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Консультация № 173839
30.10.2009, 13:42
0.00 руб.
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На поверхность металла падает монохроматическое излучение с длиной волны [$955$]1=0.13*10^-6 м. (значение работы выхода A=2.5 эВ). Найти максимальную скорость vmax фотоэлектронов и задерживающую разность потенциалов U3. Решить задачу для излучения с длиной волны [$955$]2=2,3*10^-12 м.
Обсуждение
31.10.2009, 11:18
общий
это ответ
Здравствуйте, Вика.
Дано:
λ1=0.13*10^-6 м
A=2.5 эВ = 4*10-19 Дж
λ2=2,3*10^-12 м
Найти:
vm; Uз
Решение:
Уравнение Эйнштейна для фотоэффекта
hc/[$955$] = A+EK max (1),
где h=6.63*10-34 Дж*с - постоянная планка, с=3*108 м/с - скорость света в вакууме,
EK max=mvm2/2 (2), m=9.1*10-31 кг - масса электрона
Далее, работа электрического поля по торможению электронов eUз=EK max (3), e=1.6*10-19 - заряд электрона
Из уравнения (1)
EK max= hc/[$955$] - A = 11,3*10-19 Дж
Из уравнения (2)
vm=[$8730$](2EK max/m) = 1.6*106 м/с
Из уравнения (3)
Uз = EK max/e ~ 7.1 В
Это для первой длины волны.
Дано:
λ1=0.13*10^-6 м
A=2.5 эВ = 4*10-19 Дж
λ2=2,3*10^-12 м
Найти:
vm; Uз
Решение:
Уравнение Эйнштейна для фотоэффекта
hc/[$955$] = A+EK max (1),
где h=6.63*10-34 Дж*с - постоянная планка, с=3*108 м/с - скорость света в вакууме,
EK max=mvm2/2 (2), m=9.1*10-31 кг - масса электрона
Далее, работа электрического поля по торможению электронов eUз=EK max (3), e=1.6*10-19 - заряд электрона
Из уравнения (1)
EK max= hc/[$955$] - A = 11,3*10-19 Дж
Из уравнения (2)
vm=[$8730$](2EK max/m) = 1.6*106 м/с
Из уравнения (3)
Uз = EK max/e ~ 7.1 В
Это для первой длины волны.
Об авторе:
С уважением
shvetski
С уважением
shvetski
31.10.2009, 11:22
общий
Что касается случая для [$955$]2, то при подстановке соответствующие значения получаются следующие
EK max=8.6*10-14 Дж
vm=4.4*108 м/с - скорость больше скорости света! - что это значит - я пока не понимаю
Uз=537,5 кВ
Об авторе:
С уважением
shvetski
С уважением
shvetski
31.10.2009, 21:40
общий
Химик CH:
А формула та же?
vm=√(2EK max/m) ???
У меня, как не пересчитываю, все 4,3 получается
[$8730$](2*8,6*10-14/9,1*10-31)=[$8730$](17,2/9,1 *1031-14)=[$8730$](18,9*1016)=4,3*108
Может масса электрона изменяется, поэтому скорость будет меньше...?
Уважаемый Химик, поясните, пожалуйста, как у Вас получилось 2,75?
Спасибо
shvetski :)
А формула та же?
vm=√(2EK max/m) ???
У меня, как не пересчитываю, все 4,3 получается
[$8730$](2*8,6*10-14/9,1*10-31)=[$8730$](17,2/9,1 *1031-14)=[$8730$](18,9*1016)=4,3*108
Может масса электрона изменяется, поэтому скорость будет меньше...?
Уважаемый Химик, поясните, пожалуйста, как у Вас получилось 2,75?
Спасибо
shvetski :)
Об авторе:
С уважением
shvetski
С уважением
shvetski
31.10.2009, 22:12
общий
Shvetski:
Не найдя нормальной релятивистской формулы зависимости скорости от энергии, я взял формулу Ek=m*v2/(2*[$8730$](1-v2/c2))
и выразил из неё скорость
(фактически, это релятивистская поправка к массе)
Не найдя нормальной релятивистской формулы зависимости скорости от энергии, я взял формулу Ek=m*v2/(2*[$8730$](1-v2/c2))
и выразил из неё скорость
(фактически, это релятивистская поправка к массе)
31.10.2009, 23:43
общий
Химик CH:
Принимается...
Но, здесь... похоже на циклическую ссылку - масса через скорость, скорость через массу, но другого ничего в голову не приходит...
Красиво, однако, у вас получилось.
Принимается...
Но, здесь... похоже на циклическую ссылку - масса через скорость, скорость через массу, но другого ничего в голову не приходит...
Красиво, однако, у вас получилось.
Об авторе:
С уважением
shvetski
С уважением
shvetski
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