Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Ă…. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

Answer 1

Answer:

ANSWER

According to the question,

let

λ

1

hc

=13.6(1)

2

[

2

2

1

−

3

2

1

]⋯⋯(i)

λ

2

hc

=13.6(2)

2

[

2

2

1

−

4

2

1

]⋯⋯(ii)

Dividing (ii) by (i) we get

λ

1

λ

2

=

4×(

16

3

)

1×(

36

5

)

λ

2

=

36×4×3

λ

1

×5×16

=

27

5

λ

1

=1215A

Answer 2

Question:

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Ă. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is ?

Solution:

$\dfrac{1}{\lambda_{H_2}} = RZ_H^2\left[\dfrac{1}{4} - \dfrac{1}{9}\right] = R(1)^2\left[\dfrac{5}{36\right]}$

$\dfrac{1}{\lambda_{He}} = RZ_{He}^2\left[\dfrac{1}{4} - \dfrac{1}{16}\right] = R(4)\left[\dfrac{3}{16\right]}$

$\dfrac{\lambda_{He}}{\lambda_{H_2}} = \dfrac{1}{4}\left[\dfrac{16}{3} \times\dfrac{5}{36}\right] = \dfrac{5}{27}$

${\lambda_{He}} = \dfrac{5}{27} \times 6561 = 1215 \AA$