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kiruha [24]
2 years ago
9

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

Chemistry
1 answer:
MariettaO [177]2 years ago
6 0

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

<em>where;</em>

<em />\Phi<em> is the work function of the metal</em>

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.

Learn more here:brainly.com/question/14085546

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<u>Explanation:</u>

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Mathematically,

\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}

Or,

\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}         ......(1)

We are given:

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