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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 6 of 67
Marks: +1, -0
Find energy of each of the photons which
(i) corresponds to light of frequency 3 × 101510^{15} Hz.
(ii) have wavelength of 0.50 Å.
Solution:  
(i) We know that, E = hυ
where E = energy of photons, h = Planck’s constant, υ = frequency of light
∴ E = hυ = 6.626 × 10−3410^{-34} × 3 × 101510^{15} = 1.99 × 10−1810^{-18} J
(ii) E = hcλ\frac{hc}{\lambda}
where c = velocity of light, h = Planck’s constant, λ = wavelength
E = 6.626×10−34×3×1080.50×10−10\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.50 \times 10^{-10}} = 3.98 × 10−15 J10^{-15}\,\mathrm{J}
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