Question

A photon with a wavelength of less than 50.4 nm can ionize a helium atom.

Answer 1

Answer:

• 4.38 × 10⁻³⁵ J

Explanation:

The ionization potential is the energy required to separate an electron from an atom in gas state and, so, form an ion.

Since it is said that a photon with wavelength of less than 50.4 nm can ionize a Helium atom, the energy of such photon is the ionization potential (energy) of Helium.

The energy of a photon in terms of its wavelength is given by Planck - Einstein's equation:

• E = h . f = h / (λ . c)

Where,

• E is the energy of the photon,

• λ is the wavelength,

• f is the frequency,

• h is the Planck's constant = 6.626 × 10⁻³⁴ J.s⁻¹

• c is the speed of light ≈ 3.0 × 10 ⁸ m/s

Here you know the wavelength, λ = 50.4 nm. You must convert it to meter: 50.4 nm × 10 ⁻⁹ m.

Now you can substitute and compute:

• E = h / (λ . c) = 6.626 × 10⁻³⁴ J.s⁻¹ / (50.4 × 10⁻⁹ m × 3.0 × 10 ⁸ m/s)

• E = 0.0438 × 10⁻³³ J = 4.38 × 10⁻³⁵ J ← answer