Question

What would be the wavelength and name of series respectively for the emission transition for H-atom if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm?

• A. 434 nm, Balmer
• B. 434 pm, Paschen
• C. 545 pm, Pfund
• D. 600 nm, Lyman

Answer 1

The correct option is A 434 nm, Balmer

The radii of the $n^{th}$ stationary state for a hydrogen-like specie is expressed as :

$r_n=\frac{n^2{a}_0}{Z}$

where $Z=$atomic number and $a_0=52.9pm$ radius of Bohr orbit.

For hydrogen atom, Z=1

Given that transition is from orbit radius = 1.3225 nm to 211.6 pm

Orbit with radius = 1.3225 nm=1322.5 pm

$r_n=52.9\times n^2=1322.5$

$n^2=25$ or $n=n_i=5$

Similarly for Orbit with radius = 211.6 pm

$r_n=52.9\times n^2=211.6$

$n^2=4$ or $n=n_f=2$

thus transition is from $n_i=5$ to $n_f=2$

Transition energy from $n_{i}to{n}_f$ is given as:

$\frac{1}{\lambda }=R_H[\frac{1}{n_f^2}-\frac{1}{n_i^2}]$

where $R_H=109677c{m}^{-1}$ and $n_i=5,n_f=2$

$\frac{1}{\lambda }=109677[\frac{1}{2^2}-\frac{1}{5^2}]c{m}^{-1}$

$\frac{1}{\lambda }=109677[\frac{1}{4}-\frac{1}{25}]c{m}^{-1}$

$\frac{1}{\lambda }=109677\times 0.21c{m}^{-1}$

$\frac{1}{\lambda }=23032.17c{m}^{-1}$

$\lambda =4.342\times {10}^{-5}cm=434.2nm$

Since transition is from higher energy state to n=2, it belongs to Balmer series and wavelength of the transition is 434 nm