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"source": [
"# Chapter 26: Energy levels and Spectra"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex26.1:pg-1208"
]
},
{
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"execution_count": 1,
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"name": "stdout",
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"text": [
"The ionization energy of the hydrogen atom is E= 13.6 eV\n"
]
}
],
"source": [
" #Example 26_1\n",
" \n",
" \n",
"#To find the ionization energy of the hydrogen atom\n",
"e=13.6 #units in eV\n",
"print \"The ionization energy of the hydrogen atom is E=\",round(e,1),\" eV\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex26.2:pg-1209"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The wavelength of fourth line in Paschen series is= 1002.0 nm\n"
]
}
],
"source": [
" #Example 26_2\n",
" \n",
" \n",
" #To find the wavelength of fourth line in Paschen series\n",
"n1=3 #Units in constant\n",
"n2=7 #Units in constant\n",
"r=1.0974*10**7 #units in meter**-1\n",
"lamda=round((1/r)*((n1**2*n2**2)/(n2**2-n1**2))*10**9) #Units in nm\n",
"print \"The wavelength of fourth line in Paschen series is=\",round(lamda),\" nm\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex26.3:pg-1209"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The energy associated with line 1 is E1= -54.4 eV\n",
"The energy associated with line 2 is E2= -13.6 eV\n",
"The energy associated with line 3 is E3= -6.04 eV\n",
"\n",
"The first line of balmer series is lamda= 163.0 nm and belongs to the ultraviolet region\n"
]
}
],
"source": [
" #Example 26_3\n",
" \n",
" \n",
" #To draw the energy level diagram and the find the first line of balmer type series\n",
"n=1\n",
"e1=-54.4/n**2 #units in ev\n",
"n=2\n",
"e2=-54.4/n**2 #units in ev\n",
"n=3\n",
"e3=-54.4/n**2 #units in ev\n",
"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\"\n",
"e1=1 #units in eV\n",
"e2=7.6 #Units in eV\n",
"lamda1=1240 #units in nm\n",
"lamda=(e1/e2)*lamda1 #Units in nm\n",
"print \"The first line of balmer series is lamda=\",round(lamda),\" nm and belongs to the ultraviolet region\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex26.4:pg-1209"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The longest wavelength of light capable of ionizing hydrogen atom is lamda= 91.2 nm\n"
]
}
],
"source": [
" #Example 26_4\n",
" \n",
" \n",
" #To find the longest wavelength of light capable of ionizing hydrogen atom\n",
" #First method\n",
"R=1.097*10**7 #Units in meter**-1\n",
"lamda=(1/R)*10**9 #Units in meters\n",
" #Second method\n",
"E=13.6 #units in eV\n",
"e1=1 #units in eV\n",
"lamda3=1240.0 #Units in eV\n",
"lamda2=(e1/E)*(lamda3) #Units in nm\n",
"print \"The longest wavelength of light capable of ionizing hydrogen atom is lamda=\",round(lamda2,1),\" nm\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex26.5:pg-1210"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The energy difference between n=1 and n=2 level is E= 17713.0 eV\n"
]
}
],
"source": [
" #Example 26_5\n",
" \n",
" \n",
" #To find the energy difference between the n is 1 and n is 2 level\n",
"e1=1 #Units in eV\n",
"lamda2=1240 #Units in eV\n",
"lamda3=0.07 #Units in eV\n",
"e2=lamda2/lamda3 #Units in eV\n",
"e=e2-e1 #Units in eV\n",
"print \"The energy difference between n=1 and n=2 level is E=\",round(e),\" eV\"\n",
" #In textbook answer is prinred wrong as E=18000 eV the correct answer is E=17713 eV \n"
]
}
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