Question 51 of 62intermediate🔧 ApplyNumerical3 marks

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×101.5 \times 10⁷ m s⁻¹, calculate the energy with which it is bound to the nucleus.

Correct Answer

The energy with which the electron is bound to the nucleus is 1.223×10−151.223 \times 10^{-15} J.

Exercise: EXERCISES | Q: 2.54 | (Page 41)
For More Understanding

Explanation

According to the photoelectric effect principle, the energy of the incident photon is used to overcome the binding energy of the electron, and the remaining energy is transferred as the kinetic energy of the ejected electron. Therefore, Binding Energy = Energy of photon - Kinetic energy of the electron.

Solution Steps

  1. Step 1: Calculate the energy of the incident photon Using the formula E=hcλE = \frac{hc}{\lambda} Given: λ=150 pm=150×10−12 m\lambda = 150 \text{ pm} = 150 \times 10^{-12} \text{ m}, h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}, c=3×108 m s−1c = 3 \times 10^8 \text{ m s}^{-1}

    E=6.626×10−34 J s×3×108 m s−1150×10−12 m=1.3252×10−15 JE = \frac{6.626 \times 10^{-34} \text{ J s} \times 3 \times 10^8 \text{ m s}^{-1}}{150 \times 10^{-12} \text{ m}} = 1.3252 \times 10^{-15} \text{ J}

  2. Step 2: Calculate the kinetic energy of the ejected electron Using the formula Ke=12mev2K_e = \frac{1}{2} m_e v^2 Given: v=1.5×107 m s−1v = 1.5 \times 10^7 \text{ m s}^{-1}, mass of electron me=9.11×10−31 kgm_e = 9.11 \times 10^{-31} \text{ kg}

    Ke=12×9.11×10−31 kg×(1.5×107 m s−1)2=1.0249×10−16 JK_e = \frac{1}{2} \times 9.11 \times 10^{-31} \text{ kg} \times (1.5 \times 10^7 \text{ m s}^{-1})^2 = 1.0249 \times 10^{-16} \text{ J}

  3. Step 3: Calculate the binding energy The binding energy (EbE_b) is the difference between the photon energy and the kinetic energy of the electron:

    Eb=E−KeE_b = E - K_e Eb=1.3252×10−15 J−0.10249×10−15 J=1.2227×10−15 JE_b = 1.3252 \times 10^{-15} \text{ J} - 0.10249 \times 10^{-15} \text{ J} = 1.2227 \times 10^{-15} \text{ J}
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