The diagram on the left shows an arrangement used to measure the masses of ions.
ID: 1521653 • Letter: T
Question
The diagram on the left shows an arrangement used to measure the masses of ions. An ion of mass m and charge q is produced essentially at rest as rest source S. The ion is accelerated by a potential difference of V volts and allowed to enter the magnetic field B. In the field it moves in a semicircle, striking a photographic plate at a distance x from the entry- slit. The ion inters the magnetic at a velocity of 1.7 x 10^6 m/s. The mass of the ion is 1.06x 10-26 kg. Calculate the value of x. the diameter of the circular path of the ion in the in the magnetic field B. B -1.5T q = 1.602 x 10^-19 C 5)Explanation / Answer
ANSWER
From energy equation we know that,
(1/2)mv2 = qV
v2= 2qV/m (1)
In the magnetic field
F = qvB = mv2/r
m = qrB/v (2)
From (1) and (2) we get that,
m = qrB. (m/2qV)
m2 = q2 r2 B2 m/(2qV)
m = q B2 r2/(2V)
So, r = x
m = q B2 x2/(8V) (3)
But the equation above can be expressed as,
m = qBx/2v
Thus,
x = 2mv/qB
x = 2*(1.06x10-26)*(1.7x106) / (1.602x10-19)*(1.5)
x = 0.1499 m ANS. Part a)
To solve part b) express equation (3) on terms of voltage,
V = q B2 x2/(8m)
V = (1.602x10-19)*(1.5)2*(0.1499)2 / 8*(1.06x10-26)
V = 95510.79 V ANS.
Regards!!!