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When light of wavelength 470nmfalls on the surface of potassium metal, electrons are emmited with a velocity of 6.4 × 10⁴ m/s what is the minimum energy required per mole to remove an electron from potassium metal?

When light of wavelength 470nmfalls on the surface of potassium metal, electrons are emmited with a velocity of 6.4 × 10⁴ m/s what is the minimum energy required per mole to remove an electron from potassium metal?

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Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
The active vitality of the shot out electrons and the vitality occurrence photons are connected as: 
E_kin = E_p - W 
W is the work, i.e. the base vitality required to expel an electron from the surface of the metal. 
The active vitality of the shot out 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 vitality of the episode 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 
Subsequently the base vitality to evacuate one electron is: 
W = E_p - E_kin 
= 4.227×10⁻¹⁹ J - 1.866×10⁻²⁰ J 
= 4.040×10⁻¹⁹ J 
The vitality per mole of electrons is: 
W_m = W ∙ N_a 
= 4.040×10 J ∙ 6.022×10²³ mol¹ 
= 24.33 kJ∙mol
 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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