Question

Calculate the wave number of the shortest wave appearing in the Balmer’s series of hydrogen spectrum. (Given R=1.097X107M-1)

Answer 1

Answer:

Explanation:

R

H

​

=Rydberg constant={ 109737cm }^{ -1 }109737cm

−1

The transition from n=2n=2 to n=\inftyn=∞ will have shortest wavelength

wave number=\overrightarrow { v } =\frac { 1 }{ \lambda  } =R_{ H }\left( \frac { 1 }{ 4 } -\frac { 1 }{ \infty  }  \right) \\ =\frac { R_{ H } }{ 4 } =27434.25{ cm }^{ -1 }(approx)

v

=

λ

1

​

=R

H

​

(

4

1

​

−

∞

1

​

 )

=

4

R

H

​

​

=27434.25cm

−1

(approx)