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Karnataka Board PUCPUC Science Class 11

A Group of Hydrogen Atoms Are Prepared in N = 4 States. List the Wavelength that Are Emitted as the Atoms Make Transitions and Return to N = 2 States.

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Question

A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.

Sum
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Solution

There will be three wavelengths.
(i) For the transition from (n = 4) to (= 3) state
(ii) For the transition from (n = 3) to (= 2) state
(iii) For the transition from (n = 4) to (= 2) state

Let `(lamda_1)` be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state.​ 
Here,
n1 = 2
n2 = 4
Now, the wavelength `(lamda_1)`  will be
`1/lamda_1 = R (1/n_1^2 - 1/n_2^2)`

`R = 1.097 xx 10^7 m^-1`
`1/lamda_1 = 1.097xx 10^7 xx (1/4 - 1/16)`

`rArr 1/lamda_1 = 1.097 xx 10^7 ((4-1)/(16))`

`rArr 1/lamda_1 = (1.097xx10^7xx3)/16`

`rArr lamda_1 = (16xx10^-7)/(3xx1.097)`
= 4.8617 × 10-7
= 486.1 × 10-9
= 487 nm

When an atom makes transition from (n = 4) to (= 3), the wavelength (λ2) is given by
Here again 
`n_1 = 3`
`n_2 = 4`

`1/lamda_2 = 1.097 xx 10^7 (1/9 - 1/16)`

`rArr 1/lamda_2 = 1.097 xx 10^7 ((16 -9)/144)`

`rArr 1/lamda_2 = (1.097xx 10^7 xx 7)/144`

`rArr lamda_2 = 144/(1.097 xx 10^7 xx 7)`

                      = 1875 nm

Similarly, wavelength (λ2) for the transition from (n = 3) to (= 2) is given by
When the transition is n1 = 2 to n2 = 3:
`1/lamda_3 = 1.097 xx 10^7 (1/4 - 1/9)`

`rArr 1/lamda_3 = 1.097 xx 10^7 ((9-4)/36) `

`rArr 1/lamda_3 = (1.097xx10^7)/36`

`rArr lamda_3 = (36xx10^7xx5)/36`

`rArr lamda_3 = (36xx10^7)/((1.097)xx5) = 656  nm`

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Chapter 43: Bohr’s Model and Physics of Atom - Exercises [Page 384]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 43 Bohr’s Model and Physics of Atom
Exercises | Q 9 | Page 384

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