Atomic structure

source : Socratic.org

## Atomic structure numerical-

### Question 1) The electron energy in H – atom is given by E = -21.7 x 10^{-12}/n^{2} erg. Calculate the energy required to remove an electron completely from n = 2 orbit. What is the longest wavelength ( in cm ) of light that be used to cause this transition ?

### Solution )

*E = -21.7 x 10 ^{-12}/n^{2} erg*

*n=2 , then energy of electron ,*

*E 2= -21.7 x 10 ^{-12}/(2 x 2)*

*E 2= – 5.425 x 10 ^{-12} erg*

*Energy of electron completely remove from second orbit, n = ∞*

*E ∞= -21.7 x 10 ^{-12} /∞ ^{2} = 0*

*E ∞ = 0*

*Energy required to remove electron from second orbit completely is,*

*ΔE = E ∞ – E2 = 5.425 x 10*^{-12} erg Ans.

^{-12}erg Ans.

**λ = hc / **ΔE = 6.625 x 10^{-27} x 3 x 10^{10} / 5.425 x 10^{-12} = 3.66 x 10^{-5} cm.

*longest wavelength of light which can remove an electron completely from n = 2 orbit is 3.66 x 10*^{-5} cm. Ans.

^{-5}cm. Ans.

### Question 2) Light of wavelength 12818 Å is emitted when the electron of a hydrogen atom drops from 5^{th} to 3^{rd} orbit. Find the wavelength of the photon emitted when the electron falls from 3^{rd} to 2^{nd} orbit?

### Solution)

*n1 = 3 , n2 = 5 , λ = 12818 Å*

*1 / ***λ = R**_{H} [( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]

**λ = R**

_{H}[( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]*1 / 12818*** =****R**_{H }**[( 1/ 3**^{2}) – ( 1/5 ^{2})]

**=**

**R**

_{H }**[( 1/ 3**

^{2}) – ( 1/5^{2})]*12818 = 25 x 9/16 x ***R**_{H } eq 1

**R**

_{H }eq 1*If n1 = 2 , n2 = 3 , λ = ? *

*1 / ***λ = R**_{H} [( 1/ 2^{2}) – ( 1/ 3 ^{2})]

**λ = R**

_{H}[( 1/ 2^{2}) – ( 1/ 3^{2})]**λ ** = 36 / 5 x **R**_{H } eq 2

**λ**= 36 / 5 x

**R**

_{H }eq 2*Divide eq (2) by eq (1)*

**λ / **12818 = 36 x 16 **R**_{H }/ 5 x 9 x 25 **R**_{H }

**λ /**12818 = 36 x 16

**R**/ 5 x 9 x 25

_{H }**R**

_{H }**λ= 12818 x 64 / 125 **

**λ= 12818 x 64 / 125**

**λ ****= 6562.8 Å Ans.**

**λ**

**= 6562.8 Å Ans.**

### Question 3) The ionisation energy of H- atom is 13.6 eV. What will be the ionisation energy of He^{+} and Li^{++} ions ?

### Solution )

*Ionisation energy = – energy of first orbit*

*Energy of first orbit of H- atom = – 13.6 eV*

*Energy of first orbit of He+ = – 13.6 Z ^{2}*

*Z of He ^{+} = 2*

*Energy of first orbit of He ^{+} = -13.6 x 2 x 2= -54.4 eV*

*Ionisation energy of He*^{+} = – (-54.4)= 54.4 eV Ans.

^{+}= – (-54.4)= 54.4 eV Ans.

*Energy of first orbit of Li ^{++} = – 13.6 Z2*

*Z of Li ^{++} = 3*

*Energy of first orbit of Li ^{++} = -13.6 x 3 x 3 = – 122.4 eV*

*Ionisation energy of Li*^{++} = – (-122.4)= 122.4 eV Ans.

^{++}= – (-122.4)= 122.4 eV Ans.

### Question 4) The wavelength of a certain line in Balmer series is observed to be 4341 Å . To what value of ‘n’ does this correspond ? ( **R**_{H} = 109678 cm^{-1})

**R**

_{H}= 109678 cm^{-1})### Solution )

*For Balmer series , n1 = 2*

**λ = 4341 Å**

**λ = 4341 Å**

*1 / ***λ = R**_{H} [( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]

**λ = R**

_{H}[( 1/ n_{1}^{2}) – ( 1/ n_{2}^{2})]*1 / 4341 *** = 109678 [( 1/ 2**^{2}) – ( 1/ n_{2}^{2})]

**= 109678 [( 1/ 2**

^{2}) – ( 1/ n_{2}^{2})]**( 1/ n**_{2}^{2}) =(1 / 4 )** – (1 / ****109678 x 4341) **= 0.04

**( 1/ n**=(1 / 4 )

_{2}^{2})**– (1 /**

**109678 x 4341)**= 0.04

**n**_{2}^{2} = 1 / 0.04 = 25

**n**

_{2}^{2}= 1 / 0.04 = 25**n**_{2} = 5 Ans.

**n**

_{2}= 5 Ans.