Атом водорода находится в состоянии, описываемом волновой функцией Ψ ( r , θ , ϕ ) = A e x p ( − μ r a ) Ψ(r,θ,ϕ)=Aexp(−μ a r ​ ), здесь A A – неизвестная константа, a a ‒ боровский радиус, μ = μ= 2. Найдите вероятность обнаружить атом в невозбужденном состоянии. Приведите ответ с округлением до сотых. Указание ∫ ∞ 0 x 2 e x p ( − n x ) d x = 2 ⋅ n − 3 ∫ 0 ∞ ​ x 2 exp(−nx)dx=2⋅n −3

Для нахождения вероятности обнаружить атом в невозбужденном состоянии нужно найти квадрат модуля волновой функции и проинтегрировать его по всему пространству:

P = ∫∫∫ |Ψ(r,θ,ϕ)|^2 r^2 sinθ dr dθ dϕ

P = ∫∫∫ A^2 exp(-2r/a) r^2 sinθ dr dθ dϕ

P = A^2 ∫∫∫ r^2 exp(-2r/a) sinθ dr dθ dϕ

Используя указание, получаем:

P = A^2 * 2 * 2^(-3) * a^3

P = A^2 * a^3 / 4

Так как вероятность должна быть равна 1, то:

1 = A^2 * a^3 / 4

A^2 = 4 / a^3

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