Question

For an electron in the 1s state of hydrogen, what is the probability of being in a spherical shell of thickness 0.010aB at distance (a) ½ aB, (b) aB, and (c) 2aB from the proton?

Accepted Answer

Understand the problem: The probability of finding an electron in a spherical shell is determined by the radial probability density function, which is derived from the square of the radial wavefunction multiplied by the volume element. For the hydrogen atom in the 1s state, the radial wavefunction is given by $$ R_{1s}(r) = \frac{2}{a_B^{3/2}} e^{-r/a_B} $$, where $$ a_B  is $$the Bohr radius. Write the expression for the radial probability density: The radial probability density is $$ P(r) = |R_{1s}(r)|^2 \cdot 4\pi r^2 $$, where $$ |R_{1s}(r)|^2  is $$the square of the radial wavefunction. Substituting $$ R_{1s}(r) $$, we get $$ P(r) = \frac{4}{a_B^3} e^{-2r/a_B} \cdot 4\pi r^2 . $$Set up the probability for a spherical shell: The probability of finding the electron in a thin spherical shell of thickness $$ dr  at a $$distance $$ r  is$$ $$ dP = P(r) \cdot dr . $$Substituting $$ P(r) $$, we have $$ dP = \frac{16\pi}{a_B^3} r^2 e^{-2r/a_B} dr . $$Evaluate the probability for each case: For each distance $$ r $$ ($$ r = \frac{1}{2}a_B, a_B, 2a_B $$), substitute the value of $$ r $$ into the expression $$ dP = \frac{16\pi}{a_B^3} r^2 e^{-2r/a_B} dr $$ and use $$ dr = 0.010a_B  to $$calculate the probability for the corresponding spherical shell. Perform the calculations: For each case, simplify the expression by substituting $$ r $$ and $$ dr $$ into $$ dP . $$This will yield the probability for the electron to be in the specified spherical shell at the given distance. Note that the final numerical evaluation can be done separately if needed.

For an electron in the 1s state of hydrogen, what is the probability of being in a spherical shell of thickness 0.010a_B at distance (a) ½ a_B, (b) a_B, and (c) 2a_B from the proton?

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Key Concepts

Quantum Mechanics

Hydrogen Atom Model

Probability Density