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Structure of Atom question

2021 · 25 Feb · Shift 2 · Q18
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Structure of Atom question

2021 · 25 Feb · Shift 2 · Q18

JEE MainChemistryStructure of AtomNumerical+4 / −1
Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization enegy of metal A in kJ mol −-− 1 is ‾\underline{\hspace{2cm}}​. (Rounded off to the nearest integer) [h = 6.63 ×\times× 10 −-− 34 Js, c = 3.00 ×\times× 108 ms −-− 1, NA = 6.02 ×\times× 1023 mol −-− 1]
Numerical answer
View written solutionFree

Correct answer: 181

  1. Use the photon energy formula

The radiation is just sufficient to ionise the atom, so the ionization energy per atom is equal to the energy of one photon:

E=hcλE = \frac{hc}{\lambda}E=λhc​

Given:

h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s c=3.00×108 m s−1c = 3.00 \times 10^8\,\text{m s}^{-1}c=3.00×108m s−1 λ=663 nm=663×10−9 m\lambda = 663\,\text{nm} = 663 \times 10^{-9}\,\text{m}λ=663nm=663×10−9m

  1. Calculate energy per photon

E=(6.63×10−34)(3.00×108)663×10−9E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{663 \times 10^{-9}}E=663×10−9(6.63×10−34)(3.00×108)​

First, numerator:

6.63×10−34×3.00×108=19.89×10−26=1.989×10−25 J6.63 \times 10^{-34} \times 3.00 \times 10^8 = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}\,\text{J}6.63×10−34×3.00×108=19.89×10−26=1.989×10−25J

Now divide by 663×10−9663 \times 10^{-9}663×10−9:

E=1.989×10−25663×10−9E = \frac{1.989 \times 10^{-25}}{663 \times 10^{-9}}E=663×10−91.989×10−25​

E=3.00×10−19 J per atomE = 3.00 \times 10^{-19}\,\text{J per atom}E=3.00×10−19J per atom

  1. Convert to per mole

Multiply by Avogadro's number:

Emol=3.00×10−19×6.02×1023E_{\text{mol}} = 3.00 \times 10^{-19} \times 6.02 \times 10^{23}Emol​=3.00×10−19×6.02×1023

Emol=18.06×104 J mol−1E_{\text{mol}} = 18.06 \times 10^4\,\text{J mol}^{-1}Emol​=18.06×104J mol−1

Emol=1.806×105 J mol−1E_{\text{mol}} = 1.806 \times 10^5\,\text{J mol}^{-1}Emol​=1.806×105J mol−1

  1. Convert to kJ mol−1^{-1}−1

Emol=180.6 kJ mol−1E_{\text{mol}} = 180.6\,\text{kJ mol}^{-1}Emol​=180.6kJ mol−1

  1. Round off to nearest integer

181\boxed{181}181​

  1. Comparison with stored answer

Stored correct answer = 181181181

Our derived answer = 181181181

So, the answer agrees with the stored correct answer.

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