Solution: Given: First operating voltage = 40 kV = 40 Â 10
3 V; tube
current = 25 mA =25 Â 10
À3
A; second operating voltage = 50 kV = 50 Â 10
3 V;
power (input) = ?, tube current = ?
We know that power is defined as
P ¼ VI ¼ 40 Â 10
3
 25  10
À3
¼ 1000watt
In second case, P = 1000 W and V = 50 kV = 50 Â 10
3 V: Therefore,
I =
P
V
¼
1000
50 Â 10 3 ¼ 0.02A = 20 mA
Example 6 For platinum metal, the energy levels K, L and M lie roughly at 78, 12
and 3 eV, respectively. Determine the approximate wavelength of characteristic
K a and K b lines.
Solution: Given: Energy levels (Fig. 7.4) of platinum lie at E 1 ¼ 78 keV;
E 2 ¼ 12keV; E 3 ¼ 3keV; K a ¼ ?; K b ¼ ?
Draw the energy level diagram as shown in Fig. 7.4. From this, we can obtain
hm
ð Þ K a = E L À E K ¼ À 12 + 78 = 66 keV
ðiÞ
and hm
ð Þ K b = E M À E K ¼ À 3 + 78 = 75 keV
ðiiÞ
Equation (i) can be written as:
Fig. 7.4 Energy level
diagram for platinum
268
7 Diffraction of Waves and Particles by Crystal