1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
|
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 3: Review of Electromagnetic waves"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.1: Atomic_Spacing_of_Nacl.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"clear \n",
"clc\n",
"disp('Exa-3.1');\n",
"w=0.250; theta=26.3;n=1 // n=1 for hydrogen atom and rest all are given values\n",
"d=n*w/(2*sind(theta)); // bragg's law\n",
"printf('Hence the atomic spacing is %.3f nm.',d);"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.2: Time_taken_to_release_an_electron.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"clear \n",
"clc\n",
"disp('Exa-3.2');\n",
"I=120;r=0.1*10^-9;Eev=2.3 //I-intensity in W/m^2 r in m & E in electron volt\n",
"A=%pi*r^2;K=1.6*10^-19; // A=area and K is conversion factor from ev to joules\n",
"t= Eev*K/(I*A); //time interval\n",
"printf('The value of time interval was found out to be %.1f sec',t);"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.3: Solution_for_a_and_b.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"clear \n",
"clc\n",
"disp('Exa-3.3(a)');\n",
"w=650*10^-9;h=6.63*10^-34;c=3*10^8; //given values and constant taken in comfortable units\n",
"E=h*c/w; printf('The Energy of the electron is %.3e J ',E);\n",
"E=E/(1.6*10^-19);printf('which is equivalent to %f eV\n',E);\n",
"printf('The momentum of electron is p=E/c i.e %.2f/c \n',E);\n",
"disp('Exa-3.3(b)');\n",
"E2=2.40; //given energy of photon.\n",
"w2=h*c*10^9/(E2*1.6*10^-19); //converting the energy in to eV and nm \n",
"printf('The wavelength of the photon is %.2f nm',w2);"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.4: Solution_for_a_b_and_c.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"clear \n",
"clc\n",
"disp('Exa-3.4(a)');\n",
"hc=1240; phi=4.52 //both the values are in eV\n",
"w1=hc/phi; \n",
"printf('The cutoff wavelength of the tungsten metal is %.3fnm\n ',w1);\n",
"disp('Exa-3.4(b)');\n",
"w2=198; //given value of wavelength \n",
"Kmax=(hc/w2)-phi;printf('The max value of kinetic energy is %.3f eV\n',Kmax);\n",
"disp('Exa-3.4(c)');\n",
"Vs=Kmax; printf('The numerical value of the max kinetic energy is same as stopping potential in volts.Hence %.2f V',Vs);\n",
" "
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.5: Solution_for_a_b_and_c.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"clear \n",
"clc\n",
"disp('Exa-3.5(a)');\n",
"T1=293; Kw=2.898*10^-3;\n",
"w1=Kw/T1;\n",
"printf('The wavelength at which emits maximum radiation is %.2f um.\n',w1*10^6);\n",
"disp('Exa-3.5(b)');\n",
"w2=650*10^-9; \n",
"T2=Kw/w2;\n",
"printf('The temperature of the object must be raised to %.0f K.\n',T2);\n",
"disp('Exa-3.5(c)');\n",
"x=(T2/T1)^4; printf('Thus the thermal radiation at higher temperature is %.2e times the room (lower) tempertaure.\n',x);"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.6: Solution_for_a_b_c_and_d.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"clear \n",
"clc\n",
"disp('Exa-3.6(a)');\n",
"w1=0.24;wc=0.00243;theta=60; //given values w=wavelength(lambeda)\n",
"w2=w1+(wc*(1-cosd(theta))); \n",
"printf('The wavelength of x-rays after scattering is %.4f nm\n',w2);\n",
"disp('Exa-3.6(b)');\n",
"hc=1240;\n",
"E2=hc/w2;E1=hc/w1; printf('The energy of scattered x-rays is %.0f eV\n',E2);\n",
"disp('Exa-3.6(c)');\n",
"K= E1-E2; //The kinetic energy is the difference in the energy before and after the collision;\n",
"printf('The kinetic energy of the x-rays is %.3f eV\n',K);\n",
"disp('Exa-3.6(d)');\n",
"phi2=atand(E2*sind(theta)/(E1-E2*cosd(theta)))\n",
"printf('The direction of the scattered eletron is %.1f degrees',phi2);"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Scilab",
"language": "scilab",
"name": "scilab"
},
"language_info": {
"file_extension": ".sce",
"help_links": [
{
"text": "MetaKernel Magics",
"url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
}
],
"mimetype": "text/x-octave",
"name": "scilab",
"version": "0.7.1"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
