Question

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Answer 1

The radius of the n^th orbit of hydrogen-like particles is given by,

`"r" = (0.529" n"^2)/"Z"` Å

`"r" = (52.9 " n"^2)/"Z" "pm"`

For radius (r₁) = 1.3225 nm

= 1.32225 × 10^-9 m

= 1322.25 × 10^-12 m

= 1322.25 pm

`"n"_1^2 = ("r"_1"Z")/(52.9)`

`"n"_1^2 = (1322.25"Z")/52.9`

Similarly

`"n"_2^2 = (211.6"Z")/52.9`

`"n"_1^2/"n"_2^2 = 1322.5/211.6`

`"n"_1^2/"n"_2^2 = 6.25`

`"n"_1/"n"_2 = 2.5 `

`"n"_1/"n"_2 = 25/10 = 5/2`

Thus, the transition is from the 5^th orbit to the 2^nd orbit. It belongs to the Balmer series.

Wave number (`bar "v"`) for the transition is given by,

`1.097 xx 10^7 " m"^(-1) (1/2^2- 1/5^2)`

` = 1.097 xx 10^7 " m"^(-1)(21/100)`

= 2.303 × 10⁶ m^-1

∴Wavelength (λ) associated with the emission transition is given by,

`lambda = 1/v`

`= 1/(2.303 xx 10^6 " m"^(-1))`

= 0.434 ×10^–6 m

λ = 434 nm

​This transition belongs to Balmer series and comes in the visible region of the spectrum.

Answer 2

Radius of nth orbit of H-like particles = 0.529n²/Z Å = 52.9n²/Z pm

r₁ = 1.3225 nm = 1322.5 pm = 52.9n₁²

r₂ = 211.6 pm = 52.9n₂²/Z 

∴ r₁/r_{2 =} 1322.5 pm/211.6 pm = n₁²/n₂²

⇒ n₁²/n₂² = 6.25

⇒ n₁/n₂ = 2.5

If n₂ = 2, n₁ = 5.

Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.

ṽ = 1.097×10⁷ m^-1 (1/2² - 1/5²) = 1.097×10⁷×21/100 m^-1

λ = 1/ṽ = 100/(1.097×21×10⁷) m = 434×10^-9 m = 434 nm

It lies in visible range.