An electron moves from an energy level of -8.6 eV to another level of -3.5 eV. Determine the frequency of the absorbed radiation. [1 eV = 1.6 × \(10^{19}\) J, h = 6.6 × \(10^{34}\) Js]
- \( 2.93 \times 10^{15}\)Hz
- \( 3.59 \times 10^{16}\)Hz
- \( 5.39 \times 10^{15}\)Hz
-
\( 1.24 \times 10^{15}\)Hz
The correct answer is: D
Explanation
h = \( 6.6 \times 10^{-34}\)Js, 1eV = \( 1.6 \times 10^{-19}\)J, \(E_1\) = - 8.6eV, \(E_2\) = - 3.5eV, f = ?
ΔE = hf
f = \(\frac{ΔE}{h}\)
ΔE = \(E_2\) - \(E_1\) = - 3.5eV - (- 8.6eV) = 5.1eV
but 5.1eV = 5.1 X \( 1.6 \times 10^{-19}\) = \( 8.16 \times 10^{-19}\)J
f = \(\frac{8.16 \times 10^{-19}}{ 6.6 \times 10^{-34}}\)
Therefore, f = \( 1.24 \times 10^{15}\)Hz