Please answer and SHOW ALL WORK A sample of unknown material was tested with a M
ID: 2302777 • Letter: P
Question
Please answer and SHOW ALL WORK
A sample of unknown material was tested with a Mass Spectrometer In a chamber S, a gas discharge is taking place. Ions are produced essentially at rest in S. The ions' charge are known as q = +1.6 times10 -19 C but their masses are unknown. The ions are accelerated by potential difference V =1000 volts and allowed to enter a magnetic field = 0.10T e, which is perpendicular to the paper in the figure. In the field the ions move in a semicircle, striking photographic plate(detector). We found the ions deposited at two positions: d 1 = 31.56cm,and d 2 = 36.44cm . So you know there two ions with different masses. Please determine the direction of . It is out of the paper or into the paper? Then calculate the kinetic energy of the ions, the mass of the ions m 1 and m 2. Then convert the unit of its mass from kilogram to atomic unit u, u =1.6605 x10 -27 kg. m = X u. Determine the values of X for each ion. Use the table given below to determine the elements of these ions.Explanation / Answer
r1 = d1/2 = 31.56/2 = 15.78 cm = 0.1578 m
r2 = d2/2 = 36.44/2 = 18.22 cm = 0.1822 m
workdone on ion, W = q*V
W = 1.6*10^-19*1000
= 1.6*10^-16 J
we know, W = chnage in kinetic enrgy
W = 0.5*m1*v1^2
v1 = sqrt(2*W/m1)^2
= 1.7888*10^-8/sqrt(m1)
we know, m*v^2 = q*v*B (in magnetic filed)
r = m*v/(B*q)
so,
for first particle,
r1 = m1*v1/(B*q)
r1 = m1*1.7888*10^-8/(sqrt(m1)*B*q)
0.1578 = sqrt(m2)1.7888*10^-8/(0.1*1.6*10^-19)
m1 = (0.1578*0.1*1.6*10^-19/1.7888*10^-8)^2
= 1.992*10^-26
= 19.91*10^-27 kg
x1 = 19.91*10^-27/1.6605*106-27
= 12 <<<<<<<<----------------Answer
for second particle,
r2 = m2*v2/(B*q)
r2 = m2*1.7888*10^-8/(sqrt(m1)*B*q)
0.1822 = sqrt(m2)1.7888*10^-8/(0.1*1.6*10^-19)
m2 = (0.1822*0.1*1.6*10^-19/1.7888*10^-8)^2
= 2.656*10^-26
= 26.56*10^-27 kg
x2 = 26.56*10^-27/1.6605*106-27
= 16 <<<<<<<<----------------Answer