when light of wavelength 470 nm falls on the surface of potassium metal, electron are emitted with a velocity of 6.4*10^4 m/s. what us the minimum energy required per mole to remove an electron from potassium metal
Answers
Answer:
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 J ∙ 6.022×10²³ mol¹
= 2.433×10⁴ J∙mol
= 24.33 kJ∙mol
Explanation:
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