Neutron dose

At the maze entrance, the neutron dose equivalent is usually the dominant component for high energy accelerators above 10 MV. The neutron dose at any point in the maze depends on several factors, including the distance from the inner maze point A to the isocentre (d₁), the surface area S of the treatment room, the inner maze entrance cross-sectional area (A_r) and the cross-sectional area of the maze (S₁). It is also a function of the energy, the gantry angle and the field size of the photon beam.

At the inner maze point A (shown in Fig. 8), Eq. (23) gives the total neutron fluence, which is a function of d₁, S and Q_N. To reduce the neutron fluence at A, a longer distance d₁, or a smaller inner maze entrance area A_r, may be chosen when designing the room. A larger room size will also reduce the neutron dose at A.

D _A

d j =5 7 10. ¥ ^-16¥j ¥10

-2

6.2

D_c=W¥D_j

Neutron dose decreases along the maze with a tenth value length (T_N) of 5–7.5 m for many medical accelerator rooms. The value of T_N depends on the cross-sectional area of the maze. A smaller cross-sectional area will allow more interactions between the neutron and the wall, and thus reduces the dose measured at the maze entrance, and the tenth value length.

The neutron dose in the maze is highest at the gantry angle when the head of the accelerator is closest to the inner maze entrance. This is because the neutrons produced leave the head in all directions as the high Z head shielding material has little effect in stopping them. The lowest dose is found when the gantry head is farthest away, even when the photon beam may be pointing at the direction of the inner maze entrance. It is not unusual to see a difference of a factor of 2 in dose between the two gantry angles. When the beam is pointing downward, the dose is slightly higher than the average of the two extreme cases [29]. For shielding calculation purposes, it is considered appropriate to use the neutron dose data with the beam pointing down.

Smaller field sizes will result in a higher neutron dose at any point down the maze at a distance more than 1 m from point A. Comparing the doses at the same point with the beam set at the largest and smallest field sizes, the difference is about 10–20% for a maze of the design described in this publication [29]. Therefore, for conservative reasons, collimators are assumed to be at the fully closed position when making neutron dose estimates.

McGinley and Butker [30] evaluated the neutron dose equivalent at the maze entrance of a number of high energy medical accelerator facilities, and compared their results with the empirical method developed by Kersey [31].

They found that Kersey’s method in general produced higher dose estimates, and therefore it is conservative in nature for purposes of shielding requirement calculations. Kersey’s method gives the neutron dose equivalent at the maze entrance as follows:

(26) where

D_n is the neutron dose equivalent at the maze entrance, in Sv per X ray·Gy at the isocentre.

H₁ is the neutron dose equivalent at 1 m from the X ray source (target) in mSv per X ray·Gy at the isocentre. Values of H₁ are tabulated in Table 10.

A_r and S₁ are cross-sectional areas, in m², of the inner maze entrance and the maze, respectively.

D_n =H₁¥10^-3¥(A_r/S₁) ( /¥ 1 d₁)²¥10^-d2/5

d₁ is the distance, in m, from the isocentre to the inner maze point A as defined above.

d₂ is the distance, in m, from the inner maze point A to the outer entrance of the maze.

For a maze with an additional bend as shown in Fig. 9, the dose at the maze entrance is given by the equation below [14]:

(27) where

d₂ is the distance, in m, from point A to point B in Fig. 9;

d₃ is the distance, in m, from point B to the maze entrance.

From this equation it is evident that the addition of a bend in the maze design reduces the neutron dose at the maze entrance by a factor of 1/3 for the same total maze length. This is because the majority of the neutrons will encounter more collision interactions with the maze wall before exiting the maze entrance. This reduction will not hold if one of the maze bends is too short, or the cross-sectional area of the maze or the maze entrance is too large.

The reader is cautioned to evaluate the specific situation for the validity of the equation.

For accelerator facilities having a structural design similar to the one shown in Fig. 8, the neutron dose equivalent may be obtained using an alternative method developed by Wu and McGinley [29]. They found that the neutron dose decreases along the maze with a tenth value length proportional to the square root of the cross-sectional area of the maze:

(28)

where

T_N is the tenth value length, in m;

S₁ is the cross-sectional area of the maze, in m.

Wu and McGinley [29] also found that the neutron dose equivalent at a point along the maze is given by the equation:

D_n =H₁¥10^-3¥(A_r/S_l) ( /¥ 1 d₂)²¥(10^-d2/5) (¥ 10^-d3/5) ( / )¥ 1 3

T_N =2 06. ¥ S₁

(29)

where

D_n is the neutron dose equivalent at the maze entrance, in Sv per X ray·Gy at isocentre;

ϕ_A is the neutron fluence given by Eq. (23).

Equations (26), (27) and (29) all give reliable dose equivalent estimates.

Equation (27) usually produces more conservative estimates for shielding purposes. For treatment room designs of exceptional size, or mazes of exceptional width or length, Eq. (29) will produce more accurate results. The reader is advised to evaluate the merits of both equations before choosing the value to obtain the estimated weekly dose due to neutrons.

The weekly dose due to neutrons is given by the following equation:

D_E = W × D_N (30)

A

B C

d

d

d

FIG. 9. Room with two bends in the maze showing distances used to determine the capture gamma dose.

D_{n A} A S_r

d d

T_N

= ¥ ¥ ¥ ¥ ¥ +

È Î Í

- -Ê

ËÁ ˆ

¯˜ -Ê ËÁ

ˆ

¯˜

2 4 10. ¹⁵ j / ₁ 1 64 10. ^1.9 10

2 2

ÍÍÍ

˘

˚

˙˙

˙

where

D_E is the weekly dose equivalent due to neutrons, in Sv·week^–1; W is the weekly workload (Gy·m²).

The total weekly dose D_W at the external maze entrance is the sum of all three components: D_d from Eq. (18), D_c from Eq. (25) and D_E from Eq. (30):

D_W = D_d + D_c + D_E (31)