Suppose a particle of ionizing radiation deposits 1.50 MeV in the gas of a Geige

Question

Suppose a particle of ionizing radiation deposits 1.50 MeV in the gas of a Geiger tube, all of which goes to creating ion pairs. Each ion pair requires 30.0 eV of energy. (a) The applied voltage sweeps the ions out of the gas in 1.10 µs. What is the current in Amps? (b) This current is smaller than the actual, since the applied voltage in the Geiger tube accelerates the separated ions, which then create other ion pairs in subsequent collisions. What is the current if this last effect multiplies the number of ion pairs by 980? A

Explanation / Answer

E = 1.5*10^6 eV
energy, per ion pair creation = 30 eV
Number of ion pairs, n = 50000

a) I = Q / t
Q = n*e = 50000*1.6*10^-19 = 8*10^-15 C
t = 1.1*10^-6 s
I = 7.2727*10^-9 A
b) I' = I*980 = 7.1272*10^-6 A

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

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