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Question: According to bohr theory,the electronic energy of hydrogen atom in the nth bohr orbit is given by Ea...

According to bohr theory,the electronic energy of hydrogen atom in the nth bohr orbit is given by Ea = ( -21.76×10^-19)/n^2 joules. Calculate the longest wavelength of light that will be needed to remove an electron from the third bohr orbit of the He+ ion

Answer

205.53 nm

Explanation

Solution

To solve this problem, we need to follow these steps:

  1. Understand the energy formula for hydrogen-like species: The given energy formula En=−21.76×10−19n2E_n = \frac{-21.76 \times 10^{-19}}{n^2} joules is for a hydrogen atom (Z=1). For a hydrogen-like ion with atomic number Z, the energy in the nth orbit is modified by multiplying by Z2Z^2. So, for a hydrogen-like ion: En=−21.76×10−19×Z2n2 joulesE_n = \frac{-21.76 \times 10^{-19} \times Z^2}{n^2} \text{ joules}

  2. Identify the given parameters for He+ ion:

    • For He+ ion, the atomic number (Z) = 2.
    • The electron is in the third Bohr orbit, so the principal quantum number (n) = 3.

  3. Calculate the energy of the electron in the third orbit of He+ ion (E3E_3): E3=−21.76×10−19×(2)2(3)2E_3 = \frac{-21.76 \times 10^{-19} \times (2)^2}{(3)^2} E3=−21.76×10−19×49E_3 = \frac{-21.76 \times 10^{-19} \times 4}{9} E3=−87.04×10−199E_3 = \frac{-87.04 \times 10^{-19}}{9} E3=−9.67111...×10−19 joulesE_3 = -9.67111... \times 10^{-19} \text{ joules}

  4. Determine the energy required to remove the electron: To remove an electron from an orbit, it needs to be excited to an infinite distance from the nucleus (n=∞), where its energy is considered to be zero (E∞=0E_\infty = 0). The energy required for this process is the ionization energy from that orbit, which is the negative of the electron's energy in that orbit. Energy required (EphotonE_{photon}) = E∞−E3=0−E3=−E3E_\infty - E_3 = 0 - E_3 = -E_3 Ephoton=−(−9.67111...×10−19 J)E_{photon} = -(-9.67111... \times 10^{-19} \text{ J}) Ephoton=9.67111...×10−19 joulesE_{photon} = 9.67111... \times 10^{-19} \text{ joules} The "longest wavelength" corresponds to the minimum energy required to remove the electron, which is exactly this ionization energy.

  5. Calculate the wavelength of light using the Planck-Einstein relation: The energy of a photon is given by E=hcλE = \frac{hc}{\lambda}, where:

    • h = Planck's constant = 6.626×10−34 J s6.626 \times 10^{-34} \text{ J s}
    • c = speed of light = 3.0×108 m/s3.0 \times 10^8 \text{ m/s}
    • λ\lambda = wavelength of light

    Rearranging the formula to find λ\lambda: λ=hcEphoton\lambda = \frac{hc}{E_{photon}} λ=(6.626×10−34 J s)×(3.0×108 m/s)9.67111...×10−19 J\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m/s})}{9.67111... \times 10^{-19} \text{ J}} λ=19.878×10−269.67111...×10−19 m\lambda = \frac{19.878 \times 10^{-26}}{9.67111... \times 10^{-19}} \text{ m} λ=2.0553×10−7 m\lambda = 2.0553 \times 10^{-7} \text{ m}

  6. Convert the wavelength to nanometers (nm): Since 1 m=109 nm1 \text{ m} = 10^9 \text{ nm}: λ=2.0553×10−7×109 nm\lambda = 2.0553 \times 10^{-7} \times 10^9 \text{ nm} λ=205.53 nm\lambda = 205.53 \text{ nm}

The longest wavelength of light needed to remove an electron from the third Bohr orbit of the He+ ion is approximately 205.53 nm.