Question

How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?

Answer 1

The wavelengths in the given distance (50−100 nm) must lie within the Lyman sequence.  The number of wavelengths between 50 100 3.

Explanation:

The Balmer array has wavelengths of 364 nm (for $n_2$ = 3) to 655 nm ( $n_2$ = ∞).

Therefore, the specified wavelength range (380-780 nm) lies in the Balmer array.

The Balmer series wavelength can be derived from

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

Here, R =  Rydberg's constant = $1,097 \times 107 m^{-1}$

The transition wavelength from n = 3 to n = 2

$\frac{1}{\lambda_{1}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=656.3 \mathrm{nm}$

The transition wavelength from = 4 = 2

$\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=486.1 \mathrm{nm}$

The transition wavelength from = 5 = 2

$\frac{1}{\lambda_{3}}=R\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)=434.0 \mathrm{mm}$

The transition wavelength from = 6 = 2

$\frac{1}{\lambda_{4}}=R\left(\frac{1}{2^{2}}-\frac{1}{6^{2}}\right)=410.2 \mathrm{nm}$

The transition wavelength from = 7 = 2

$\frac{1}{\lambda_{5}}=R\left(\frac{1}{2^{2}}-\frac{1}{7^{2}}\right)=397.0 \mathrm{nm}$

So the emitted  wavelengths by the visible range of atomic hydrogen (380−780 ) 5.

The Lyman sequence contains wavelengths ranging from 91 n ( $n_2$ = 2) 121 ( $n_2$ = ∞).

So the wavelengths in the given distance (50−100 nm) must lie within the Lyman sequence.

The Lyman series wavelength can be found by   $\frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)$

The transition wavelength from n = 2 to n = 1

$\frac{1}{\lambda_{1}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=122 \mathrm{nm}$

The transition wavelength from n = 3 to n = 1

$\frac{1}{\lambda_{2}}=R\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)=103 \mathrm{nm}$

The transition wavelength from n = 4 to n = 1

$\frac{1}{\lambda_{3}}=R\left(\frac{1}{1^{2}}-\frac{1}{4^{2}}\right)=97.3 \mathrm{nm}$

The transition wavelength from n =5  to n = 1  

$\frac{1}{\lambda_{4}}=R\left(\frac{1}{1^{2}}-\frac{1}{5^{2}}\right)=95.0 \mathrm{nm}$

The transition wavelength from n = 6  to n = 1

$\frac{1}{\lambda_{5}}=R\left(\frac{1}{1^{2}}-\frac{1}{6^{2}}\right)=93.8 \mathrm{nm}$

It can be remembered, then, that the number of wavelengths between 50 100 3.