Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

744 CHAPTER 14
According to rule 2, the maximum total energy deposited during the 0.5-second
viewing time is given in Table 14-8:
H(MPE) = 1.8 t 0.75 × 10 −3 J/cm 2 = 1.8 (0.5) 0.75 × 10 −3 = 1.1 × 10 −3 J/cm 2
is the MPE for the entire pulse train. Since the PRF is 1 MHz, the total number of
pulses entering the eye during 0.5 second is
6 pulses
n = 0.5s× 1 × 10
s
= 5 × 105 pulses.
The limiting radiant exposure per pulse, therefore, is
1.1 × 10
H (MPE)rule 2 =
−3 J
cm2 5 × 105 pulses
= 2.2 × 10−9 J/cm2
pulse .
The irradiance (E ) limit that corresponds to a given H(MPE) for a repetitively
pulsed beam for an exposure time t seconds is given by
E (MPE)group = PRF × t × H(MPE) J/cm2
, (14.12)
pulse
where PRF is the pulse repetition frequency. When we substitute the appropriate
values into Eq. (14.12), we obtain
5 pulses
J/cm2
E (MPE) = 5 × 10 × 0.5 second × 2.2 × 10−9
s pulse
−4 W mW
= 5.5 × 10 = 0.55 .
cm2 cm2 According to rule 3, the MPE per pulse must be reduced by the factor Cp = n –0.25 .
Since the number of pulses n during the exposure period = 5 × 10 5 , the reduced
MPE per pulse is calculated from
H (MPE)rule 3 = n −0.25 × MPEsingle pulse
= 5 × 10 5 −0.25 × 5 × 10 −7
and the corresponding irradiance, from Eq. (14.11) is:
J
J/cm2
= 1.9 × 10−8
cm2 pulse ,
5 pulses
J/cm2
E (MPE)rule 3 = 5 × 10 × 0.5 second × 1.9 × 10−8
s pulse
−3 W mW
= 4.75 × 10 = 4.75 .
cm2 cm2 The lowest calculated MPE, 2.2 × 10 −9 J/cm 2 per pulse, corresponding to 0.55
mW/cm 2 , is the MPE that applies to this case.