A 500kV is applied across an X-ray tube. Calculate the maximum velocity of the electrons produced [Me = 9.1 x 10\(^{-31}\)kg, e = 1.6 x 10\(^{-19}\)C]

delightoke

2 years ago

The kinetic energy gained by an electron accelerated through a potential difference is given by the equation:

KE = eV

where KE is the kinetic energy, e is the charge of an electron and V is the potential difference.

The maximum velocity of an electron produced by an X-ray tube is given by the equation:

v = √(2KE/me)

where v is the velocity, KE is the kinetic energy, me is the mass of an electron.

Substituting the values given, we have:

KE = eV = (1.6 x 10^-19 C) x (500 x 10^3 V) = 8 x 10^-14 J

v = √(2KE/me) = √[(2 x 8 x 10^-14 J)/(9.1 x 10^-31 kg)]

v = 4.2 x 10^8 m/s

Therefore, the maximum velocity of the electrons produced is 4.2 x 10^8 m/s. The answer is (a) 4.2 x 10^8 m/s.


#copied from supergb

Chiamakaokpe

3 years ago

I don't Understand at all

godsgift914

5 years ago

Please can someone show me how calculate it

Algebraic922

10 years ago

The ans is B

Onyeblessil

10 years ago

The equation is: 1/2MeV^2=e*potential energy(500kV). The ans is D

Oosi001

4 months ago

WHERE IS THE EXPLANATION

ikeanwuli

2 years ago

KE = eV = (1.6 x 10^-19 C) x (500 x 10^3 V) = 8 x 10^-14 J

v = √(2KE/me) = √[(2 x 8 x 10^-14 J)/(9.1 x 10^-31 kg)]

v = 4.2 x 10^8 m/s