Question 19 of 62intermediate🔧 ApplyNumerical3 marks

The electron energy in hydrogen atom is given by Eₙ = (–2.18×102.18 \times 10⁻¹⁸)/n² 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?

Correct Answer

(i) Energy required = 0.545×10−180.545 \times 10^{-18} J

(ii) Longest wavelength = 3.647×10−53.647 \times 10^{-5} cm

Exercise: EXERCISES | Q: 2.19 | (Page 41)
For More Understanding

Explanation

To remove an electron completely from the n=2n = 2 orbit, it must transition to n=∞n = \infty, where the energy is zero. The energy required is the difference between E∞E_\infty and E2E_2. The longest wavelength of light corresponds to the exact energy required for this ionization, calculated using λ=hcΔE\lambda = \frac{hc}{\Delta E}.

Solution Steps

  1. Step 1: Calculate the energy of the electron in the n=2n = 2 orbit. Using the given formula En=−2.18×10−18n2E_n = \frac{-2.18 \times 10^{-18}}{n^2} J: E2=−2.18×10−18×122=−0.545×10−18E_2 = -2.18 \times 10^{-18} \times \frac{1}{2^2} = -0.545 \times 10^{-18} J.

  2. Step 2: Calculate the energy required to remove the electron completely. When the electron is completely removed, it is free from the nucleus, corresponding to n=∞n = \infty. Thus, E∞=0E_\infty = 0. Energy required (ΔE\Delta E) = E∞−E2E_\infty - E_2 ΔE=0−(−0.545×10−18)=0.545×10−18\Delta E = 0 - (-0.545 \times 10^{-18}) = 0.545 \times 10^{-18} J.

  3. Step 3: Calculate the longest wavelength of light. The longest wavelength corresponds to the minimum energy required for this transition. Using ΔE=hcλ\Delta E = \frac{hc}{\lambda}:

    λ=hcΔE\lambda = \frac{hc}{\Delta E}

    Where h=6.626×10−34h = 6.626 \times 10^{-34} J s and c=3.0×108c = 3.0 \times 10^8 m s−1^{-1}. λ=6.626×10−34×3.0×1080.545×10−18\lambda = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{0.545 \times 10^{-18}} m λ=19.878×10−260.545×10−18\lambda = \frac{19.878 \times 10^{-26}}{0.545 \times 10^{-18}} m λ=36.47×10−8\lambda = 36.47 \times 10^{-8} m.

  4. Step 4: Convert the wavelength to cm. Since 11 m = 100100 cm, λ=36.47×10−8×100\lambda = 36.47 \times 10^{-8} \times 100 cm = 3.647×10−53.647 \times 10^{-5} cm.

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