Question

Calculate the wave number of the spectral line of shortest wavelength trasition in balmer series of atomic hydrogen

Answer 1

1/lemda= R (1/n^2-1/n2^2)
n=2 ,n2=3 ,4,5,6

Answer 2

Answer:

The shortest wavelength in the 'balmer series' of a hydrogen atom is $27419.5\ \mathrm{cm}^{-1}$

Explanation:

The 'shortest wavelength' in the 'balmer series' of a 'hydrogen' atom is  

$\frac{1}{\lambda}=R\left\{\frac{1}{n 1^{2}}-\frac{1}{n 2^{2}}\right\}$

Here $n {1} =2, n {2} =\infty$ - substituting the values in the above equation

$\frac{1}{\lambda}=R\left\{\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right\}$ where R = 109678 for hydrogen atom

$\frac{1}{\lambda}=109678\left\{\frac{1}{2^{2}}-\frac{1}{{\infty}^{2}}\right\}$

$\frac{1}{\lambda}=109678\left\{\frac{1}{2^{2}}\right\}$

$\frac{1}{\lambda}=\frac{109678}{4}$

$=27419.5\ \mathrm{cm}^{-1}$