Atomic structure source : Socratic.org

## Atomic structure numerical-

### Solution )

E = -21.7 x 10-12/n2 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.

λ = hc / ΔE = 6.625 x 10-27 x 3 x 1010 /  5.425 x 10-12 = 3.66 x 10-5 cm.

### Solution)

n1 = 3 , n2 = 5 , λ = 12818 Å

#### 12818 = 25 x 9/16 x RH                      eq 1

If  n1 = 2 , n2 = 3 , λ = ?

#### λ  =  36 / 5 x RH                               eq 2

Divide eq (2) by eq (1)

### 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  Z2

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.

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

### Solution )

For Balmer series , n1 = 2