Question

calculate the longest wavelength in the balmer sereies of hydrogen atom? ​

Answer 1

hlo mate ✌️

Wavelength of photon emitted due to transition in H-atom $$\dfrac{1}{\lambda} = R \bigg( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \bigg)$$

Shortest wavelength is emitted in Balmer series if the transition of electron takes place from $$n_2 = \infty$$ to $$n_1 = 2$$.

hope it help u.

Answer 2

Wavelength and energy are inversely proportional so as the energy increase the wavelength decrease and vice versa. All the balmer series of hydrogen atom has n=2

Calculating the wavelenght using rydberg equation and the rydberg's equation is:

For longer wavelength:

\

= 1.52*10^{6}∗10

6

Therefore \frac{1}{lambda} = 1.52*10^{6}m :

$</p><p>= \frac{1}{lambda} = (1.907*10^{7})(\frac{1}{2^{2} } -\frac{1}{∞^{2} } ) </p><p>lambda \:$

therefore lambda = 3.65*10^{-7} m3.65∗10

−7

m =365nm

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