Задача №1146(1115) Сборник задач по физике А.П.Рымкевич
Дано:
\({U}_{з1}=0.5\,\text{В}\)
\({U}_{з2}=2\,\text{В}\)
\({\nu}_{1}=3.9\cdot {10}^{14}\,\text{Гц}\)
\({\nu}_{2}=7.5\cdot {10}^{14}\,\text{Гц}\)
\(e=1.6\cdot {10}^{-19}\,\text{Кл}\)
\(h=?\)
Решение:
$$h\cdot\nu=A+e\cdot {U}_{з}$$
$$h\cdot({\nu}_{2}-{\nu}_{1})=e\cdot ({U}_{з2}-{U}_{з1})\Rightarrow$$
$$\Rightarrow h=\frac{e\cdot ({U}_{з2}-{U}_{з1})}{{\nu}_{2}-{\nu}_{1}}$$
$$h=\frac{1.6\cdot {10}^{-19}\,\text{Кл}\cdot (2\,\text{В}-0.5\,\text{В})}{7.5\cdot {10}^{14}\,\text{Гц}-3.9\cdot {10}^{14}\,\text{Гц}}$$
$$h=6.667\cdot {10}^{-34}\,\text{Дж·с}$$
$$h=\frac{1\,\text{э}\cdot (2\,\text{В}-0.5\,\text{В})}{7.5\cdot {10}^{14}\,\text{Гц}-3.9\cdot {10}^{14}\,\text{Гц}}$$
$$h=4.167\cdot {10}^{-15}\,\text{эВ·с}$$
Ответ: постоянная Планка равна \(6.667\cdot {10}^{-34}\) Дж·с или \(4.167\cdot {10}^{-15}\) эВ·с.