A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 × 10¹⁴ Hz. The velocity of ejected electron is ____________ × 10⁵ ms⁻¹ (Nearest integer)
[Use : h = 6.63 × 10⁻³⁴ Js, mₑ = 9.0 × 10⁻³¹ kg]

Solution

Chemistry ma’am · 4 steps

The photon pays the threshold energy first, and only what is left over becomes the electron's kinetic energy.

1. Photon frequency. Convert the wavelength to a frequency with ν = c/λ. ν = (3 × 10⁸)/(500 × 10⁻⁹) = 6.0 × 10¹⁴ Hz, which is above the threshold, so electrons are ejected.

   1/2mₑ v² = h(ν - ν₀)

2. Kinetic energy. ν - ν₀ = 6.0 × 10¹⁴ - 4.3 × 10¹⁴ = 1.7 × 10¹⁴ Hz. So 1/2mₑ v² = 6.63 × 10⁻³⁴ × 1.7 × 10¹⁴ = 1.127 × 10⁻¹⁹ J.
3. Solve for v. v² = (2 × 1.127 × 10⁻¹⁹)/(9.0 × 10⁻³¹) = 2.50 × 10¹¹ m² s⁻². Taking the square root gives v = 5.0 × 10⁵ m s⁻¹.
4. So. The ejected electron moves at about 5 × 10⁵ m s⁻¹, so the answer is 5.

The common slip: 9. Using the full photon energy hν as kinetic energy, without subtracting hν₀, gives v ≈ 9.4 × 10⁵ m s⁻¹, so 9 instead of 5.

The idea to keep. In every photoelectric question, first subtract the threshold energy hν₀ from hν. Only the remainder is 1/2mv², and remember the 2 when you solve for v.

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