29.07 Radiation of a certain wavelength causes electrons with a maximum kinetic

Question

29.07 Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.74 eV to be ejected from a metal whose work function is 1.79 eV. What will be the maximum kinetic energy (in eV) with which this same radiation ejects electrons from another metal whose work function is 1.58 eV?

Number Units eVphotons/secnmTJ Light is shining perpendicularly on the surface of the earth with an intensity of 680 W/m2. Assuming that all the photons in the light have the same wavelength (in vacuum) of 497 nm, determine the number of photons per second per square meter that reach the earth. Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.74 eV to be ejected from a metal whose work function is 1.79 eV. What will be the maximum kinetic energy (in eV) with which this same radiation ejects electrons from another metal whose work function is 1.58 eV?

Explanation / Answer

Energy density incident = 680 J/ s -m2 Energy incident per unit areapersecond E = 680J Energy of singlephoton e = h*c/ ?

= 6.626 *10-34 * 3 * 108 / (697 *10-9) e = 2.85* 10-19 J no. of photonsincident n = E/ e = 680 / (2.85 *10-19) n = 2.38* 1021 photons / s   

2) Let? be the work function, the minimum energy required toremove an electron the metal Er = ?+ 1/2 * m * v2 where Er is theenergy of an incident photon We knowEr = 1.79 eV + .74 eV = 2.53 eV For thesecond metal 1/2 * m * v2 = Er - 1.58 eV = 0.95 eV 2) Let? be the work function, the minimum energy required toremove an electron the metal Er = ?+ 1/2 * m * v2 where Er is theenergy of an incident photon We knowEr = 1.79 eV + .74 eV = 2.53 eV For thesecond metal 1/2 * m * v2 = Er - 1.58 eV = 0.95 eV

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