Get Answers to all your Questions

header-bg qa

The wavelength of the first line of Lyman series for the hydrogen atom is equal to that of the second line of Balmer series for a hydrogen-like ion. The atomic number Z of hydrogen-like ion is.

  • Option 1)

    3

  • Option 2)

    4

  • Option 3)

    1

  • Option 4)

    2

 

Answers (1)

best_answer

As we learnt in 

Lymen series -

frac{1}{lambda }= Rleft ( frac{1}{1^{2}}- frac{1}{n^{2}} 
ight )

left ( uv : region 
ight )

- wherein

n=2,3,4,-----

When electron jump from higher orbital to 1st energy level

i.e. ground state

    

 

 AND
 

 

Balmer series -

frac{1}{lambda }= Rleft ( frac{1}{2^{2}}- frac{1}{n^{2}} 
ight )

left ( : visible : light 
ight )

- wherein

n=3,4,5-----

When electron jump from higher orbital to n=2 energy level

 

 

 Lymen series of hydrogen

\frac{1}{\lambda } = R \left ( \frac{1}{1^{2}}-\frac{1}{2^{2}} \right ) \: \: = \frac{3R}{4}

Second line of Balmer series

\frac{1}{\lambda } = R Z^{2} \left ( \frac{1}{4}-\frac{1}{16} \right ) \: \: = \frac{3R}{16}Z^{2}

 

According to question,

\frac{3R}{4}= \: \frac{3R Z^{2}}{16} \: \: \: \: \Rightarrow Z^{2}=4 \ \: or \: \: Z=2


Option 1)

3

Incorrect

Option 2)

4

Incorrect

Option 3)

1

Incorrect

Option 4)

2

Correct

Posted by

Plabita

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks