Question

When a light of wavelength 417nm fall on the surface of potassium metal , electron are emitted

with a velocity of 6.4 x 10 4 m/ s2 what is the minimum energy required per mole to remove an

electron from potassium metal​

Answer 1

Answer:

We know that the kinetic energy of the ejected electrons and the energy incident photons are related as:  

E_kin = E_p - W  

where W is the work function (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