Question

The argon atoms are excited into an excited state before emitting the 488.0 nm laser. It is known that the energy of the first ionization energy of argon is 1520 kJ mol-1. What is the energy level of the excited state (in unit eV) lies below the vacuum energy level (0 eV)

Answer 1

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

$E = \dfrac{hc}{\lambda}$

$E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}$

$E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}$

$E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}$

$E =4.057 \times 10^{-19} \ J$

Converting Joules (J) to eV ; we get,

$E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}$

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV