Предмет: Физика, автор: ContrDog

Определить работу выхода электрона с поверхности фотокатода, если при облучении фотоэлемента светом с длиной волны л(лямбда)=210 нм фототок прекращается при задерживающем напряжении 2,7 В

Ответы

Автор ответа: tana20101997

h*c/L=A+e*U
A=h*c/L -e*U=6,62*10^-34*3*10^8/210*10^-9 -1,6*10^-19*2,7=5,14*10^-19 Дж

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