Question

-
In hydrogen atom the wavelength of photon released
when transition of electron takes place from orbit
of radius 1322.5 pm to 0.2116 nm is :-
(1) 3048 Å
(2) 4342 Å
(3) 5248 Å
(4) 6328 Å​

Answer 1

The wavelength of photon released is $4342\ A$

(2) is correct option.

Explanation:

Given that,

Radius $R_{1}=1322.5\ pm$

Radius $R_{2}=0.2116\ nm$

We need to calculate the excited state

Using formula of radius of hydrogen

$R_{1}=r_{0}\times n_{1}^2$

Put the value into the formula

$1322.5\times10^{-12}=0.529\times10^{-10}\times n_{1}^2$

$n_{1}^2=\dfrac{1322.5\times10^{-12}}{0.529\times10^{-10}}$

$n_{1}^2=25$

Again, We need to calculate the excited state

Using formula of radius of hydrogen

$R_{1}=r_{0}\times n_{1}^2$

Put the value into the formula

$0.2116\times10^{-9}=0.529\times10^{-10}\times n_{1}^2$

$n_{2}^2=\dfrac{0.2116\times10^{-9}}{0.529\times10^{-10}}$

$n_{2}^2=4$

We need to calculate the wavelength

Using formula of wavelength

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

$\dfrac{1}{\lambda}=1.09677\times10^{7}\times (\dfrac{1}{4}-\dfrac{1}{25})$

$\dfrac{1}{\lambda}=2303217\ m$

$\lambda=4.3417\times10^{-7}\ m$

$\lambda=4342\ A$

Hence, The wavelength of photon released is $4342\ A$

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Topic : wavelength

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