Aphoton with a wavelength of less than 50.4 nm can ionize a helium atom.
what is the ionization potential of helium?
answer in j

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⁻¹c 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

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

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