Question

The wavenumber for the first Balmer line for Li2+ is 136,800 cm-1
, what is the
wavenumber of the first line of the Balmer series of hydrogen atom? What will be the
energy change and frequency of the transition?

Answer 1

Answer:

the answer of this question is 888,31

Answer 2

Answer:

Atomic Structure

Explanation:

• First balmer line for Li2+ wave number :

For finding the wave number we will use Rydberg formula:

Let, Wave number is,

1 =1/λ = $R_{H} = ( \frac{1}{n_{1} ^{2} } - \frac{1}{n_{2}^{2} } )$

The first line of Balmer series is $n_{1}$ = 2 , $n_{2}$ =3

$v_{h}$ = 1/λ H = RH x 12 (1/4 – 1/9)

Now, we consider the Li ion,

V li2+ = 1/ wavelength Li2+ = RH x 32 (1/4-1/9)

       = 9x Vh =9 X15200

    = 136800 cm-1

• The energy change and the frequency of transition:

Here, the energy change will occurs in due to the transition that will be related to the frequency of the Electromagnetic wave.

By using the planck’s equation

E = h.

Where,

E is energy

H is hertz

V is the frequency of electromagnetic radiation

Now, the frequency of the wave will be related to the wavelength and the speed of the light.

The equation for speed of light is

C=

Where v is index of refraction.

is wavelength

C is  Weber's constant .

The speed of light is measured in c =$3 * 10^{8}$m/s.

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