Question

Using the expression for the energy of an electron in the n^th orbit, Show that `1/lambda = "R"_"H"(1/"n"^2 - 1/"m"^2),` Where symbols have their usual meaning. 

Answer 1

1. Let, E_m = Energy of an electron in m^th higher orbit
   E_n = Energy of electron in the n^th lower orbit 
2. According to Bohr’s third postulate,
   E_m - E_n = hv
   ∴ v = `("E"_"m" - "E"_"n")/"h"` ….(1)
3. But E_m = `-("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^2"m"^2)` ….(2)
   E_n = `-("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^2"n"^2)` ….(3) 
4. From equations (1), (2) and (3),
   v = `(-("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^2"m"^2) - (-("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^2"n"^2)))/"h"`
   ∴ v = `("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^3")` `[-1/"m"^2 + 1/"n"^2]`
   ∴ `"c"/lambda = ("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^3")` `[1/"n"^2 - 1/"m"^2]` ......[∵ v = `"c"/lambda`]
   where, c = speed of electromagnetic radiation
   ∴ `1/lambda =  ("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^3"c")` `[1/"n"^2 - 1/"m"^2]`
5. But, `("m"_"e""e"^4)/(8epsilon_0^2"h"^3"c") = "R"_"H"` = Rydberg’s constant
   = `1.097 × 10^7 "m"^-1`
   ∴ `1/lambda = "R"_"H""Z"^2 [1/"n"^2 - 1/"m"^2]` ….(4)
   For hydrogen Z = 1,
   ∴ `1/lambda = "R"_"H"[1/"n"^2 - 1/"m"^2]` ….(5)