The wavelength of the third line of the Balmer series for a hydrogen atom is

Question

The wavelength of the third line of the Balmer series for a hydrogen atom is (A) (21/100 RH) (B) (100/21 RH) (C) (21 RH/100 R) (D) (100 textR textH/21)

Tardigrade Admin · Accepted Answer

(1/λ )=RH((1/n12) - (1/n22)) for the H-atom For the Balmer series, n1=2 and n2=5 (for the third line). (1/λ )=RH((1/22) - (1/52))=(21/100)RH So λ text= ( text100/ text21RH)

Q. The wavelength of the third line of the Balmer series for a hydrogen atom is

$\frac{21}{100 R _{H}}$

$\frac{100}{21 R _{H}}$

$\frac{21 R _{H}}{100 R}$

$\frac{100 R _{H}}{21}$

$\frac{1}{λ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ for the H-atom

For the Balmer series,

$n_{1} = 2$ and $n_{2} = 5$ (for the third line).

$\frac{1}{λ} = R_{H} (\frac{1}{2 ^{2}} - \frac{1}{5 ^{2}}) = \frac{21}{100} R_{H}$

So $λ = \frac{100}{21R _{H}}$

Q. The wavelength of the third line of the Balmer series for a hydrogen atom is

100RH​21​

21RH​100​

100R21RH​​

21100RH​​

λ1​=RH​(n12​1​−n22​1​) for the H-atom For the Balmer series, n1​=2 and n2​=5 (for the third line). λ1​=RH​(221​−521​)=10021​RH​ So λ=21RH​100​

$\frac{21}{100 R _{H}}$

$\frac{100}{21 R _{H}}$

$\frac{21 R _{H}}{100 R}$

$\frac{100 R _{H}}{21}$

$\frac{1}{λ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ for the H-atom

For the Balmer series,

$n_{1} = 2$ and $n_{2} = 5$ (for the third line).

$\frac{1}{λ} = R_{H} (\frac{1}{2 ^{2}} - \frac{1}{5 ^{2}}) = \frac{21}{100} R_{H}$

So $λ = \frac{100}{21R _{H}}$