Question

the ratio of the frequency corresponding to the third line in the Lyman series of hydrogen atomic spectrum to that of the first line in the Balmer series of Li2+ spectum is
a.4/5
b.5/4
c.4/3
d.3/4
​

Answer 1

Answer:

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Answer 2

Answer:

The ratio of frequency is 3/4 i.e.option(d).

Explanation:

First lets how the frequency is given atomic spectrum,

$\bar{\nu}=R_HZ^2(\frac{1}{n^2_1}-\frac{1}{n^2_2})$             (1)

Where,

$\bar{\nu}$=frequency

RH=rydberg constant

Z=atomic number of the atom

n₁,n₂=transition states

For Hydrogen atom

Z=1

For the third line of the Lyman series,

n₁=1 and n₂=4

By substituting the value of Z,n₁, and n₂ in equation (1) we get;

$\bar{\nu_1}=R_H(1)^2(\frac{1}{(1)^2}-\frac{1}{(4)^2})=R_H(\frac{1}{1}-\frac{1}{16})=\frac{15}{16}R_H$       (2)

For Lithium atom

Z=3

For the first line of the Balmer series,

n₁=2 and n₂=3

$\bar{\nu_2}=R_H(3)^2(\frac{1}{(2)^2}-\frac{1}{(3)^2})=9R_H(\frac{1}{4}-\frac{1}{9})=9R_H\frac{5}{36}=\frac{5}{4}R_H$     (3)

By taking the ratio of equations (2) and (3) we get;

$\frac{\bar{\nu_1}}{\bar{\nu_2}}=\frac{\frac{15R_H}{16}}{\frac{5R_H}{4}}=\frac{3}{4}$              (4)

Hence, the ratio of the frequency corresponding to the third line in the Lyman series of hydrogen atomic spectrum to that of the first line in the Balmer series of Li2+ spectrum is 3/4 i.e.option(d).