Question 18 of 62intermediate🔧 ApplyNumerical3 marks

What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18×102.18 \times 10⁻¹¹ ergs.

Correct Answer

(i) Energy required to shift the electron = 2.09×10−182.09 \times 10^{-18} J

(ii) Wavelength of the light emitted = 9.5×10−89.5 \times 10^{-8} m or 95 nm

Exercise: EXERCISES | Q: 2.18 | (Page 41)
For More Understanding

Explanation

The ground state energy is given in ergs, which must first be converted to Joules using the relation 1 erg=10−7 J1 \text{ erg} = 10^{-7} \text{ J}. The energy of an electron in the nthn^{th} orbit is given by En=E1n2E_n = \frac{E_1}{n^2}. The energy required to shift the electron from n=1n=1 to n=5n=5 is the difference ΔE=E5−E1\Delta E = E_5 - E_1. When the electron returns to the ground state, it emits a photon of the same energy ΔE\Delta E. The wavelength of this emitted light is calculated using the relation λ=hcΔE\lambda = \frac{hc}{\Delta E}, where h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s} and c=3.0×108 ms−1c = 3.0 \times 10^8 \text{ ms}^{-1}.

Solution Steps

  1. Step 1: Convert ground state energy to Joules Given, E1=−2.18×10−11E_1 = -2.18 \times 10^{-11} ergs. Since 1 erg=10−7 J1 \text{ erg} = 10^{-7} \text{ J}, E1=−2.18×10−11×10−7 J=−2.18×10−18 JE_1 = -2.18 \times 10^{-11} \times 10^{-7} \text{ J} = -2.18 \times 10^{-18} \text{ J}.

  2. Step 2: Calculate energy of the fifth orbit Using En=−RH(1n2)E_n = -R_H \left( \frac{1}{n^2} \right), E5=E152=−2.18×10−18 J25=−0.0872×10−18 JE_5 = \frac{E_1}{5^2} = \frac{-2.18 \times 10^{-18} \text{ J}}{25} = -0.0872 \times 10^{-18} \text{ J}.

  3. Step 3: Calculate energy required to shift the electron

    ΔE=E5−E1=−0.0872×10−18 J−(−2.18×10−18 J)\Delta E = E_5 - E_1 = -0.0872 \times 10^{-18} \text{ J} - (-2.18 \times 10^{-18} \text{ J})

    ΔE=2.0928×10−18 J≈2.09×10−18 J\Delta E = 2.0928 \times 10^{-18} \text{ J} \approx 2.09 \times 10^{-18} \text{ J}.

  4. Step 4: Calculate the wavelength of emitted light When the electron returns to the ground state, it emits energy ΔE=2.0928×10−18 J\Delta E = 2.0928 \times 10^{-18} \text{ J}. Using λ=hcΔE\lambda = \frac{hc}{\Delta E},

    λ=6.626×10−34 J s×3.0×108 ms−12.0928×10−18 J\lambda = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.0 \times 10^8 \text{ ms}^{-1}}{2.0928 \times 10^{-18} \text{ J}}

    λ=9.496×10−8 m≈9.5×10−8 m\lambda = 9.496 \times 10^{-8} \text{ m} \approx 9.5 \times 10^{-8} \text{ m} (or 95 nm).

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