- Deuterium Element
- Hydrogen Deuterium Singly Ionized Helium Cell
- Hydrogen Deuterium Singly Ionised Helium Atom
- Hydrogen Deuterium Singly Ionized Helium Chemical
In which of the following systems will the wavelength corresponding to n = 2 to n = 1 be minimum?
Options
Hydrogen atom
Deuterium atom
Singly ionized helium
Doubly ionized lithium
Hydrogen Deuterium singly ionised Helium and doubly ionised lithium all have one electron around the nucleus. Consider an electron transition from to. If the wavelengths of emitted radiation are and respectively then approximately which one of the following is correct? Hydrogen (H), deuterium (D), singly ionized helium (Het) and doubly ionized lithium (Li) all have one electron around the nucleus. Consider n=2 to n=1 transition. The wavelengths of emitted radiations are 2, 1, 2, and 2, respectively. Hydrogen, deuterium, and singly ionized helium are all examples of one-electron atoms. The deuterium nucleus has the same charge as the hydrogen nucleus, and almost exactly twice the mass.
Solution
Doubly ionized lithium
The wavelength corresponding the transition fromn2 to n1 is given by
`1/lamda =RZ^2 (1/n_1^2 - 1/n_2^2)`]
Here,
R = Rydberg constant
Z = Atomic number of the ion
From the given formula, it can be observed that the wavelength is inversely proportional to the square of the atomic number.
Therefore, the wavelength corresponding to n = 2 to n = 1 will be minimum in doubly ionized lithium ion because for lithium, Z = 3.
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$lambda_1$ =$lambda_2 =4 lambda_3 = 9lambda_4$
B$4lambda_1 = 2lambda_2 = 2lambda_3 =lambda_4$
C$lambda_1 = 2lambda_2 =2 surd 2 lambda_3 = 3 surd 2 lambda_4$
$lambda_1 = lambda_2 =2lambda_3 =surd 2lambda_4$
Solution:
$Z_{1:}=:1,:Z_2:=:1,:Z_3:=:2:and:Z_4:=3.:$
$:frac{1}{lambda }:=:RZ^2left(frac{1}{1^2}:-:frac{1}{2^2}right):or:::::::lambda :=frac{4}{3RZ^2}:$
$:or:::::lambda Z^2:=:constant::$
$:So::lambda _1left(1right)^2:=:lambda _2:left(1right)^2:=:lambda _3left(2right)^2:=:lambda aleft(3^2right):$
$:or:lambda _1:=:lambda _1:lambda _2:=:4lambda _3:=:9lambda _{4.}$
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