A particular hydrogen like ion emits radiation of frequency $2.92 \times 10^{15}$ Hz when it makes transition from n=3 to n=1. The frequency in Hz of radiation emitted in transition from n=2 to n=1 will be :

Question

The nuclides of $^{197}_{79}\mathrm{Au}$ and $^{196}_{80}\mathrm{Hg}$ are called ______ of each other

Answer 1

To determine the frequency of the radiation emitted when the electron transitions from $ n=2 $ to $ n=1 $ for a hydrogen-like ion, we can use the Rydberg formula for the frequency of emitted radiation:

v = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

where:

$ v $ is the frequency of radiation emitted.
$ R $ is the Rydberg constant ($ R = 3.29 \times 10^{15} $ Hz for hydrogen atom).
$ Z $ is the atomic number of the atom/ion.
$ n_1 $ and $ n_2 $ are the principal quantum numbers of the initial and final states.

Given that the transition from $ n=3 $ to $ n=1 $ has a frequency of $ 2.92 \times 10^{15} $ Hz, we can write this equation as:

2.92 \times 10^{15} = RZ^2\left(\frac{1}{1^2} - \frac{1}{3^2}\right)

Simplifying:

2.92 \times 10^{15} = RZ^2\left(1 - \frac{1}{9}\right) = RZ^2\left(\frac{8}{9}\right)

We need to find the frequency for the transition from $ n=2 $ to $ n=1 $:

v = RZ^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = RZ^2\left(1 - \frac{1}{4}\right) = RZ^2\left(\frac{3}{4}\right)

Dividing the two expressions:

\frac{v}{2.92 \times 10^{15}} = \frac{\left(\frac{3}{4}\right)}{\left(\frac{8}{9}\right)}

This simplifies to:

v = 2.92 \times 10^{15} \times \frac{3}{4} \times \frac{9}{8}

Simplifying the calculation:

v = 2.92 \times 10^{15} \times \frac{27}{32}

Calculating the value:

v = \frac{78.84}{32} \times 10^{15} = 2.46 \times 10^{15} \text{ Hz}

Therefore, the frequency of the radiation emitted when transitioning from $ n=2 $ to $ n=1 $ is $ 2.46 \times 10^{15} $ Hz. The correct answer is $2.46 \times 10^{15}$.

Answer 2