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-rw-r--r--3648/CH26/EX26.1/Ex26_1.sce6
-rw-r--r--3648/CH26/EX26.1/Ex26_1.txt1
-rw-r--r--3648/CH26/EX26.2/Ex26_2.sce9
-rw-r--r--3648/CH26/EX26.2/Ex26_2.txt1
-rw-r--r--3648/CH26/EX26.3/Ex26_3.sce16
-rw-r--r--3648/CH26/EX26.3/Ex26_3.txt4
-rw-r--r--3648/CH26/EX26.4/Ex26_4.sce13
-rw-r--r--3648/CH26/EX26.4/Ex26_4.txt1
-rw-r--r--3648/CH26/EX26.5/Ex26_5.sce11
-rw-r--r--3648/CH26/EX26.5/Ex26_5.txt1
10 files changed, 63 insertions, 0 deletions
diff --git a/3648/CH26/EX26.1/Ex26_1.sce b/3648/CH26/EX26.1/Ex26_1.sce
new file mode 100644
index 000000000..386b94777
--- /dev/null
+++ b/3648/CH26/EX26.1/Ex26_1.sce
@@ -0,0 +1,6 @@
+//Example 26_1
+clc();
+clear;
+//To find the ionization energy of the hydrogen atom
+e=13.6 //units in eV
+printf("The ionization energy of the hydrogen atom is E=%.1f eV",e)
diff --git a/3648/CH26/EX26.1/Ex26_1.txt b/3648/CH26/EX26.1/Ex26_1.txt
new file mode 100644
index 000000000..51a7ab4b0
--- /dev/null
+++ b/3648/CH26/EX26.1/Ex26_1.txt
@@ -0,0 +1 @@
+ The ionization energy of the hydrogen atom is E=13.6 eV \ No newline at end of file
diff --git a/3648/CH26/EX26.2/Ex26_2.sce b/3648/CH26/EX26.2/Ex26_2.sce
new file mode 100644
index 000000000..15fa0b6e3
--- /dev/null
+++ b/3648/CH26/EX26.2/Ex26_2.sce
@@ -0,0 +1,9 @@
+//Example 26_2
+clc();
+clear;
+//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
+printf("The wavelength of fourth line in Paschen series is=%d nm",lamda)
diff --git a/3648/CH26/EX26.2/Ex26_2.txt b/3648/CH26/EX26.2/Ex26_2.txt
new file mode 100644
index 000000000..f549949cc
--- /dev/null
+++ b/3648/CH26/EX26.2/Ex26_2.txt
@@ -0,0 +1 @@
+The wavelength of fourth line in Paschen series is=1005 nm \ No newline at end of file
diff --git a/3648/CH26/EX26.3/Ex26_3.sce b/3648/CH26/EX26.3/Ex26_3.sce
new file mode 100644
index 000000000..e08d3b89b
--- /dev/null
+++ b/3648/CH26/EX26.3/Ex26_3.sce
@@ -0,0 +1,16 @@
+//Example 26_3
+clc();
+clear;
+//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
+printf("The energy associated with line 1 is E1=%.1f eV\nThe energy associated with line 2 is E2=%.1f eV\nThe energy associated with line 3 is E3=%.2f eV\n",e1,e2,e3)
+e1=1 //units in eV
+e2=7.6 //Units in eV
+lamda1=1240 //units in nm
+lamda=(e1/e2)*lamda1 //Units in nm
+printf("The first line of balmer series is lamda=%d nm and belongs to the ultraviolet region",lamda)
diff --git a/3648/CH26/EX26.3/Ex26_3.txt b/3648/CH26/EX26.3/Ex26_3.txt
new file mode 100644
index 000000000..c3211fb11
--- /dev/null
+++ b/3648/CH26/EX26.3/Ex26_3.txt
@@ -0,0 +1,4 @@
+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 nm and belongs to the ultraviolet region \ No newline at end of file
diff --git a/3648/CH26/EX26.4/Ex26_4.sce b/3648/CH26/EX26.4/Ex26_4.sce
new file mode 100644
index 000000000..fad109339
--- /dev/null
+++ b/3648/CH26/EX26.4/Ex26_4.sce
@@ -0,0 +1,13 @@
+//Example 26_4
+clc();
+clear;
+//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 //Units in eV
+lamda2=(e1/E)*(lamda3) //Units in nm
+printf("The longest wavelength of light capable of ionizing hydrogen atom is lamda=%.1f nm",lamda2)
diff --git a/3648/CH26/EX26.4/Ex26_4.txt b/3648/CH26/EX26.4/Ex26_4.txt
new file mode 100644
index 000000000..d16ad5b97
--- /dev/null
+++ b/3648/CH26/EX26.4/Ex26_4.txt
@@ -0,0 +1 @@
+The longest wavelength of light capable of ionizing hydrogen atom is lamda=91.2 nm \ No newline at end of file
diff --git a/3648/CH26/EX26.5/Ex26_5.sce b/3648/CH26/EX26.5/Ex26_5.sce
new file mode 100644
index 000000000..38a87f075
--- /dev/null
+++ b/3648/CH26/EX26.5/Ex26_5.sce
@@ -0,0 +1,11 @@
+//Example 26_5
+clc();
+clear;
+//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
+printf("The energy difference between n=1 and n=2 level is E=%d eV",e)
+//In textbook answer is prinred wrong as E=18000 eV the correct answer is E=17713 eV
diff --git a/3648/CH26/EX26.5/Ex26_5.txt b/3648/CH26/EX26.5/Ex26_5.txt
new file mode 100644
index 000000000..db4129404
--- /dev/null
+++ b/3648/CH26/EX26.5/Ex26_5.txt
@@ -0,0 +1 @@
+The energy difference between n=1 and n=2 level is E=17713 eV \ No newline at end of file