Solveeit Logo
Login

Question

Question: If \(\upsilon_{1}\) is the frequency of the series limit of Lyman series,\(\upsilon_{2}\) is the fre...

If υ1\upsilon_{1} is the frequency of the series limit of Lyman series,υ2\upsilon_{2} is the frequency of the first line of Lyman series and υ3\upsilon_{3} is the frequency of the series limit of the Balmer sereis, then

A

υ1−υ2=υ3\upsilon_{1} - \upsilon_{2} = \upsilon_{3}

B

υ1=υ2−υ3\upsilon_{1} = \upsilon_{2} - \upsilon_{3}

C

1υ2=1υ1+1υ3\frac{1}{\upsilon_{2}} = \frac{1}{\upsilon_{1}} + \frac{1}{\upsilon_{3}}

D

1υ1=1υ2+1υ3\frac{1}{\upsilon_{1}} = \frac{1}{\upsilon_{2}} + \frac{1}{\upsilon_{3}}

Answer

υ1−υ2=υ3\upsilon_{1} - \upsilon_{2} = \upsilon_{3}

Explanation

Solution

For Lyman series

υ=R[112−1n2]\upsilon = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{n^{2}} \right\rbrack

Where n = 2, 3, 4……….

For the series limit of Lyman series n=∞n = \infty

υ1=Rc[112−1∞2]=Rc\upsilon_{1} = Rc\left\lbrack \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right\rbrack = Rc …… (i)

For the first line of Lyman series n = 2

υ2=Rc[112−122]=34Rc\upsilon_{2} = Rc\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack = \frac{3}{4}Rc …… (ii)

For Balmer series

υ=R[112−1n2]\upsilon = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{n^{2}} \right\rbrack

Where n = 3, 4, 5……….

For the series limit of Balmer series n=∞n = \infty

υ3=Rc[122−1∞2]=Rc4\upsilon_{3} = Rc\left\lbrack \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right\rbrack = \frac{Rc}{4} …… (iii)

From equation (i), (ii) and (iii) we get

υ1=υ2+υ3\upsilon_{1} = \upsilon_{2} + \upsilon_{3} or υ1−υ2=υ3\upsilon_{1} - \upsilon_{2} = \upsilon_{3}