First slide

Hydrogen spectrum

Question

An electron in a Bohr orbit of hydrogen atom with the quantum number n₂ has an angular momentum $4.2176 \times 10^{- 34} {kgm}^{2} s^{- 1}$ . If the electron drops from this level to the next lower level, the wavelength of this line is

Moderate

A

18 nm
B

187 .6 pm
C

1876 $Å$
D

1.876 x 10⁴ $Å$

Solution

$Angular momentum, L = 4.2176 \times 10^{- 34} = \frac{n_{2} h}{2 π}$
$\Rightarrow n_{2} = 4$
For the transition from n₂= 4 to n₁ = 3, the wavelength of spectral line $λ$ .
$\begin{array}{rcl} \frac{1}{λ} & = & \frac{13.6}{h c} (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = \frac{13.6 eV}{1240 eVnm} (\frac{7}{9 \times 16}) \\ λ & = & \frac{1240 \times 144}{13.6 \times 7} = 1876 nm = 18760 Å \\ = & 1.876 \times 10^{4} Å \end{array}$

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