X-Ray Production in Beiser's Physics Textbook: Typo or Not?

[asdf1]

in my modern physics textbook 6/E by arthur beiser,
the formula for x-ray production is written as
(namdamin)=1.24*10^(-6)/V V*m
is that a typo? how can namda have units of V*m?

 

[HallsofIvy]

Perhaps it would help if you explained exactly what you are talking about!

As far as I know "x-ray production" doesn't have units! What exactly does
"nambdamin" mean? (Was that supposed to be "lambda"?) Oh, and what is V? volts?

 

[asdf1]

x-ray production doesn't have units?
jeepers~
i must be misunderstanding modern physics...
:P
i don't know how to type the symbols, but yes, "nambda min" means the "lambda" minimun...
V stands for volts...

 

[lightgrav]

the second V (that's multiplied by meters) is a UNIT, Volts.
But the first V (that you divide by) is a VARIABLE ... Voltage.

These are NOT the same thing, but the units cancel.

"the lamda minimum ..."
lamda of WHAT? V of WHAT?

 

[Gokul43201]

in my modern physics textbook 6/E by arthur beiser,
the formula for x-ray production is written as
(namdamin)=1.24*10^(-6)/V V*m
is that a typo? how can namda have units of V*m?

You are talking about the minimum wavength of x-rays produced by accelerating electrons through a voltage V and slamming them against a target. Naturally, the maximum energy of an x-ray will be the final KE of the electron before it hits the target.

That gives

$$KE(electron) = e*V = \frac{hc}{\lambda_{min}}$$
$$\implies~ \lambda_{min} = \frac{hc}{eV} = \frac {hc}{e} \cdot \frac{1}{V}$$

The quantity $hc/e$ has a value of $1.24 *10^{-6}~Vm$

When you divide by the units of the applied voltage V (ie: volts), you are left with units of meters, the correct unit for a wavelength !

 

[asdf1]

thank you very much! :)