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Physics Question on Atoms

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 series. 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

Frequency, v=RC[1n12−1n22]v=R C\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]
v1=RC[1−1∞]=RCv_{1}=R C\left[1-\frac{1}{\infty}\right]=R C
v2=RC[1−14]=34RCv_{2}=R C\left[1-\frac{1}{4}\right]=\frac{3}{4} R C
v3=RC[14−1∞]=RC4v_{3}=R C\left[\frac{1}{4}-\frac{1}{\infty}\right]=\frac{R C}{4}
⇒v1−v2=v3\Rightarrow v_{1}-v_{2}=v_{3}