# Chapter 26: Energy levels and Spectra¶

## Ex26.1:pg-1208¶

In [1]:
  #Example 26_1
import math

#To find the ionization energy of the hydrogen atom
e=13.6        #units in eV
print "The ionization energy of the hydrogen atom is E=",round(e,1)," eV"

The ionization energy of the hydrogen atom is E= 13.6  eV


## Ex26.2:pg-1209¶

In [2]:
  #Example 26_2
import math
#To find the wavelength of fourth line in Paschen series
n1=3     #Units in constant
n2=7     #Units in constant
r=1.0974*10**7         #units in meter**-1
lamda=round((1/r)*((n1**2*n2**2)/(n2**2-n1**2))*10**9)       #Units in nm
print "The wavelength of fourth line in Paschen series is=",round(lamda)," nm"

The wavelength of fourth line in Paschen series is= 1002.0  nm


## Ex26.3:pg-1209¶

In [3]:
  #Example 26_3
import math

#To draw the energy level diagram and the find the first line of balmer type series
n=1
e1=-54.4/n**2      #units in ev
n=2
e2=-54.4/n**2      #units in ev
n=3
e3=-54.4/n**2      #units in ev
print "The energy associated with line 1 is E1=",round(e1,1)," eV\nThe energy associated with line 2 is E2=",round(e2,1)," eV\nThe energy associated with line 3 is E3=",round(e3,2)," eV\n"
e1=1       #units in eV
e2=7.6       #Units in eV
lamda1=1240         #units in nm
lamda=(e1/e2)*lamda1        #Units in nm
print "The first line of balmer series is lamda=",round(lamda)," nm and belongs to the ultraviolet region"

The energy associated with line 1 is E1= -54.4  eV
The energy associated with line 2 is E2= -13.6  eV
The energy associated with line 3 is E3= -6.04  eV

The first line of balmer series is lamda= 163.0  nm and belongs to the ultraviolet region


## Ex26.4:pg-1209¶

In [4]:
  #Example 26_4
import math

#To find the longest wavelength of light capable of ionizing hydrogen atom
#First method
R=1.097*10**7          #Units in meter**-1
lamda=(1/R)*10**9                 #Units in meters
#Second method
E=13.6        #units in eV
e1=1        #units in eV
lamda3=1240.0         #Units in eV
lamda2=(e1/E)*(lamda3)        #Units in nm
print "The longest wavelength of light capable of ionizing hydrogen atom is lamda=",round(lamda2,1)," nm"

The longest wavelength of light capable of ionizing hydrogen atom is lamda= 91.2  nm


## Ex26.5:pg-1210¶

In [5]:
  #Example 26_5
import math

#To find  the energy difference between the n is 1 and n is 2 level
e1=1        #Units in eV
lamda2=1240        #Units in eV
lamda3=0.07       #Units in eV
e2=lamda2/lamda3        #Units in eV
e=e2-e1            #Units in eV
print "The energy difference between n=1 and n=2 level is E=",round(e)," eV"
#In textbook answer is prinred wrong as E=18000 eV the correct answer is E=17713 eV

The energy difference between n=1 and n=2 level is E= 17713.0  eV