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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.
Solution
Expression for wavenumber:
- Let, Em = Energy of an electron in mth higher orbit
En = Energy of an electron in an nth lower orbit - According to Bohr’s third postulate,
Em − En = hν
∴ ν = `("E"_"m" - "E"_"n")/"h"` ….(1) - 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) - 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]` - 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. - 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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