If the threshold frequency for tungsten is 1.3 x 1015Hz, what is its work function? [H = 6.6 x 10-34 Js]

Analyze the question You are given the threshold frequency of tungsten (f₀) and Planck's constant (h), and you are asked to find the work function (φ).

Concept The work function is the minimum energy needed to free an electron from a metal's surface. It is related to the threshold frequency by:

φ = h f₀

Below the threshold frequency, no photoelectrons are emitted. At exactly the threshold frequency, the photon energy equals the work function.

Solution Given:

• f₀ = 1.3 × 10¹⁵ Hz
• h = 6.6 × 10⁻³⁴ Js

φ = h f₀ φ = (6.6 × 10⁻³⁴) × (1.3 × 10¹⁵)

Multiply the numbers: 6.6 × 1.3 = 8.58

Add the powers of 10: 10⁻³⁴ × 10¹⁵ = 10⁻¹⁹

φ = 8.58 × 10⁻¹⁹ J

Answer This matches option A.

Checking the other options: B and C have the wrong powers of 10, and D has the wrong power and no unit.

Exam tip For multiplication like this, multiply the numbers first and then add the exponents. Here, -34 + 15 = -19. Since all the options share the digits 8.58 or 8.5, the exponent is what decides the answer, so check it carefully. Also remember that work function questions are nearly always φ = hf₀, so keep that formula ready.