A nurse plans care for a client who has an external fixator on the lower leg. Which intervention would the nurse include in the plan of care to decrease the client’s risk for infection?
Washing the frame of the fixator once a day.
Releasing fixator tension for 30 minutes twice a day.
Avoiding moving the extremity by holding the fixator.
Scheduling for pin care to be provided every shift.
The Correct Answer is D
Choice A reason: Washing the fixator frame may maintain hygiene but does not directly address the primary infection risk at pin sites, where skin breaks allow bacterial entry. Pin sites are more critical, as infections like osteomyelitis can develop from poor pin care.
Choice B reason: Releasing fixator tension disrupts the device’s stability, risking bone misalignment and delayed healing. It does not reduce infection risk and may increase tissue trauma, potentially creating more entry points for bacteria at the pin sites.
Choice C reason: Avoiding movement by holding the fixator prevents normal joint mobility, risking stiffness, but does not address infection risk. Pin sites remain the primary infection source, as bacteria can enter through skin breaks, requiring specific cleaning protocols.
Choice D reason: Regular pin care (every shift) with sterile technique (e.g., chlorhexidine) removes debris and bacteria from pin sites, reducing the risk of osteomyelitis. External fixators penetrate skin, creating infection-prone areas, making frequent, meticulous pin care the most effective intervention to prevent infection.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is ["0.6"]
Explanation
Step 1 is: Convert pounds to kilograms 33 ÷ 2.2 = 15 Result = 15 kg
Step 2 is: Multiply weight by dosage 15 × 20 = 300 Result = 300 mg/day
Step 3 is: Divide total daily dose into 4 doses 300 ÷ 4 = 75 Result = 75 mg per dose
Step 4 is: (125 ÷ 1) = 125 mg/mL
Step 5 is: (75 ÷ 125) = 0.6 Result = 0.6 mL per dose
Final answer = 0.6 mL per dose
Correct Answer is A
Explanation
Choice A reason: A seizure lasting over 5 minutes is considered status epilepticus, a medical emergency requiring immediate intervention to prevent brain hypoxia or injury. Prolonged seizures increase glutamate release, causing excitotoxicity, neuronal damage, and systemic complications like respiratory failure, necessitating urgent 911 response.
Choice B reason: Postictal sleepiness and lethargy are common after generalized seizures due to neuronal exhaustion and temporary metabolic changes in the brain. This is expected and not an emergency unless prolonged or accompanied by other concerning symptoms like respiratory distress.
Choice C reason: Falling at seizure onset is typical in generalized seizures due to loss of muscle control. Unless injury (e.g., head trauma) is suspected, it does not warrant an immediate 911 call, as it is a common seizure manifestation.
Choice D reason: Postictal confusion and slurred speech are normal after a generalized seizure, reflecting temporary cortical dysfunction. These resolve within minutes to hours and do not require emergency services unless persistent or associated with other critical symptoms like unresponsiveness.
Whether you are a student looking to ace your exams or a practicing nurse seeking to enhance your expertise , our nursing education contents will empower you with the confidence and competence to make a difference in the lives of patients and become a respected leader in the healthcare field.
Visit Naxlex, invest in your future and unlock endless possibilities with our unparalleled nursing education contents today
Report Wrong Answer on the Current Question
Do you disagree with the answer? If yes, what is your expected answer? Explain.
Kindly be descriptive with the issue you are facing.