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  • The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge plus q1, mius q2 is modified

The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge +q1, –q2 is modified to

\left | F \right |=\frac{q_{1}q_{2}}{\left (4\pi \epsilon_{0} \right )}\frac{1}{r^{2}},r\geq R_{0}\\ =\frac{q_{1}q_{2}}{\left (4\pi \epsilon_{0} \right )}\frac{1}{R_{0}^{2}} \left ( \frac{R_{0}}{r} \right ),r\geq R_{0}\\

Calculate in such a case, the ground state energy of an H-atom, if \epsilon = 0.1, R_{0} = 1\AA

Answers (1)

Let the case consider the case when r\leq R_{0}=1 \AA

Let \epsilon =2+\delta

F=\frac{q_{1}q_{2}}{4\pi \epsilon_{0} }.\frac{R_{0}^{\delta}}{r^{2+\delta}}= \frac{xR_{0}^{\delta}}{r^{2+\delta}}

where F=\frac{q_{1}q_{2}}{4\pi \epsilon_{0} }=x=\left ( 1.6 \times 10^{-19} \right )^{2}\times 9 \times 10^{9}=2.04 \times 10^{-29}Nm^{2}

The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons provides the necessary centripetal force.

\frac{mv^{2}}{r}=\frac{xR_{0}^{\delta}}{r^{2+\delta}}=\; or\; v^{2}=\frac{xR_{0}^{\delta}}{mr^{1+\delta}}...........(i)\\ mvr=nh \Rightarrow r=\frac{nh}{mv}=\frac{nh}{m}\left [ \frac{m}{xR_{0}^{\delta}} \right ]^{1/2}r^{(1+\delta)/2}\\ (a\\lying \; Bohr \; second \;postulates)\\ Solving\;this\;fir\;r,we\;ger\; r_{n}=\left ( \frac{n^{2}h^{2}}{xR_{0}^{\delta}} \right )^{\frac{1}{1-\delta}}\\

Where rn is the radius of nth orbit of electron for n=1 and substituting the values of constant we get

r_{1}=\left [ \frac{h^{2}}{mxR_{0}^{2}} \right ]^{\frac{1}{1-\delta}}\\ \Rightarrow r_{1}=\left [ \frac{1.05^{2} \times 10^{-68}}{9.1 \times 10^{-31} \times 2.3 times 10^{-28}times 10^{+19}} \right ]^{\frac{1}{29}}\\ =8times 10^{-11}=0.08 nm(<0.1 nm)

This is the radius of orbit of electron in ground state of hydrogen atom again using Bohr second postulate the speed of the electron

v_{n}=\frac{nh}{mr_{n}}=nh\left ( \frac{mxR_{0}^{\delta}}{n^{2}h^{2}} \right )^{\frac{1}{1-\delta}}

For n=1 the speed of electron in ground state v_{1}=\frac{h}{mr_{1}}=1.44 \times 10^{6}m/s

The kinetic energy of electron in the ground state

KE=\frac{1}{2}mv_{1}^{2} \times -^{-19}J= eV

Potential energy of electron in the ground state till R0

U=\int_{0}^{R_{0}}F dr =\int_{0}^{R_{0}}\frac{x}{r^{2}}dr=\frac{x}{R_{0}}

Potential energy from R0 to r,  U=\int_{0}^{R_{0}}F dr =\int_{0}^{R_{0}}\frac{xR_{0}^{\delta}}{r^{2+\delta}}dr \\ U=+xR_{0}^{\delta}\int _{R_{0}}^{r} \frac{dr}{r^{2+\delta}}=+\frac{xR_{0}^{\delta}}{-1-\delta}\left [ \frac{1}{r^{1+\delta}} \right ]_{R_{0}}^{r}\\ U=\frac{xR_{0}^{\delta}}{1+\delta}\left [ \frac{1}{r^{1+\delta}}-\frac{1}{R_{0}^{1+\delta}} \right ]=\frac{x}{1+\delta}\left [ \frac{R_{0}^{\delta}}{r^{1+\delta}}-\frac{1}{R_{0}} \right ]\\ U=-x\left [ \frac{R_{0}^{\delta}}{r^{1+\delta}}-\frac{1}{R_{0}} +\frac{1+\delta}{R_{0}}\right ]\\ u=-x\left [ \frac{R_{0}^{-19}}{r^{-0.9}}-\frac{1.9}{R_{0}} \right ]\\ =\frac{2.3}{0.9}\times^{-18}\left [ (0.8)^{0.9}-1.9 \right ]J=-17.3eV\\ Hence\;total \; energy\; of\; electron\; in\; ground\; state=\left ( -17.3+5.9 \right )=-11.4eV

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