Предмет: Физика, автор: Наташунька1996

Работа выхода электронов из ртути равна 4,53эВ. Частота излучения 1,8*10^15Гц.Найдите запирающее напряжение.

Ответы

Автор ответа: Аноним

дано
Авых = 4,53эВ
v = 1,8*10^15Гц
h = 6.626*10^-34
q =1.6*10^-19
найти
U
решение
hv = Aвых +qU
U = (hv - Aвых) /q= (6.626*10^-34*1.8*10^15 - 4.53*1.6*10^-19) /(1.6*10^-19)= 2.92 B

ответ
2.92 B

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