If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is

Question

If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is (A) (6/5R) (B) (36/5R) (C) (64/7R) (D) None of these

Tardigrade Admin · Accepted Answer

As we know (1/λ)= R ((1/22) - (1/32) ) = R ((1/4) - (1/9)) (1/λ) = R ((9-4/36)) = (5R/36) ∴ λ = (36/5R )

Q. If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is

$\frac{6}{5 R}$

$\frac{36}{5 R}$

$\frac{64}{7 R}$

None of these

As we know $\frac{1}{λ} = R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}})$
$= R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ} = R (\frac{9 - 4}{36}) = \frac{5 R}{36}$
$∴ λ = \frac{36}{5 R}$

Q. If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is

5R6​

5R36​

7R64​

None of these

As we know λ1​=R(221​−321​) =R(41​−91​) λ1​=R(369−4​)=365R​ ∴λ=5R36​

$\frac{6}{5 R}$

$\frac{36}{5 R}$

$\frac{64}{7 R}$

As we know $\frac{1}{λ} = R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}})$
$= R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ} = R (\frac{9 - 4}{36}) = \frac{5 R}{36}$
$∴ λ = \frac{36}{5 R}$