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Structure of Atom - Spectrum-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The electromagnetic spectrum is the arrangement of all types of electromagnetic radiations in order of increasing or decreasing wavelengths or frequencies.

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A continuous spectrum is produced when white light passes through a prism, resulting in a series of colors (VIBGYOR) that merge into one another without any gaps.

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An emission spectrum (or line spectrum) is observed when atoms of an element are excited (e.g., by heating or electric discharge). They emit light at specific wavelengths, appearing as bright lines on a dark background.

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Max Planck proposed that atoms and molecules emit or absorb energy only in discrete quantities called 'quanta'. The energy of a quantum is given by E=hνE = h\nu.

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Bohr's atomic model explains the origin of spectral lines: when an electron jumps from a higher energy level (EhigherE_{higher}) to a lower energy level (ElowerE_{lower}), it emits a photon of energy equal to the difference between these levels: ΔE=Ehigher−Elower\Delta E = E_{higher} - E_{lower}.

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Dual nature of light: Light behaves both as a wave (characterized by wavelength λ\lambda and frequency ν\nu) and as a particle (characterized by photons).

📐Formulae

c=νλc = \nu \lambda

E=hνE = h\nu

E=hcλE = \frac{hc}{\lambda}

νˉ=1λ\bar{\nu} = \frac{1}{\lambda}

ΔE=E2−E1\Delta E = E_2 - E_1

💡Examples

Problem 1:

Calculate the frequency of a radiation having a wavelength of 600 nm600\text{ nm}. (Given: speed of light c=3×108 m/sc = 3 \times 10^8\text{ m/s})

Solution:

Given: Wavelength λ=600 nm=600×10−9 m=6×10−7 m\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m} = 6 \times 10^{-7}\text{ m} Speed of light c=3×108 m/sc = 3 \times 10^8\text{ m/s} Using the formula: ν=cλ\nu = \frac{c}{\lambda} ν=3×1086×10−7\nu = \frac{3 \times 10^8}{6 \times 10^{-7}} ν=0.5×1015 Hz\nu = 0.5 \times 10^{15}\text{ Hz} ν=5×1014 Hz\nu = 5 \times 10^{14}\text{ Hz}

Explanation:

To find the frequency, we convert the wavelength to meters and use the relationship between speed, frequency, and wavelength.

Problem 2:

Determine the energy of a photon associated with light of frequency 2×1014 Hz2 \times 10^{14}\text{ Hz}. (Take Planck's constant h=6.626×10−34 J sh = 6.626 \times 10^{-34}\text{ J s})

Solution:

Given: Frequency ν=2×1014 Hz\nu = 2 \times 10^{14}\text{ Hz} h=6.626×10−34 J sh = 6.626 \times 10^{-34}\text{ J s} Using Planck's equation: E=hνE = h\nu E=(6.626×10−34)×(2×1014)E = (6.626 \times 10^{-34}) \times (2 \times 10^{14}) E=13.252×10−20 JE = 13.252 \times 10^{-20}\text{ J} E=1.3252×10−19 JE = 1.3252 \times 10^{-19}\text{ J}

Explanation:

The energy of a single photon is directly proportional to its frequency, calculated using Planck's constant.

Problem 3:

In an atom, an electron transitions from an energy level of −2.42×10−19 J-2.42 \times 10^{-19}\text{ J} to a lower level of −5.45×10−19 J-5.45 \times 10^{-19}\text{ J}. Calculate the energy of the emitted photon.

Solution:

Energy of higher level E2=−2.42×10−19 JE_2 = -2.42 \times 10^{-19}\text{ J} Energy of lower level E1=−5.45×10−19 JE_1 = -5.45 \times 10^{-19}\text{ J} ΔE=E2−E1\Delta E = E_2 - E_1 ΔE=(−2.42×10−19)−(−5.45×10−19)\Delta E = (-2.42 \times 10^{-19}) - (-5.45 \times 10^{-19}) ΔE=(−2.42+5.45)×10−19\Delta E = (-2.42 + 5.45) \times 10^{-19} ΔE=3.03×10−19 J\Delta E = 3.03 \times 10^{-19}\text{ J}

Explanation:

The energy of the emitted photon is the difference between the two stationary states of the electron.