Предмет: Физика, автор: Akuna

Поток света выбивает из катода электроны с максимальной кинетической энергией 1эВ . При уменьшении длины волны падающего света в 2 раза максимальная кинетическая энергия выбитых электронов увеличивается в 3 раза. Чему равна работа выхода электронов с поверхности катода в Дж?

Ответы

Автор ответа: sokolovvovikn
0
При уменьшении длины волны в 2 раза энергия фотона увеличивается в 2 раза ( E=h*c?L)  Ef2=2*Ef1   Ek2=3*Ek1
Ef1=A+Ek1(1)
Ef2=A+Ek2=> 2*Ef1=A+3*Ek2(2)
Вычитаем из 2 1 : Ef=2*Ek=2 эВ
A=Ef1-Ek1=2-1=1 эВ
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