Respuesta :
Answer:
(a) [tex]r_{d}=\sqrt{2}\times r_{P}[/tex]
(b) [tex]r_{\alpha}=\sqrt{2}\times r_{P}[/tex]
Explanation:
charge on proton, q = e
mass pf proton, m = mp
charge on deuteron, q' = e
mass pf deuteron, m' = 2mp
charge on alpha particle, q'' = 2e
mass pf alpha particle, m'' = 4mp
Potential difference = V
Magnetic field = B
The kinetic energy is given by eV.
So, 1/2 mv^2 = e V
where, v be the velocity of the particle.
[tex]v=\sqrt{\frac{2eV}{m}}[/tex]
The formula for the radius of circular path is given by
[tex]r = \frac{mv}{Bq}[/tex]
By substituting the value of v
[tex]r = \frac{\sqrt{2eVm}}{Bq}[/tex]
[tex]r\alpha \frac{\sqrt{m}}{q}[/tex]
The radius of path of proton is given by
[tex]r_{p}\alpha \frac{\sqrt{m_{p}}}{e}[/tex].... (1)
(a) radius of deuteron
[tex]r_{d}\alpha \frac{\sqrt{2m_{p}}}{e}[/tex]
By comparing equation (1), we get
[tex]r_{d}=\sqrt{2}\times r_{P}[/tex]
(b) radius of alpha particle
[tex]r_{\alpha}\alpha \frac{\sqrt{4m_{p}}}{2e}[/tex]
By comparing equation (1), we get
[tex]r_{\alpha}=\sqrt{2}\times r_{P}[/tex]
