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  • In the Bohr's model of hydrogen - like atom the force between the nucleus and the electrons is modified as, <img alt="\large F= \frac{e^{2}}{4\pi \epsilon _{0}}\left ( \frac{1}{r^{2}}

In the Bohr's model of hydrogen - like atom the force between the nucleus and the electrons is modified as,

\large F= \frac{e^{2}}{4\pi \epsilon _{0}}\left ( \frac{1}{r^{2}}+\frac{\beta }{r^{3}} \right )

Where \large \beta    is a constant . For the atom, the radius of the nth orbit in the terms of the Bohr radius

\large \left ( a_{0}=\frac{\epsilon _{0}h^{2}}{m\pi e^{2}} \right ) 

 

Option: 1

r_{n}= a_{0}\: \: n-\beta
 


Option: 2

r_{n}= a_{0}\: \: n^{2}+\beta


Option: 3

r_{n}= a_{0}\: \: n^{2}-\beta

 


Option: 4

r_{n}= a_{0}\: \: n+\beta


Answers (1)

F=\frac{mv^2}{r}=\frac{e^{2}}{4 \pi \varepsilon_{0}}\left(\frac{1}{r^{3}}+\frac{\beta}{r^{3}}\right)

\therefore v^{2}=\frac{e^{2}}{4 \pi \varepsilon_{0} m }\left(\frac{1}{r}+\frac{\beta}{r^{2}}\right)

From Bohr's postulate

mvr=\frac{nh}{2\pi}

\therefore v=\frac{nh}{2\pi mr}

comparing both we get

\frac{n^{2} h^{2}}{4 \pi^{2} m^{2} r^{2}}=\frac{e^{2}}{4 \pi \varepsilon_{0} m }\left(\frac{1}{r}+\frac{\beta}{r^{2}}\right)

So,

r_{n}= a_{0} n^{2}-\beta

 

Posted by

Ramraj Saini

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