91Ó°ÊÓ

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$\frac{1}{2}{a}_B$.

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

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

and the probability density is

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

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

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

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

The probability is

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

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

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$a_B$

Likewise,

The radial wave function for$r=a_B$is

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

The probability density for $r=a_B$is

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

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

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

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$2a_B$

For $r=2a_B$,

The radial wave function is

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

The probability density is

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

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