Question

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) $\frac{1}{2}{a}_B$, (b) $a_B$, and (c) $2a_B$from the proton?

Short answer

(a) 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 $\frac{1}{2}{a}_B$is $3.7\times {10}^{-3}$.

(b) 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 $a_B$is $5.4\times {10}^{-3}$.

(c) 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 $2a_B$is $2.9\times {10}^{-3}$.

Step-by-step solution

Part (a) Step 1 : Given Information

We have given an electron in the $1s$state of hydrogen,

We have to find the probability of being in a spherical shell of thickness $0.010a_B$at distance of$\frac{1}{2}{a}_B$.

Part (a) Step 2 : Simplification

We know that , the radial wave function of hydrogen in the $1s$state is

$R_{1s}\left(r\right)=\frac{1}{\sqrt{\pi {a^3}_B}}{e}^{-r/a_B}$

and the probability density is

$P_r\left(r\right)=4\pi r^2{\left|R_{nl}\left(r\right)\right|}^2$

The radial wave function and probability density for $r=\frac{1}{2}{a}_B$is

role="math" localid="1650362505827" $\Rightarrow R_{1s}\left(\frac{1}{2}{a}_B\right)=\frac{1}{\sqrt{\pi {a^3}_B}}{e}^{-\frac{1}{2}}=\frac{0.607}{\sqrt{\pi {a^3}_B}}$

$\Rightarrow P_r\left(\frac{1}{2}{a}_B\right)=4\pi {\left(\frac{a_B}{2}\right)}^2{\left|\frac{0.607}{\sqrt{\pi {a_B}^3}}\right|}^2=\frac{0.368}{a_B}$

The probability is

Prob(in $\delta r$at $r$)$=P_r\left(r\right)\delta r=P_r\left(\frac{1}{2}{a}_B\right)\left(0.010a_B\right)=\left(\frac{0.368}{a_B}\right)\left(0.010a_B\right)=3.7\times {10}^{-3}$

The probability of being in a spherical shell of thickness $0.010a_B$at distance of $\frac{1}{2}{a}_B$is$3.7\times {10}^{-3}$.

Part (b) Step 1 : Given Information

We have given an electron in the $1s$state of hydrogen,

We have to find the probability of being in a spherical shell of thickness $0.010a_B$at distance of$a_B$

Part (b) Step 2 : Simplification

Likewise,

The radial wave function for$r=a_B$is

$\Rightarrow R_{1s}\left(a_B\right)=\frac{1}{\sqrt{\pi {a^3}_B}}{e}^{-1}=\frac{0.368}{\sqrt{\pi {a^3}_B}}$

The probability density for $r=a_B$is

$\Rightarrow P_r\left(a_B\right)=4\pi {a_B}^2{\left|\frac{0.368}{\sqrt{\pi {a_B}^3}}\right|}^2=\frac{0.541}{a_B}$

The prob(in $\delta r$at r)$=P_r\left(r\right)\delta r=P_r\left(a_B\right)\left(0.010a_B\right)=5.4\times {10}^{-3}.$

The probability of being in a spherical shell of thickness $0.010a_B$at distance of $a_B$is $5.4\times {10}^{-3}.$

Part (c) Step 1 : Given Information

We have given an electron in the $1s$state of hydrogen,

We have to find the probability of being in a spherical shell of thickness $0.010a_B$at distance of$2a_B$

Part (c) Step 2 : Simplification

For $r=2a_B$,

The radial wave function is

$\Rightarrow R_{1s}\left(2a_B\right)=\frac{1}{\sqrt{\pi {a^3}_B}}{e}^{-2}=\frac{0.135}{\sqrt{\pi {a^3}_B}}$

The probability density is

$\Rightarrow P_r\left(2a_B\right)=4\pi {\left(2a_B\right)}^2{\left|\frac{0.135}{\sqrt{\pi {a_B}^3}}\right|}^2=\frac{0.293}{a_B}$

The prob(in$\delta r$at $r$)$=P_r\left(r\right)\delta r=P_r\left(2a_B\right)\left(0.010a_B\right)=2.9\times {10}^{-3}.$

The probability of being in a spherical shell of thickness $0.010a_B$at distance of $2a_B$is$2.9\times {10}^{-3}.$