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WAEC Physics 2023 Paper

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Question : 28

Total: 50

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} \text{ J}, h=6.6 \times

10^{34} \text{ Js}

$2.93 \times 10^{15}\text{ Hz}$
$3.59 \times 10^{16}\text{ Hz}$
$5.39 \times 10^{15}\text{ Hz}$
$1.24 \times 10^{15}\text{ Hz}$

Solution:

Concept:

Energy absorbed by the electron is equal to the photon energy, which is given by

E = h f

.

Explanation:

The energy difference between the two levels is

\Delta E = E_2 - E_1 = (-3.5 \ \mathrm{eV}) - (-8.6 \ \mathrm{eV}) = 5.1 \ \mathrm{eV}

.

Convert this energy into joules:

\Delta E = 5.1 \times 1.6 \times 10^{-19} = 8.16 \times 10^{-19} \ \mathrm{J}

.

Using

f = \frac{\Delta E}{h}

,

f = \frac{8.16 \times 10^{-19}}{6.6 \times 10^{-34}} = 1.236 \times 10^{15} \ \mathrm{Hz}

.

Therefore the frequency of absorbed radiation is approximately

1.24 \times 10^{15} \ \mathrm{Hz}

.

Answer:

1.24 \times 10^{15} \ \mathrm{Hz}

(Option D).

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