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JAMB Physics 2024 Paper

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Question : 61

Total: 125

An electron falls from an energy level of -5.44eV to another energy level, E. If the emitted photon is of wavelength 5.68 x 10

^{-6}

m, calculate the energy change. [ Plank's constant = 6.63 x 10

^{-34}

Js, emitted radiation speed = 3.0 x 10

^8

^{-1}

]

1.49 x 10 $^{-20}$ J
1.49 x 10 $^{-19}$ J
3.49 x 10 $^{-20}$ J
3.49 x 10 $^{-19}$ J

Solution:

Concept:

The energy change when an electron jumps between energy levels is equal to the energy of the emitted photon.

Explanation:

The energy of the emitted photon is given by

E = \frac{hc}{\lambda}

.

Given:

h = 6.63 \times 10^{-34}\ \mathrm{J\,s}

and

c = 3.0 \times 10^8\ \mathrm{m\,s^{-1}}

.

Given wavelength

\lambda = 5.68 \times 10^{-6}\ \mathrm{m}

.

Substitute the values:

E = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{5.68 \times 10^{-6}}

E = \frac{1.989 \times 10^{-25}}{5.68 \times 10^{-6}}

E = 3.50 \times 10^{-20}\ \mathrm{J}

This photon energy equals the energy change of the electron.

Answer:

The energy change is

3.50 \times 10^{-20}\ \mathrm{J}

, which is closest to option C:

3.49 \times 10^{-20}\ \mathrm{J}

.

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