In which of the following systems will the wavelength corresponding to n = 2 to n = 1 be minimum?

• Deuterium Atom

Options

• Hydrogen atom

• Deuterium atom

• Singly ionized helium

• Doubly ionized lithium

1. Hydrogen (1H1), deuterium (1H2), singly ionised helium (2He4+) and doubly ionised lithium (3Li8)2+ all have one electron around the nucleus. Consider an electron transition from n =2 to n=1.
2. Hydrogen (1H^1), Deuterium (1H^2), singly ionised Helium (2He^4)^ + and double ionised lithium (3Li^6)^ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1.

Solution

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.

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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Deuterium Atom

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