Question

Calculate the energy separations in joules, kilojoules per mole, electronvolts, and reciprocal centimeters (cm^-1) between the levels:

a) n = 1 and n = 2, and
b) n = 5 and n = 6

Assume that the length of the box is 1.0 nm and the particle you are dealing with is an electron with m_c = 9.109 times 10^-31 kg.

Answer 1

Answer:

$1.7918386517\times 10^{-19}\ J$, $1.11989915731\ eV$, $9022.34970645\ /cm$, $108.053470193\ kJ/mol$

$6.5700750562\times 10^{-19}\ J$, $4.10629691012\ eV[/tex}, [tex]33081.9489235\ /cm$, $396.196057373\ kJ/mol$

Explanation:

Energy is given by

$E=\dfrac{\pi^2 h^2}{2mL^2}(n_2^2-n_1^2)\\\Rightarrow E=\dfrac{\pi^2 (1.05\times 10^{-34})^2}{2\times 9.109\times 10^{-31}\times (1\times 10^{-9})^2}(2^2-1^2)\\\Rightarrow E=1.7918386517\times 10^{-19}\ J$

Energy is $1.7918386517\times 10^{-19}\ J$

In eV

$\dfrac{1.7918386517\times 10^{-19}}{1.6\times 10^{-19}}=1.11989915731\ eV$

Energy is $1.11989915731\ eV$

In cm⁻¹

$\dfrac{1.7918386517\times 10^{-19}}{1.986\times 10^{-23}}=9022.34970645\ /cm$

Energy is $9022.34970645\ /cm$

In kJ/mol

$1.11989915731\times 96.485=108.053470193\ kJ/mol$

Energy is $108.053470193\ kJ/mol$

Energy is given by

$E=\dfrac{\pi^2 h^2}{2mL^2}(n_2^2-n_1^2)\\\Rightarrow E=\dfrac{\pi^2 (1.05\times 10^{-34})^2}{2\times 9.109\times 10^{-31}\times (1\times 10^{-9})^2}(6^2-5^2)\\\Rightarrow E=6.5700750562\times 10^{-19}\ J$

Energy is $6.5700750562\times 10^{-19}\ J$

In eV

$\dfrac{6.5700750562\times 10^{-19}}{1.6\times 10^{-19}}=4.10629691012\ eV$

Energy is $4.10629691012\ eV$

In cm⁻¹

$\dfrac{6.5700750562\times 10^{-19}}{1.986\times 10^{-23}}=33081.9489235\ /cm$

Energy is $33081.9489235\ /cm$

In kJ/mol

$4.10629691012\times 96.485=396.196057373\ kJ/mol$

Energy is $396.196057373\ kJ/mol$