Question

calculate the wavelength of limiting spectral line in balmer series of h atom.
Limiting spectral line=spectral line with shortest wavelength​

Answer 1

Answer:

656.3

Explanation:

Wavelength of photon emitted due to transition in H-atom  

λ

1

​  

=R(  

n  

1

2

​  

 

1

​  

−  

n  

2

2

​  

 

1

​  

)

Shortest wavelength is emitted in Balmer series if the transition of electron takes place from n  

2

​  

=∞ to n  

1

​  

=2.

∴ Shortest wavelength in Balmer series  

λ  

s

​  

 

1

​  

=R(  

2  

2

 

1

​  

−  

∞

1

​  

)

Or λ  

s

​  

=  

R

4

​  

 

⟹ λ  

s

​  

=  

1.097×10  

7

 

4

​  

=364.6nm

Longest wavelength is emitted in Balmer series if the transition of electron takes place from n  

2

​  

=3 to n  

1

​  

=2.

∴ Longest wavelength in Balmer series  

λ  

L

​  

 

1

​  

=R(  

2  

2

 

1

​  

−  

3  

2

 

1

​  

)

Or λ  

s

​  

=  

5R

36

​  

 

⟹ λ  

s

​  

=  

5×1.097×10  

7

 

36

​  

=656.3nm