Предмет: Физика, автор: kotakot

СРОЧНЯК ПАЦАНЫ! ДОФИГА БАЛЛОВ
Для измерения энергии микрочастиц используют системную единицу - электрон-вольт (эВ). Такую энергию приобретает электрон, пройдя разность потенциалов 1 В. Какую кинетическую энергию приобрел электрон, пройдя ускоряющую разность потенциалов 2кВ??

Желательно через дано :О

Вот формулы по которым она должна решаться(точнее какаято одна из них):

A=q(ф2-ф1)=qU
E=Ud
C=qU=e0e*sd

W=q^22C

Ответы

Автор ответа: Luna92

1

Заряженные частицы с единичным зарядом имеют свою кинетич. энергию Eкин=3/2kТ, она приобретается в случае пробега разностью потенциалов U, то 3/2kT=eU.

А в случае U=2*10^3 В, получается, E=eU => E=2*10^3 эВ=2 кэВ.

Вроде так.

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