Question

calculate the wave number of the line in lyman series when an electrons jumps from orbit 3 to orbit 1?​

Answer 1

Explanation:

According to Rydberg formula:

v

ˉ

=R(

n

1

2

1

−

n

2

2

1

)

Here, R=109677=109678cm

−1

and

v

ˉ

=82200cm

−1

thus,

109678

82200

=(

1

2

1

−

n

2

2

1

)

or

4

3

=(1−

n

2

2

1

)

or

n

2

2

1

=1−

4

3

=

4

1

or n

2

=2.

∴ The electron jumps from second orbit(n

2

) to ground state(n

1

).

Answer 2

Answer:

The wavenumber of the line in the Lyman series when an electron jump from the orbit $3$ to the orbit $1$ is $9.7511\ *\ 10^{6}\ m^{-1}$·

Explanation:

It is mentioned that the electron jump from orbit $3$ to orbit $1.$

To calculate the wavenumber for the emission of an electron, we have the relationship that

     $v\ =$  $R$   $[\frac{1}{(n^f)^2}\ \ -\ \ \frac{1}{(n_i)^2} ]$

where,

$v\ =$   The Wave number

$R$  $=$ The Rydberg constant $=\ 1.097\ *\ 10^{7}$ $m^{-1}$

Since this is the line in the lyman series and the electrons jump from orbit $3$ to orbit $1.$

$n_i\ =\ 3$  

$n_f\ =\ 1$

On substituting these values in the equation we get the wavenumber as,

⇒ $v\ =$  $R$   $[\frac{1}{(n^f)^2}\ \ -\ \ \frac{1}{(n_i)^2} ]$

⇒ $v\ =$  $1.097\ *\ 10^{7}\ [\ \frac{1}{1^2} \ \ -\ \ \frac{1}{3^2}\ ]\ \ m^{-1}$

⇒ $v\ =$  $1.097\ *\ 10^{7}\ [\ \frac{1}{1} \ \ -\ \ \frac{1}{9}\ ]\ \ m^{-1}$

⇒ $v\ =$  $1.097\ *\ 10^{7}\ [\ \frac{9\ -\ 1}{9} \ ]\ \ m^{-1}$

⇒ $v\ =$  $9.7511\ *\ 10^{6}\ m^{-1}$

Therefore,

The wavenumber of the line in the Lyman series when an electron jump from the orbit $3$ to the orbit $1$ is $9.7511\ *\ 10^{6}\ m^{-1}$·