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Question: Electromagnetic radiation of wavelength \(242{\text{ nm}}\) is just sufficient to ionise the sodium ...

Electromagnetic radiation of wavelength 242 nm242{\text{ nm}} is just sufficient to ionise the sodium atom. What is the ionisation energy of sodium per atom?
A) 494.5×10−6 J/atom494.5 \times {10^{ - 6}}{\text{ J/atom}}
B) 8196.5×10−10 J/atom8196.5 \times {10^{ - 10}}{\text{ J/atom}}
C) 5.85×10−15 J/atom5.85 \times {10^{ - 15}}{\text{ J/atom}}
D) 8.214×10−19 J/atom8.214 \times {10^{ - 19}}{\text{ J/atom}}

Explanation

Solution

The ionisation energy of the atom is inversely proportional to the wavelength and thus, directly proportional to the frequency. This is because wavelength and frequency are inversely proportional to each other. As the wavelength increases, the energy of the atom decreases.

Formula Used: E=hcλE = \dfrac{{hc}}{\lambda }

Complete answer:
Convert the units of wavelength from nm{\text{nm}} to m{\text{m}} using the relation as follows:
1 nm=1×10−9 m1{\text{ nm}} = 1 \times {10^{ - 9}}{\text{ m}}
Thus, λ=242 nm×1×10−9 m1 nm\lambda = 242{\text{ }}{{{\text{nm}}}} \times \dfrac{{1 \times {{10}^{ - 9}}{\text{ m}}}}{{1{\text{ }}{{{\text{nm}}}}}}
λ=242×10−9 m\lambda = 242 \times {10^{ - 9}}{\text{ m}}
Thus, the wavelength is 242×10−9 m242 \times {10^{ - 9}}{\text{ m}}.

Calculate the energy of the photon using the Planck-Einstein equation as follows:
E=hcλE = \dfrac{{hc}}{\lambda }
Where, EE is the energy of the photon,
hh is the Planck’s constant,
cc is the velocity of light in vacuum,
λ\lambda is the wavelength.

Substitute 6.626×10−34 J s6.626 \times {10^{ - 34}}{\text{ J s}} for the Planck’s constant, 3×108 m s−13 \times {10^8}{\text{ m }}{{\text{s}}^{ - 1}} for the velocity of light in vacuum, 242×10−9 m242 \times {10^{ - 9}}{\text{ m}} for the wavelength. Thus,
⇒E=6.626×10−34 J s×3×108 m s−1242×10−9 m\Rightarrow E = \dfrac{{6.626 \times {{10}^{ - 34}}{\text{ J }}{{\text{s}}} \times 3 \times {{10}^8}{\text{ }}{{\text{m}}}{\text{ }}{{{{\text{s}}^{ - 1}}}}}}{{242 \times {{10}^{ - 9}}{\text{ }}{{\text{m}}}}}
⇒E=8.214×10−19 J/atom\Rightarrow E = 8.214 \times {10^{ - 19}}{\text{ J/atom}}
Thus, ionisation energy of sodium per atom having wavelength 242 nm{\text{242 nm}} is 8.214×10−19 J/atom8.214 \times {10^{ - 19}}{\text{ J/atom}}.

Thus, the correct option is (D) 8.214×10−19 J/atom8.214 \times {10^{ - 19}}{\text{ J/atom}}.

Note: The amount of energy required to remove one mole of electron from an ion is known as its ionisation energy. During ionisation, the neutral atom is converted to an ion. IE1{\text{I}}{{\text{E}}_1} is the first ionization energy. The energy required to remove one mole of electron from the ion with charge +1 is known as the first ionization energy. Similarly, IE2{\text{I}}{{\text{E}}_2} is the second ionization energy. The energy required to remove one mole of electron from the ion with charge +2 is known as the second ionization energy. And, IE3{\text{I}}{{\text{E}}_3} is the third ionization energy. The energy required to remove one mole of electron from the ion with charge +3 is known as the third ionization energy.