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Chapter 19: Problem 2561

Ionization energy of isolated phosphorous atom \(10 \mathrm{eV}\). Ionization energy of same atom in Si is nearly \(\mathrm{eV}\) (Relative Permittivity of silicon \(=12\) ) (A) \(0.1\) (B) \(0.2\) (C) \(0.3\) (D) \(0.4\)

Short Answer

Expert verified
The ionization energy of the same phosphorous atom in Si is closest to \(0.4 \mathrm{eV}\) (Option D).

Step by step solution

01

Known values

We are given: 1. Ionization energy of isolated phosphorous atom (E_p) = 10 eV. 2. Relative permittivity of silicon (ε_r) = 12.
02

Formula for ionization energy in a solid

Using the concept of interaction between the charged particles (nucleus and electron) in a solid, and after simplifying considering the permittivity in a solid medium, the formula for ionization energy in a solid is given by: Ionization energy in a solid (E_s) = Ionization energy of isolated atom (E_p) / (4πε_0ε_r) where ε_0 is the vacuum permittivity, which is equal to \(8.85 × 10^{-12} \mathrm{F/m}\).
03

Calculate the ionization energy in silicon

Substitute the given values into the equation: E_s = \( \frac{10 \mathrm{eV}}{ 4 \pi \times 8.85 \times 10^{-12} \mathrm{F/m} \times 12}\) Since we are only interested in the ratio between the ionization energies, we can omit the denominator, as it will get canceled out during the division. E_s = \( \frac{10 \mathrm{eV}}{12}\) E_s = 0.833 eV
04

Find the closest answer

Comparing the calculated ionization energy in silicon with the given options: (A) 0.1 eV (B) 0.2 eV (C) 0.3 eV (D) 0.4 eV E_s = 0.833 eV is closest to the option (D) 0.4 eV. Therefore, the answer is (D) 0.4 eV.

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