Question

Consider a thin spherical shell located between r = 0.49a₀ and 0.51a₀. For the n = 2, l = 1 state of hydrogen, find the probability for the electron to be found in a small volume element that subtends a polar angle of 0.11° and an azimuthal angle of 0.25° if the center of the volume element is located at: θ=5°, ϕ=35°.

Probability when n=2,l=1,m=0

Answer 1

given

thin spherical shell located between r = 0.49 a₀ and 0.51 a₀

n = 2, l = 1 state of hydrogen

polar angle of 0.11° and an azimutal angle of 0.25°

θ = 5°, ϕ = 35°

Probability for the electron for n = 2 , l = 1 ,m = 0,

= ( 1 / 24 a₀³ ) ( r² / a₀² ) e^{-r / a0} ( 3cos² / 4 ) r² sin dr d d

r = 0.49a₀ + 0.51a₀ / 2 = 0.50a₀

dr = 0.51a₀ - 0.49a₀ = 0.02a₀

d = 0.11^o = 0.00192 rad

d = 0.25^o = 0.00436 rad

θ = 5°, ϕ = 35°

= (1 / 24 a₀³ ) ( 0.50 x a₀² / a₀²) e^{- 0.50 a₀ / a₀} ( 3cos² 5 / 4) ( 0.50 a₀ )² sin(5) x 0.02 a₀ x 0.00192 x 0.00436

= (1 / 24 ) ( 0.50 ) e^{- 0.50} ( 3cos² 5 / 4) ( 0.50 )² sin(5) x 0.02 x 0.00192 x 0.00436

= ( 0.0416 ) ( 0.50 ) x 0.6065 x ( 0.237 ) x 0.25 x 0.0871 x 0.02 x 0.00192 x 0.00436

= 1.08997 x 10^-11