Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1angstrom , and (ii) R = 10 angstrom

If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1Å, and (ii) R = 10 Å.

Answers (1)

The electron present in H-atom revolves around the point size proton of a defined radius rB (Bohr’s radius) in a circular path when the atom is in the ground state.

As\;mvr_{B}=h\; and\; \frac{Mv^{2}}{r_{B}}=\frac{-1 \times e \times e}{4 \pi \varepsilon_{0}r_{B}^{2} }\\ \frac{m}{r_{B}}\left ( \frac{-h^{2}}{m^{2}r_{B}^{2}} \right )=\frac{e^{2}}{4 \pi \epsilon_{0}r_{B}^{2}}\\ r_{B}=\frac{4 \pi \epsilon_{0}}{e^{2}}\frac{h^{2}}{m}=0.53\AA\\ K.E=\frac{1}{2}mv^{2}=\left (\frac{m}{2} \right )\left (\frac{h}{mr_{B}} \right )^{2}\\ =\frac{h^{2}}{2mr_{B}^{2}}=13.6eV

total energy of the electron i.e

E=K+U=+13.6eV-27.2eV=-13.6eV

(i)when r=0.1Å:R<rB (as rB =0.51 Å )and the ground state energy is the same as obtained earlier for point size proton i.e -13.6 eV

(ii)when r=10Å:R>>rB  the electrons move inside the proton (assumed to be a sphere of radius R ) with new Bohr's radius r'B

clearly

r'_{B}=\frac{4 \pi \epsilon_{0}}{m(e)(e')}
[Replacing e2 by  (e)(e') where e' is the charge on the sphere of radius r'B  ]

Since 

e'=\left [\frac{e}{(4 \pi/3)R^{3}} \right ]\left [ \left ( \frac{4\pi}{3} \right ) {r_{B}^{3}}'\right ]=\frac{{r_{B}^{3}}'}{R^{3}}\\\\ {r_{B}^{3}}'=\frac{4 \pi \epsilon_{0}h^{2}}{m_{e}\left ( e{r_{B}^{3}}'R^{3} \right )}=\left (\frac{4 \pi \epsilon_{0}h^{2}}{m_{e}^{2}} \right )\left ( \frac{R^{3}}{{r_{B}^{3}}'} \right )\\\\ or\;\;{r_{B}^{3}}'= \left (\frac{4 \pi \epsilon_{0}h^{2} }{me^{2}} \right )R^{3}\\\\ =(0.51\AA)(10\AA)=510\AA

r'B =4.8Å which is less  than R(=10Å)

KE of the electron

K'=\frac{1}{2}mv^{2}=\left ( \frac{m}{2} \right )\left ( \frac{h^{2}}{m^{2}{r_{B}^{2}}'} \right )=\frac{h}{m{r_{B}^{2}}'}\\ =\left (\frac{h^{2}}{m{r_{B}^{2}}'} \right )\left (\frac{ r_{B}}{{r_{B}}'} \right )=13.6eV\left ( \frac{0.51\AA}{4.8\AA} \right )=0.16eV

The potential at a point inside the change proton,

i.e

V=\frac{k_{e}}{R}\left (3-\frac{{r_{B}^{2}}'}{R^{2}} \right )=k_{e}e\left ( \frac{3R^{2}-{r_{B}^{2}}'}{R^{3}} \right )

The potential energy of electron and proton u=-eV

=-e(k_{e}e)\left [ \frac{3R^{2}-{r_{B}^{2}}'}{R^{3}} \right ]\\ =-\left ( \frac{e^{2}}{4 \pi \varepsilon _{0}r_{B}} \right )\left [ \frac{r_{B} \left ( 3R^{2}-{r_{B}^{2}}' \right ) }{R^{3}} \right ]\\ =-\left ( 24.2eV \right )\left [ \frac{(0.51\AA)(300\AA-23.03\AA)}{1000\AA} \right ]\\\\ =3.83eV

The total energy of the electron

E=K+U=0.16eV-3.83eV=-3.67eV

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question