Why does an electrostatic chopper reduce peak current in a He⁻ beam?

[elduderino]

Homework Statement

Problematic Part of the problem:

Consider a negative ion beam of He- ions. The given beam characteristics are the beam energy (E) and the beam current (I).

An electrostatic chopper is placed on the beam line, which pulses the DC beam with a pulse width of say $$t_1$$ nanoseconds.

What are the new beam characterisitcs?

Homework Equations

Normal Distribution:
$$f(x)= \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{t^2}{2\sigma^2}}$$

The Attempt at a Solution

I would have said the new current profile would be, for an incident DC beam of beam current I0

$$I(t)=I_0\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{t^2}{2\sigma^2}}$$

if the beam emerges from the chopper at t=0. Also, since the FWHM of the beam is given to be $$t_1$$ the standard deviation would be

$$\sigma=\frac{t_1}{2.354}$$

(from http://en.wikipedia.org/wiki/Full_width_at_half_maximum)

This seems pretty straightforward, however, this appears to be wrong because according to this, the peak current has changed, and become .046 times its initial value

$$I(0)=I_0 f(0)$$ where f(x) is the normal distribution...can anyone tell me what I am doing wrong.

 

[elduderino]

anyone?

you can just tell me how a dc beam is chopped to a gaussian pulse...