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Radiation Coming from Transition N = 2 to N = 1 of Hydrogen Atoms Falls on Helium Ions in N = 1 and N = 2 States. What Are the Possible

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प्रश्न

Radiation coming from transition n = 2 to n = 1 of hydrogen atoms falls on helium ions in n = 1 and n = 2 states. What are the possible transitions of helium ions as they absorbs energy from the radiation?

योग
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उत्तर

Energy of radiation (E) from the hydrogen atom is given by

`E = 13.6 (1/n_1^2 - 1/n_2^2 )`

Hydrogen atoms go through transition, n = 1 to n = 2.

The energy released is given by

`E = 13.6 (1/1-1/4)`

`= 13.6xx3/4 = 10.2  eV`
For He,

Atomic no, Z = 2

Let us check the energy required for the

transition in helium ions from n = 1 to n = 2.

`therefore` n1 =1 to n= 2

Energy (E1) of this transition is given by

`E_1 = Z^2 13.6 (1/n_1^2 - 1/n_2^2)`

= `4xx13.6(1 - 1/4)`

= 40.8 eV

E1 > E,

Hence, this transition of helium ions is not possible.

Let us check the energy required for the transition in helium ion from n = 1 to n = 3.
`therefore n_1 =1` to `n_2 = 3`

Energy (E2) for this transition is given by
E2 =`Z^2 xx 13.6 (1/n_2^2 - 1/n_1^2)`

= `4xx13.6xx(1/1- 1/9)`

= 48.3 eV

It is clear that E2 > E.

Hence, this transition of helium ions is not possible.

Similarly, transition from n1 = 1 to n2 = 4 is also not possible.

Let us check the energy required for the transition in helium ion from n = 2 to n=3

∴ n= 2 to n2 = 3

Energy (E3) for this transition is given by

`E_3 = 13.6xx4(1/4 - 1/9)`

= `(20xx13.6)/36 = 7.56  ev`

Let us check the energy required for the transition in helium ion from n = 2 to n = 3.

∴ n​​= 2 to n2 = 4

Energy (E_4) for this transition is given by

`E_4 = 13.6xx4 (1/4 - 1/16)`

`= 13.6xx3/4 = 10.2  eV`

We find that

E3 < E

E4 = E

Hence, possible transitions are from n = 2 to n = 3 and n = 2 to n = 4.

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अध्याय 43: Bohr’s Model and Physics of Atom - Exercises [पृष्ठ ३८५]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 43 Bohr’s Model and Physics of Atom
Exercises | Q 28 | पृष्ठ ३८५

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