Question

The Balmer series in hydrogen atom produces only infrared emission. O True O False

Answer 1

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

$n_{i}=2$.

$[tex]n_{f}=3,4,5,.....\infty$.

Now the value of wavelength can be calculated as,

$\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}$.

Here, $R=109677 cm^{-1}$.

And $n_{f}=3$.

Now,

$\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3})$.

Therefore,

$\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm$

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

Answer 2

Answer:

False

Explanation:

The Balmer series produce visible light in the hydrogen atom spectrum.

Balmer series corresponds to a transition from higher energy level to an energy level corresponding to principal quantum number, n = 2.

This transition results in the hydrogen atom spectra of 4 different wavelengths of the visible light which includes 434 nm, 410 nm, 486 nm and 656 nm.

Its not Balmer series that produces all wavelengths of light of infrared emission but its Paschen series that produces all wavelengths in the infra red region of the hydrogen atom spectrum