consider a 1.2m long fluorescent tube with 120V (mains) between the cathode and

Question

consider a 1.2m long fluorescent tube with 120V (mains) between the cathode and anode (the 2 ends).

In reality an electron only travels about 0.03m down the tube before hitting an atom. If the voltage change is proportional to distance travelled down the tube, and the electron starts from rest, how much KE does it have just before hitting the atom?

Question 14 options:

4.8 E-19 J

7.6 E-16 J

5.7 E-19 J

3.3 E-23 J

None of the above

A)

4.8 E-19 J

B)

7.6 E-16 J

C)

5.7 E-19 J

D)

3.3 E-23 J

E)

None of the above

Explanation / Answer

voltage is directly proportional to distance

and voltage at 1.2 m ling tube is 120 V so

V = 120 * x / 1.2

work = charge * voltage

work = 1.6 * 10^-19 * 120 * x / 1.2

at distance 0.03 m work done will be

work done = 1.6 * 10^-19 * 120 * 0.03 / 1.2

work done = 4.8 * 10^-19 J

workdone = change in kinetic energy

initial kinetic energy = 0 since it starts from rest so,

kinetic energy = 4.8 * 10^-19 J

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.