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Question
Obtain an expression for the radius of Bohr orbit for H-atom.
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Solution
Let us consider an electron revolving around the nucleus in a circular orbit of radius ‘r’.
According to Bohr’s first postulate, the centripetal force is equal to the electrostatic force of attraction. That is
`"mv"^2/"r"=1/(4piepsilon_o)xx"e"^2/"r"^2`
`"Or,""v"^2="e"^2/(4piepsilon_o"mr")` -------------------(1)
According to the Bohr's second postulate:
`"Angular momentum"= "n""h"/(2pi)`
`"mvr"="n""h"/(2pi)`
Or, `"v"="nh"/(2pi"mr")` -----------------(2)
Or, `"v"^2=("n"^2"h"^2)/(4pi^2"m"^2"r"^2)` ---------------------(3)
Comparing eqn (1) and eqn (3), we get
`"e"^2/(4piepsilon_o"mr")=("n"^2"h"^2)/(4pi^2"m"^2"r"^2)`
`"Or,""r"=(("h"^2epsilon_o)/(pi"me"^2))"n"^2` ----------------------(4)
This equation gives the radius of the nth Bohr orbit.
`"For n"=1,"r"_1=(("h"^2epsilon_o)/(pi"me"^2))=0.537" ---------------(5)"`
`"In general,"" r"_n=(("h"^2epsilon_o)/(pi"me"^2))"n"^2`
The above equation gives the radius of Bohr orbit.
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