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
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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
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Answer:
[tex]your \: answer \: is \: \\ \green{d ) \frac{3}{4} } \\ \blue{hope \: my \: anwer \: will \: help \: you}[/tex]
Answer:
The ratio of frequency is 3/4 i.e.option(d).
Explanation:
First lets how the frequency is given atomic spectrum,
[tex]\bar{\nu}=R_HZ^2(\frac{1}{n^2_1}-\frac{1}{n^2_2})[/tex] (1)
Where,
[tex]\bar{\nu}[/tex]=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;
[tex]\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[/tex] (2)
For Lithium atom
Z=3
For the first line of the Balmer series,
n₁=2 and n₂=3
[tex]\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[/tex] (3)
By taking the ratio of equations (2) and (3) we get;
[tex]\frac{\bar{\nu_1}}{\bar{\nu_2}}=\frac{\frac{15R_H}{16}}{\frac{5R_H}{4}}=\frac{3}{4}[/tex] (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).