Increasing the kVp from 60 to 69 kVp with identical technique parameters yields receptor exposure of which value?

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Multiple Choice

Increasing the kVp from 60 to 69 kVp with identical technique parameters yields receptor exposure of which value?

Explanation:
When you raise the kilovoltage while keeping all other exposure factors the same, the beam becomes more penetrating and delivers more photons to the image receptor. A common approximation for how receptor exposure changes with kVp at fixed mA and time is that it varies with the square of kVp: I ∝ (kVp)^2. So the new exposure is the old exposure multiplied by (69/60)^2. That factor is about (1.15)^2 ≈ 1.32, a roughly 32% increase. If the exposure at 60 kVp was 15 Gy, it becomes 15 × 1.32 ≈ 19.8 Gy, which rounds to about 20 Gy. Therefore the receptor exposure is about 20 Gy. The other values would require a smaller or larger change than this 32% increase.

When you raise the kilovoltage while keeping all other exposure factors the same, the beam becomes more penetrating and delivers more photons to the image receptor. A common approximation for how receptor exposure changes with kVp at fixed mA and time is that it varies with the square of kVp: I ∝ (kVp)^2.

So the new exposure is the old exposure multiplied by (69/60)^2. That factor is about (1.15)^2 ≈ 1.32, a roughly 32% increase. If the exposure at 60 kVp was 15 Gy, it becomes 15 × 1.32 ≈ 19.8 Gy, which rounds to about 20 Gy.

Therefore the receptor exposure is about 20 Gy. The other values would require a smaller or larger change than this 32% increase.

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