Question 13 of 62intermediate🔧 ApplyNumerical2 marks

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Correct Answer

The wavelength of the emitted light is 486 nm.

Exercise: EXERCISES | Q: 2.13 | (Page 41)
For More Understanding

Explanation

When an electron in a hydrogen atom transitions from a higher energy level (ni=4n_i = 4) to a lower energy level (nf=2n_f = 2), energy is emitted in the form of a photon. The energy difference (ΔE\Delta E) is calculated using the energy level formula, and then the frequency and wavelength of the emitted photon are determined using Planck's constant and the speed of light.

Solution Steps

  1. Step 1: Identify the initial and final energy levels. Here, ni=4n_i = 4 and nf=2n_f = 2. This transition gives rise to a spectral line in the visible region of the Balmer series.

  2. Step 2: Calculate the energy change (ΔE\Delta E) using the formula:

    ΔE=2.18×10−18J[1ni2−1nf2]\Delta E = 2.18 \times 10^{-18} J \left[ \frac{1}{n_i^2} - \frac{1}{n_f^2} \right] ΔE=2.18×10−18J[142−122]=2.18×10−18J[116−14]=−4.08×10−19J\Delta E = 2.18 \times 10^{-18} J \left[ \frac{1}{4^2} - \frac{1}{2^2} \right] = 2.18 \times 10^{-18} J \left[ \frac{1}{16} - \frac{1}{4} \right] = -4.08 \times 10^{-19} J

    It is an emission energy.

  3. Step 3: Calculate the frequency of the emitted photon (taking energy in terms of magnitude) using v=ΔEhv = \frac{\Delta E}{h}:

    v=4.08×10−19J6.626×10−34Js=6.16×1014 Hzv = \frac{4.08 \times 10^{-19} J}{6.626 \times 10^{-34} J s} = 6.16 \times 10^{14} \text{ Hz}

  4. Step 4: Calculate the wavelength (λ\lambda) using λ=cv\lambda = \frac{c}{v}:

    λ=3.0×108 ms−16.16×1014 Hz=486 nm\lambda = \frac{3.0 \times 10^8 \text{ ms}^{-1}}{6.16 \times 10^{14} \text{ Hz}} = 486 \text{ nm}
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