Resultados: A partir de los resultados obtenidos se calcularΓ‘ la energΓa cinΓ©tica en cada uno de los casos para posteriormente el llenado de la tabla. πΈπ = βπ£ β π· DΓ³nde: h=constante de Planck=6.626π₯10β34 π½/π v=la frecuencia del color en este caso=π = πβπ π»π§ Ξ¦=el voltaje obtenido=mV Entonces calculando esto: πΈπ = (6.626π₯10β34 )(0.5) β (297.7 π₯10β3 )=-0.2977 J πΈπ = (6.626π₯10β34 )(0.41) β (244.02π₯10β3 )=-0.24402 J πΈπ = (6.626π₯10β34 )(0.026) β (146.4π₯10β3 )=-0.1464 J πΈπ = (6.626π₯10β34 )(.045) β (333.08 π₯10β3 )=-0.33308 J πΈπ = (6.626π₯10β34 )(0.037) β (217.8π₯10β3 )= -0.2178J πΈπ = (6.626π₯10β34 )(0.037) β (178.8 π₯10β3 )=-0.1788 J
Color Rojo Naranja Amarillo Verde Azul Violeta
Longitud de onda Ξ(m) 6200 5925 5500 5300 4650 4645
Voltaje de corte V0(Volts) 297.7 mV 244.02 mV 146.4 mV 333.08 mV 217.8 mV 178.8 mV
Frecuencia F(Hz) 0.5 0.41 0.026 .045 0.037 0.037
EnergΓa cinΓ©tica Ek(J) -0.2977 -0.24402 -0.1464 -0.33308 -0.2178 -0.1788
Frecuencia Xi(v)
-0.2977 -0.24402 -0.1464 -0.33308 -0.2178 -0.1788 -1.4178
0.5 0.41 0.026 .045 0.037 0.037 1.055
π= π=
EnergΓa cinetica Yi(Ek)
πβπ₯ππ¦πββπ₯π(βπ¦π) πβπ₯π 2 β(βπ₯π)2
=
Xi2
-0.14885 -0.10004 -3.8064x10-3 -0.01498 -8.0586x10-3 -6.6156x10-3 -0.2823506
.25 0.1681 6.76x10-4 2.025x10-3 1.369x10-3 1.369x10-3 0.423539
6(β0.2823506)β(1.055)(β1.4178)
βπ₯π 2 (βπ¦π)ββπ₯π(βπ₯ππ¦π) πβπ₯π 2 β(βπ₯π)2
xiyi
6(0.423539)β(1.055)2
= 138.86x10-3
(0.423539)β(1.4178)β(1.055)(β0.2823506)
=
6(0.423539)β(1.055)2
=-212.5871x10-3