Name: Bohr Atom Model: Frequency of Emitted Radiation
Uploaded: 2019-07-19
Duration: 3 min 54 s
Description: Electrons jump in different energy levels. Let's understand how we can calculate the frequency of emitted and absorbed radiation as electrons transition into...

Question

State Bohr

'

s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong.

(Planck' s constant = 6.6 \times 10^{- 34} J s, Rydberg' s constant = 1.097 \times 10^{7} m^{- 1})

.

Accepted Answer

According to Bohr, an electron in an atom could revolve in certain stable orbits without the emission of radiant energy and the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\frac{h}{2 π}$ where $h$ is Planck's constant $(= 6.6 \times 10^{- 34} J s)$ .

The angular momentum $(L)$ is given by

$L = mvr$ , where $m$ is mass of electron, $v$ is velocity in circular orbit, and $R$ is radius of circular orbit. Therefore,

$L = mvr = \frac{nh}{2 π}$

where, $n = 1, 2, 3, . . .$ is called the principal quantum number and the above condition is known as the Bohr $'$ s quantization condition of angular momentum.

Whenever an electron makes transition from one stationary orbit to another stationary orbit, energy is radiated in form of spectral line.

The wavelength of a spectral line is given as

$\frac{1}{λ} = RZ (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

where $R$ is Rydberg $'$ s constant and $Z$ is atomic number. The value of $R$ is $1.097 \times 10^{7} m^{- 1}$ .

For Bracket series, $(Planck' s constant = 6.6 \times 10^{- 34} J s, Rydberg' s constant = 1.097 \times 10^{7} m^{- 1})$ and $n_{2} = 5, 6, 7, . . .$

$Z = 1$ for hydrogen atom and for the shortest wavelength of the Bracket series $n_{2} = \infty$ .

Substituting all the values in the above equation, we get,

$\frac{1}{λ} = 1.097 \times 10^{7} (\frac{1}{4^{2}} - \frac{1}{\infty^{2}})$

$\Rightarrow \frac{1}{λ} = \frac{1.097 \times 10^{7}}{16} = 0.0681 \times 10^{7} m^{- 1}$

or, $λ = 14.684 \times 10^{- 7} m$

or, $λ = 1468.4 nm$ .

Hence, the shortest wavelength of the Bracket series is $1468.4 nm$ .

This wavelength belongs to infrared region of electromagnetic spectrum.

State Bohr $'$ s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. $(Planck' s constant = 6.6 \times 10^{- 34} J s, Rydberg' s constant = 1.097 \times 10^{7} m^{- 1})$ .

Important Questions on Atoms

Learn and Prepare for any exam you want !

State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. (Planck's constant=6.6×10-34 J s, Rydberg's constant=1.097×107 m-1).

Important Questions on Atoms

Learn and Prepare for any exam you want !

State Bohr $'$ s quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum does it belong. $(Planck' s constant = 6.6 \times 10^{- 34} J s, Rydberg' s constant = 1.097 \times 10^{7} m^{- 1})$ .