Question

Calculate the frequency and the wavelength of the radiation in nanometers emitted when an electron in the hydrogen atom jumps from third orbit to the ground state. In which region of the electromagnetic will this line lie? (Rydberg constant

= 109,677 c m^{- 1}

)

Accepted Answer

Calculating wavelength:
We know that,
$\frac{1}{λ} = R_{H} (\frac{1}{n_{f}} - \frac{1}{n_{i}})$
where, $n_{f}$ is the final shell and $n_{i}$ is the initial shell
Therefore,
$\frac{1}{λ} = R_{H} (\frac{1}{n_{f}} - \frac{1}{n_{i}})$

Here, $n_{f} = 1, n_{i} = 3$ and $R = 1.097 \times 10^{7} m^{- 1}$

$\frac{1}{λ} = 1.1 \times 10^{7} (\frac{1}{1^{2}} - \frac{1}{3^{2}})$
$\frac{1}{λ} = 1.1 \times 10^{7} (1 - \frac{1}{9})$
$\frac{1}{λ} = 1.1 \times 10^{7} (\frac{8}{9})$
$λ = \frac{9}{(8 \times 1.1 \times 10^{7})}$
$λ = 1.02 \times 10^{- 7} m$
Or, $λ = 1020 \overset{\circ}{A} = 102 nm (1 \overset{\circ}{A} = 0.1 nm)$

The calculated wavelength lies in the ultraviolet region.

Calculating frequency:
As we know that,
$v = \frac{c}{λ}$
Frequency $= \frac{(3 \times 10^{8} m / s)}{(1.02 \times 10^{- 7} m)}$
$= 2.94 \times 10^{15} Hz$

S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Structure of Atom, Exercise 4: PROBLEMS FOR PRACTICE

S C Kheterpal Chemistry Solutions for Exercise - S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Structure of Atom, Exercise 4: PROBLEMS FOR PRACTICE

Questions from S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Structure of Atom, Exercise 4: PROBLEMS FOR PRACTICE with Hints & Solutions

Learn and Prepare for any exam you want !