Question

What is the maximum kinetic energy of an emitted electron if the incident light has a wave length of 217 nm. For Silver.

Answer 1

The maximum kinetic energy of the emitted electron is 2.335 x 10⁻¹⁹ J.

The given parameters;

• wavelength of the incident light, λ = 217 nm

The Einstein's photoelectric equation is given as;

$E = K.E_{max} + \Phi$

where;

$\Phi$ is the work function of the metal

The work function of silver is given as;

• polycrystalline silver =  4.26 eV

The energy of the incident light is calculated as;

$E = hf = \frac{hc}{\lambda} \\\\ E= \frac{(6.626 \times 10^{-34} \times (3\times 10^{8})}{217\times 10^{-9}} \\\\E = 9.16 \times 10^{-19} \ J$

The maximum kinetic energy of the emitted electron is calculated as;

$K.E_{max} = E - \Phi\\\\K.E_{max} = (9.16\times 10^{-19}\ J) - (4.26 \times 1.602 \times 10^{-19} \ J)\\\\K.E_{max} = 2.335 \times 10^{-19} \ J$

Thus, the maximum kinetic energy of the emitted electron is 2.335 x 10⁻¹⁹ J.

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