Question

Calculate the wavelength of light (in nm) of the spectral line of Hydrogen where an electron falls from the 6th Bohr orbit to the 3rd Bohr orbit.

a) 540 nm
b) 2000 nm
c) 1090 nm
d) 1050 nm

Answer 1

Answer:

The wavelength of light (in nm) of the spectral line of Hydrogen where an electron falls from the 6th Bohr orbit to the 3rd Bohr orbit is 1090nm

Explanation:

We know that , the wavelength of the light is calculated by Rydberg's formula-

$\frac{1}{\pi} =R^2(\frac{1}{n^2_1} -\frac{1}{n^2_2})$  $[n_2>n_1]$

Here , R = Rydberg's constant $(1.097\times 10^7 m^-^1)$

          Z = atomic number (for hydrogen , Z= 1)

     $n_2 =6 , n_1=3$

$\pi =$ wavelength of light

Now , putting the values in the Rydberg's formula ,

  $\frac{1}{\pi} =1.097\times10^7m^-^1(\frac{1}{3^2} -\frac{1}{6^2} )$

     =$1.097\times 10^7m^-^1 (\frac{4-1}{36} )$

    =$1.097\times 10^7m^-^1(\frac{3}{36} )$

      =$1.097\times 10^7m^-^1(\frac{1}{12} )$

  $\frac{1}{\pi}$   = $0.0914167\times 10^7m^-^1$

$\pi=\frac{1}{0.0914167\times10^7m^-^1}$

$\pi=10.9389\times10^-^7\\\pi=1093.89\times10^-^9m$

  =1090nm

Hence , the wavelength of the light is 1090nm,, that is option D is correct.