Mass of singly charged uranium-238

[neik]

A singly charged uranium-238 ions are accelerated through a potential difference of 2.00 kV and entered a uniform magnetic field of 1.20 T directed perpendicular to their velocities. Determine the radius of their circular path

can some one tell me what is the mass of this singly charged uranium-238 ions?

:uhh:

thanks in advance.

 

[dextercioby]

$$m_{_{92}^{238}U^{+}} \ = 92\cdot m_{p}+146\cdot m_{n}+91\cdot m_{el}$$

Express the answer in Kg.

Daniel.

 

[thepatientmental]

The mass of a singly charged uranium-238 ion can be calculated using the formula m = qB^2r^2/2V, where m is the mass of the ion, q is its charge, B is the magnetic field strength, r is the radius of the circular path, and V is the potential difference. Plugging in the given values, we get:

m = (1)(1.20 T)^2r^2/2(2.00 kV)

= 1.44r^2/4.00 x 10^6

To determine the radius of the circular path, we can use the formula r = mv/qB, where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field strength. Since the ions are accelerated through a potential difference, their velocity can be calculated using the formula v = √(2qV/m). Plugging in the values, we get:

r = m√(2qV/m)/qB

= √(2V/m)/B

= √(2(2.00 kV)/(1.44/4.00 x 10^6))/1.20 T

= √(5.55 x 10^-4 m^2/kg)/1.20 T

= 0.00117 m

Therefore, the radius of the circular path for the singly charged uranium-238 ions is 0.00117 m. To determine the mass of the ions, we can plug this value back into the first formula:

m = (1)(1.20 T)^2(0.00117 m)^2/2(2.00 kV)

= 1.44(1.37 x 10^-6)/4.00 x 10^6

= 4.91 x 10^-10 kg

Hence, the mass of the singly charged uranium-238 ions is approximately 4.91 x 10^-10 kg.