Question

In hydrogen emission spectrum, for any series, the principal quantum number is n. Corresponding maximum wavelength λ is ______.
(R = Rydberg's constant)

Options

• `("R"(2"n" + 1))/("n"^2("n + 1"))`

• `("n"^2("n + 1")^2)/("R"(2"n" + 1))`

• `("n"^2("n + 1"))/("R"(2"n" + 1))`

• `("R"(2"n" + 1))/("n"^2("n + 1")^2)`

Answer 1

In hydrogen emission spectrum, for any series, the principal quantum number is n. Corresponding maximum wavelength λ is `underline(("n"^2("n + 1")^2)/("R"(2"n" + 1)))`.

Explanation:

In hydrogen emission spectrum, wavelength is given by

`1/lambda = "R"(1/"n"_1^2 - 1/"n"_2^2)`

For maximum wavelength of any principal quantum number n,

n₁ = n and n₂ = n + 1

`therefore 1/(lambda_"max") = "R"[1/"n"^2 - 1/("n + 1")^2]`

`= "R"[(("n + 1")^2 - "n"^2)/("n"^2("n + 1")^2)]`

`= "R"(("n"^2 + 2"n" + 1 - "n"^2))/("n"^2("n + 1")^2)`

`1/lambda_"max" = ("R"^2 (2"n" + 1))/("n"^2("n + 1")^2)`

`therefore lambda_"max" = ("n"^2("n + 1")^2)/("R"^2 (2"n" + 1))`