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 Js, Mass of electron = 9.1 × 10⁻³¹]
[Given: h = 6.626 × 10⁻³⁴ Js, Mass of electron = 9.1 × 10⁻³¹]
Answer: 2
Solution
Using de Broglie relation λ = h/√(2mK), the kinetic energy K = h²/(2mλ²) = (6.626 × 10⁻³⁴)²/(2 × 9.1 × 10⁻³¹ × (3.3 × 10⁻¹⁰)²) ≈ 2.215 × 10⁻¹⁸ J. The minimum energy required for escape (ionization energy of H) is 13.6 eV = 13.6 × 1.602 × 10⁻¹⁹ J ≈ 2.179 × 10⁻¹⁸ J. Energy absorbed = Ionization energy + Kinetic energy = 2.179 × 10⁻¹⁸ + 2.215 × 10⁻¹⁸ ≈ 4.394 × 10⁻¹⁸ J. Ratio = 4.394/2.179 ≈ 2.
JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer