 \times 10^{-18} \frac{1}{n^2} \, J$. The energy difference ($\Delta E$) between levels n=4 and n=1 is:$$\Delta E = E_1 - E_4 = -2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = -2.18 \times 10^{-18} \left( 1 - \frac{1}{16} \right) = -2.18 \times 10^{-18} \left( \frac{15}{16} \right)$$$$\Delta E = -2.04 \times 10^{-18} \, J$$The frequency ($f$) is found using $\Delta E = h \cdot f$, where $h = 6.63 \times 10^{-34} \, J \cdot s$. Rearranging for $f$, we get:$$f = \frac{\Delta E}{h} = \frac{-2.04 \times 10^{-18}}{6.63 \times 10^{-34}} = 3.08 \times 10^{15} \, s^{-1}$$Therefore, the correct answer is A) $3.08 \times 10^{15} \, s^{-1}$.

Calculate the Frequency of the Light Emitted by a Hydrogen