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WAEC Physics 2013 Paper

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Question : 37

Total: 46

The work function of a metal is 2.56 x 10^-19J. Calculate the frequency of a photon whose energy is required to eject from the metal an electron with kinetic energy of 3.0ev. [1 eV = 1.6 x 10^-19J, h = 6.6 x 10^-34Js]

5.56 x 10¹⁹Hz
1.66 x 10¹⁹Hz
1.12 x 10¹⁵Hz
8.42 x 10¹⁴Hz

Solution:

Concept:

The minimum photon energy needed to eject an electron is the sum of the metal's work function and the electron's required kinetic energy.

Explanation:

Step 1: Convert the electron kinetic energy to joules.

K.E. = 3.0\ \text{eV} = 3.0 \times 1.6 \times 10^{-19}\ \text{J} = 4.8 \times 10^{-19}\ \text{J}

Step 2: Add the work function to get the total photon energy.

E = K.E. + W_0

E = 4.8 \times 10^{-19} + 2.56 \times 10^{-19} = 7.36 \times 10^{-19}\ \text{J}

Step 3: Use

E = hf

and solve for frequency.

f = \frac{E}{h}

f = \frac{7.36 \times 10^{-19}}{6.6 \times 10^{-34}}

f = 1.12 \times 10^{15}\ \text{Hz}

Answer:

The correct option is C.

1.12 \times 10^{15}\ \text{Hz}

.

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