Question

The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is :
1)4/3
2)525/376
3)25
4)675/11

given answer is option 4.(Explain in detail)
​

Answer 1

Concept :-

Energy of the last line is the highest and the first line is the lowest

we know that ,

E = hc / λ

From this we can conclude that , energy is inversely proportional to wavelength

Hence , wavelength of the last line is the lowest and first line is highest

The difference in energy between successive energy level decreases

| E1 - E 2| < | E2 -  E3| < | E3 -  E4 | ......

The difference between  n = 1 and n= n decreases as we go away from the nucleus

E 2 - E 1 < E3 - E 1 < E4 - E 1

Solution :-

We know that ,

$\implies\dfrac{1}{\lambda } = R\bigg [ \dfrac{1}{n_1^{2} }- \dfrac{1}{n_2^{2} } \bigg]$

Shortest wavelength

n 1 = 1 to n 2= ∞

Now let's substitute these values in the above equation ,

$\implies\dfrac{1}{\lambda } = R\bigg [ \dfrac{1}{1^{2} }- \dfrac{1}{ \infin^{2} } \bigg]$

$\implies\dfrac{1}{\lambda } = R\bigg [ \dfrac{1}{1^{2} }- 0 \bigg]$

$\implies\dfrac{1}{\lambda } = R$

Longest  wavelength

n 2 = 6 to n 1 = 5

Now let's substitute these values in the above equation ,

$\implies\dfrac{1}{\lambda" } = R\bigg [ \dfrac{1}{5^{2} }- \dfrac{1}{ 6^{2} } \bigg]$

$\implies\dfrac{1}{\lambda" } = R\bigg [ \dfrac{1}{25}- \dfrac{1}{ 36} \bigg]$

$\implies\dfrac{1}{\lambda" } = R\bigg [ \dfrac{36 - 25 }{36 \times 25 }\bigg]$

$\implies\dfrac{1}{\lambda" } = R\bigg [ \dfrac{11 }{900 }\bigg]$

Ratio of longest wavelength to shortest wavelength

$\implies\dfrac{\lambda }{\lambda "} = \dfrac{R}{\dfrac{11 R}{900} }$

$\implies\boxed{\dfrac{\lambda }{\lambda "} = \dfrac{900}{11}}$