Question

calculate the wavelength of the third line in the brackett series of hydrogen.(R=10.97×10^6/m). Give fast answer and explain it. Class 11.​

Answer 1

$\huge{\mathcal{\underline{\pink{Solution:-}}}}$

Rydberg's constant (R) = 10.97 × 10⁶ m⁻¹

Lower energy level of Brackett series (n₁) = 4

Higher energy level of Brackett series (n₂) = 7

Wavelength(λ) = ?

$\large\rm\bold{\underline{\boxed{\frac{1}{\lambda}\:=\:R_H[\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}]}}}$

$\large\rm{\rightarrow{\frac{1}{\lambda}\:=\:10.97\times\:10^6(\frac{1}{16}-\frac{1}{49})}}$

$\large\rm{\rightarrow{\frac{1}{\lambda}\:=\:\frac{33(10.97\times\:10^6\:m^-1)}{784}}}$

$\large\rm{\rightarrow{\frac{1}{\lambda}\:=\:0.461\times\:10^6\:m}}$

$\large\rm{\rightarrow{\lambda\:=\:\frac{1}{0.461\times\:10^6\:m}}}$

$\large\rm{\rightarrow{\lambda\:=\:2.169\times\:10^-6\:m}}$

$\large\rm{\rightarrow{\lambda\:=\:2169\times\:10^-9\:m}}$

$\large\rm\bold{\underline{\boxed{1\:nano\:metre\:=\:10^-9m}}}$

$\large\rm{\rightarrow{\lambda\:=\:2169\:nm}}$

∴ The wavelength associated is 2169 nm

Answer 2

Answer:

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