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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 19 of 67
Marks: +1, -0
The electron energy in hydrogen atom is given by EnE_n = (−2.18×10−18)n2 J\frac{(-2.18 \times 10^{-18})}{n^2} \ \mathrm{J}. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Solution:  
EnE_n = −2.18×10−18n2 J\frac{-2.18 \times 10^{-18}}{n^2} \ \mathrm{J}
E2E_2 = −2.18×10−18(2)2\frac{-2.18 \times 10^{-18}}{(2)^2} = −2.18×10−184\frac{-2.18 \times 10^{-18}}{4} = - 0.545 × 10−1810^{-18} = - 5.45 × 10−19 J10^{-19} \ \mathrm{J}
Wavelength of light (λ) = hcE\frac{hc}{E} = 6.626×10−34×3×1085.45×10−19\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5.45 \times 10^{-19}} = 3.647 × 10−7 m10^{-7} \ \mathrm{m}
= 3647 × 10−1010^{-10} m = 3647 Å [Since Å = 10−1010^{-10} m]
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