X-Ray Fluorescence Analytical Techniques - CNSTN : Centre ...

c = 3 x 10 8 m/s
(i)
1 2
mV = eV
2
⇒ V = =
2eV −19
3
2× 1.6× 10 × 18× 10
m
−31
9× 10
− 19+ 3+ 31−1 2 16 10 2
⇒ V = × × ×
⇒ V = 64× 10
⇒ V = 8× 10 m ⋅
(ii)
min
7
14
−1
s
hc
eV =
λ min
hc
⇒ λ min =
eV
−34
8
6.63× 10 × 3× 10
⇒ λ min =
−19
3
1.16× 10 × 18× 10
⇒ λ ≈
III.
-11
6.9 x 10 m
An X-ray tube is operated with an anode potential of 10 kV and an anode current of 15 mA.
Calculate
i.the number of electrons hitting the anode per second,
ii.the rate of production of heat at the anode stating any assumptions made,
iii.the frequency of the emitted X-ray photon of maximum energy.
e = 1.6 x 10 -19 C
h = 6.6 x 10 -34 Js
Solution
I = 15 x 10 -3 A
V = 10 x 10 3 V
e = 1.6 x 10 -19 C
h = 6.6 x 10 -34 Js
(i)