Question

Obtain an expression for wavenumber, when an electron jumps from a higher energy orbit to a lower energy orbit. Hence show that the shortest wavelength for the Balmar series is 4/R_H.  

Answer 1

Expression for wavenumber: 

1. Let, E_m = Energy of an electron in m^th higher orbit
   E_n = Energy of an electron in an n^th lower orbit
2. According to Bohr’s third postulate,
   E_m − E_n = hν
   ∴ ν = `("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 × 10⁷ m⁻¹ 
   ∴ `1/lambda = "R"_"H""Z"^2 [1/"n"^2 - 1/"m"^2]` ….(4)
   This is the required expression.
6. For shortest wavelength for Balmer series:
   n = 2 and m = ∞
   `1/lambda = "R"_"H"[1/2^2 - 1/∞]`
   = `"R"_"H"/4`
   ∴ `lambda = 4/"R"_"H"`