Question

maximum wavelength of Balmer series of h atom ​,

Answer 1

Answer:

720 nm

Explanation:

Wave length

1/lamda = RZ^2 * 1/n1^2 - 1/n2^

= 10^7 * 5/36

= 7.2 * 10^-7

= 720 nm

Answer 2

Given :

$R_H=10^7\ m^{-1}$ .

To Find :

Maximum wavelength of Balmer series of H atom .

Solution :

We know , wavelength of Balmer series of H atom is given by :

$\dfrac{1}{\lambda}=R_H(\dfrac{1}{2^2}-\dfrac{1}{n^2})$

We know , by plank's law .

$E = \dfrac{hc}{\lambda}$

Here , wavelength and energy are inversely proportional .

Therefore , wavelength is maximum when energy transition is minimum .

Therefore , n = 3 .

$\dfrac{1}{\lambda}=10^7\times (\dfrac{1}{2^2}-\dfrac{1}{3^2})\\\\\lambda=7.2\times 10^{-7}\ m\\\\\lambda=720\times 10^{-9}\ m\\\\\lambda=720\ nm$

So , option ( 1 ) is correct .

Learn More :

Atomic structure

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