Question

a spectral line in a balmer series line of hydrogen corresponds to
6561 A(wavelength) find the energy levels involved in transition responsible for the origin of this line(R=109677 1/cm)

Answer 1

Answer:

Transition of n 3→2 6→2

Name H-α / Ba-α H-δ / Ba-δ

Wavelength (nm, air) 656.279 410.1734

Energy difference (eV) 1.89 3.03

Color Red Violet

Answer 2

Given:

A spectral line in a balmer series line of hydrogen corresponds to 6561 A(wavelength)

To find:

Energy levels involved.

Calculation:

Let the higher orbit be n_(2).

In Balmer series, electron will get transferred from n_(2) to 2nd orbit.

Applying Rydberg's Formula:

$\therefore \: \dfrac{1}{ \lambda} = R \bigg \{ \dfrac{1}{ {(2)}^{2} } - \dfrac{1}{ { (n_{2})}^{2} } \bigg \}$

$= > \: \dfrac{1}{ 6561 \times {10}^{ - 10} } = 1.09 \times {10}^{7} \bigg \{ \dfrac{1}{4} - \dfrac{1}{ { (n_{2})}^{2} } \bigg \}$

$= > \: \dfrac{1}{ 6.561 \times {10}^{ - 7} } = 1.09 \times {10}^{7} \bigg \{ \dfrac{1}{4} - \dfrac{1}{ { (n_{2})}^{2} } \bigg \}$

$= > \: 0.152 \times {10}^{7} = 1.09 \times {10}^{7} \bigg \{ \dfrac{1}{4} - \dfrac{1}{ { (n_{2})}^{2} } \bigg \}$

$= > \: 0.152 = 1.09 \times \bigg \{ \dfrac{1}{4} - \dfrac{1}{ { (n_{2})}^{2} } \bigg \}$

$= > \: 0.138 = \bigg \{ \dfrac{1}{4} - \dfrac{1}{ { (n_{2})}^{2} } \bigg \}$

$= > \: \dfrac{1}{ { (n_{2})}^{2}} = 0.25 - 0.138$

$= > \: \dfrac{1}{ { (n_{2})}^{2}} = 0.112$

$= > \: { (n_{2})}^{2} = \dfrac{1}{ 0.112}$

$= > \: { (n_{2})}^{2} =8.96$

$= > \: { (n_{2})}^{2} \approx \:9$

$= > \: (n_{2}) \approx \: \sqrt{9}$

$= > \: (n_{2}) \approx \: 3$

So, electron originated from n = 3 .