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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/RH. - Physics

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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/RH.  

Answer in Brief
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Solution

Expression for wavenumber: 

  1. Let, Em = Energy of an electron in mth higher orbit
    En = Energy of an electron in an nth lower orbit
  2. According to Bohr’s third postulate,
    Em − En = hν
    ∴ ν = `("E"_"m" - "E"_"n")/"h"` ….(1)
  3. But Em = `-("Z"^2"m"_"e""e"^4)/(8epsilon_0^2"h"^2"m"^2)` ….(2)
    En = `- ("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"`
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Bohr’s Atomic Model
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Chapter 15: Structure of Atoms and Nuclei - Long Answer

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SCERT Maharashtra Physics [English] 12 Standard HSC
Chapter 15 Structure of Atoms and Nuclei
Long Answer | Q 2

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