If the wavelength for an electron emitted from H-atom is 3.3 × 10⁻¹⁰ m, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _________ times. (Nearest integer)
[ Given : h=6.626 × 10⁻³⁴ J s ]
Mass of electron =9.1 × 10⁻³¹ kg

Solution

Chemistry ma’am · 4 steps

The electron leaves with kinetic energy, so energy absorbed is escape energy plus that kinetic energy.

1. Split the energy. To escape from the ground state, the electron first needs the ionisation energy of hydrogen, 13.6 eV = 2.18 × 10⁻¹⁸ J. Anything it absorbs beyond that becomes kinetic energy, which fixes its de Broglie wavelength.

   λ = h/√2mK ⇒ K = h²/2mλ²

2. Kinetic energy. K = (6.626 × 10⁻³⁴)²/(2 × 9.1 × 10⁻³¹ × (3.3 × 10⁻¹⁰)²) = (4.39 × 10⁻⁶⁷)/(1.98 × 10⁻⁴⁹) = 2.22 × 10⁻¹⁸ J.
3. Energy absorbed. Energy absorbed = 2.18 × 10⁻¹⁸ + 2.22 × 10⁻¹⁸ = 4.40 × 10⁻¹⁸ J. The ratio to the escape energy is (4.40 × 10⁻¹⁸)/(2.18 × 10⁻¹⁸) ≈ 2.02.
4. So. The electron absorbs about 2 times the minimum energy needed for its escape.

The common slip: 1. Comparing only the kinetic energy with the ionisation energy gives a ratio near 1.02, so 1. The absorbed energy must also include the 13.6 eV spent on escaping.

The idea to keep. Energy absorbed equals ionisation energy plus the kinetic energy the electron leaves with. Get that kinetic energy from its de Broglie wavelength using K = h²/2mλ².

Answer from the NTA final answer key. Solution by PrepWiser.