The work function of sodium metal is 4.41 × 10 –19 J. If photons of wavelength 300 nm are incident on the metal, the kinetic energy of the ejected electrons will be (h = 6.63 × 10 –34 J s; c = 3 × 10⁸ m/s) ________ × 10 –21 J.

Solution

Chemistry ma’am · 4 steps

Find the photon's energy first, then subtract the work function. What is left is kinetic energy.

1. Energy balance. In the photoelectric effect, part of the photon's energy frees the electron from the metal. That part is the work function W₀. The rest becomes the electron's kinetic energy.

   KE = hc/λ - W₀

2. Photon energy. With λ = 300 nm = 3 × 10⁻⁷ m: E = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(3 × 10⁻⁷) = 6.63 × 10⁻¹⁹ J.
3. Subtract work function. KE = 6.63 × 10⁻¹⁹ - 4.41 × 10⁻¹⁹ = 2.22 × 10⁻¹⁹ J.
4. So. Rewrite it in the asked unit: 2.22 × 10⁻¹⁹ J = 222 × 10⁻²¹ J. The answer is 222.

The common slip: 2.22. Reporting the coefficient of 10⁻¹⁹ instead of 10⁻²¹ gives 2.22, not 222. Shift the decimal two places to match the blank.

The idea to keep. Photon energy hc/λ minus the work function gives the kinetic energy. Convert nm to m first, and match the power of ten the question asks for.

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