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 \times \(10^{19}\) J, h = 6.6 \times \(10^{34}\) Js]
A
\( 2.93 \times 10^{15}\)Hz
B
\( 3.59 \times 10^{16}\)Hz
C
\( 5.39 \times 10^{15}\)Hz
D
\( 1.24 \times 10^{15}\)Hz
Verified Correct Answer: D.
Physical Principles & Numerical Solution
h = \( 6.6 \times 10^{-34}\)Js, 1eV = \( 1.6 \times 10^{-19}\)J, \(E_1\) = - 8.6eV, \(E_2\) = - 3.5eV, f = ?
\DeltaE = hf
f = \(\frac{\DeltaE}{h}\)
\DeltaE = \(E_2\) - \(E_1\) = - 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
Source: Myschool Past Questions Archive (Record #69398)