The current answer posted to this question is terrible...soyou can probably chal

Question

The current answer posted to this question is terrible...soyou can probably challenge it and get lifesaver points + challengepoints if your answer is correct!
A light source of wavelength illuminates ametal and ejects photoelectrons with a maximum kinetic energy of1.10 eV. A second light source ofwavelength /2 ejects photoelectrons with a maximumkinetic energy of 4.60 eV. What is thework function of the metal?
1 eV
The current answer posted to this question is terrible...soyou can probably challenge it and get lifesaver points + challengepoints if your answer is correct!
A light source of wavelength illuminates ametal and ejects photoelectrons with a maximum kinetic energy of1.10 eV. A second light source ofwavelength /2 ejects photoelectrons with a maximumkinetic energy of 4.60 eV. What is thework function of the metal?
1 eV

Explanation / Answer

The wavelength of the source of light is The kinetic energy of the photoelectrons ejected from thesurface of the metal is K = 1.10 eV The wavelength of the second light source is /2 The kinetic energy of the photoelectrons ejected from thesurface of the metal is K1= 4.60 eV Let the work function of the metal be We know from the relation hf = + K Here, = hfo or hf = hfo + K or hfo= hf - K = (hc/) - K-------------------(1) When the second light source is used,then hf1= hfo + K1 or (hc/(/2)) = (hc/) - K + K1 or (2hc/) - (hc/) = K1 - K or (hc/) = K1 - K or = (hc/K1 - K) Here,h = 6.63 * 10-34 J.s and c = 3 *108 m/s or = (6.63 * 10-34 * 3 *108/(4.60 - 1.10) * 1.6 * 10-19) or = 3.55 * 10-7 m Therefore,from equation (1),the work function of the metalis hfo= (6.63 * 10-34 * 3 *108/3.55 * 10-7) - 1.10 * 1.6 *10-19 or hfo= 5.60 * 10-19 - 1.76 *10-19 J = 3.84 * 10-19 J or hfo= (3.84 * 10-19/1.6 *10-19) eV = 2.4 eV or hfo= 5.60 * 10-19 - 1.76 *10-19 J = 3.84 * 10-19 J or hfo= (3.84 * 10-19/1.6 *10-19) eV = 2.4 eV

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