Question

what frequency of light must be used to eject electrons from a platinum surface with a maximum kinetic energy of 2.82×10−19 j ? express your answer to three significant figures.

Answer 1

The frequency of light needed to eject electrons from a platinum surface with a specific kinetic energy can be found using the photoelectric effect equation once we know the work function of platinum.

Step-by-step explanation:

To solve this problem, we need to use the following equation:

E = h x f - Φ

where:

E is the kinetic energy of the ejected electron (J)

h is Planck's constant (6.626 x 10⁻³⁴ J s)

f is the frequency of light (Hz)

Φ is the work function of platinum (5.65 x 10⁻¹⁹ J)

We are given the following values:

E = 2.82 x 10⁻¹⁹ J

Φ = 5.65 x 10⁻¹⁹ J

We need to find f.

Rearranging the equation, we get:

f = (E + Φ) / h

Substituting the values, we get:

f = (2.82 x 10⁻¹⁹ J + 5.65 x 10⁻¹⁹ J) / (6.626 x 10⁻³⁴ J s)

f ≈ 4.26 x 10¹⁴ Hz

Therefore, the frequency of light required to eject electrons from a platinum surface with a maximum kinetic energy of 2.82 x 10⁻¹⁹ J is approximately 4.26 x 10¹⁴ Hz (to three significant figures).