When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _______Aᵒ. (Round off to the Nearest Integer).
[ Use : √3 = 1.73, h = 6.63 × 10⁻³⁴ Js
mₑ = 9.1 × 10⁻³¹ kg; c = 3.0 × 10⁸ ms⁻¹; 1eV = 1.6 × 10⁻¹⁹ J]

Solution

Chemistry ma’am · 4 steps

First find the electron's kinetic energy from the photoelectric equation, then feed it into de Broglie.

1. Photon energy. E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(248 × 10⁻⁹) = 8.02 × 10⁻¹⁹ J. Dividing by 1.6 × 10⁻¹⁹ gives about 5.0 eV.
2. Kinetic energy. The metal keeps 3.0 eV to release the electron. The electron leaves with K = 5.0 - 3.0 = 2.0 eV = 3.2 × 10⁻¹⁹ J.

   λ = h/√(2 mₑ K)

3. Put in numbers. 2 mₑ K = 2 × 9.1 × 10⁻³¹ × 3.2 × 10⁻¹⁹ = 5.82 × 10⁻⁴⁹, and its square root is 7.63 × 10⁻²⁵ kg m s⁻¹. So λ = (6.63 × 10⁻³⁴)/(7.63 × 10⁻²⁵) = 8.69 × 10⁻¹⁰ m.
4. So. λ ≈ 8.7 Å, which rounds to 9.

The common slip: 5. Using the full photon energy of 5 eV as the kinetic energy, without subtracting the 3.0 eV threshold, gives about 5.5 Å, which rounds to 5.

The idea to keep. The de Broglie wavelength of a photoelectron depends only on its kinetic energy. Subtract the work function first, convert eV to joules, then use λ = h/√2mK.

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