A light of wavelength 470 nm falls on a metal surface electrons are emitted waith speed equals 6.4100000what is the minimum energy required per mole to remove 1 electron from pottassium?

Answer 1

The kinetic energy of the ejected electrons and the energy incident photons are related as:

E_kin = E_p - W

W is the work function, i.e. the minimum energy required to remove an electron from the surface of the metal.

The kinetic energy of the ejected electron is:

E_kin = (1/2)∙m_e∙v²

= (1/2) ∙ 9.109×10⁻³⁰ kg ∙ (6.4×10⁴ m∙s⁻¹)²

= 1.866×10⁻²⁰ J

The energy of the incident photon is:

E_p = h∙f = h∙c/λ

= 6.626×10⁻³⁴ J∙s ∙ 2.998×10⁸ m∙s⁻¹ / 470×10⁻⁹ m

= 4.227×10⁻¹⁹ J

Hence the minimum energy to remove one electron is:

W = E_p - E_kin

= 4.227×10⁻¹⁹ J - 1.866×10⁻²⁰ J

= 4.040×10⁻¹⁹ J

The energy per mole of electrons is:

W_m = W ∙ N_a

= 4.040×10⁻¹⁹ J ∙ 6.022×10²³ mol⁻¹

= 2.433×10⁴ J∙mol⁻¹

= 24.33 kJ∙mol⁻¹