## HW 1a 7b

$c=\lambda v$

Lung Sheng Liang 3J
Posts: 90
Joined: Wed Sep 30, 2020 9:33 pm

### HW 1a 7b

When an electron beam strikes a block of copper, x rays with a frequency of 2.0 x 10^18 Hz are emitted. What is the frequency of the wavelengths in picometers?

Does light striking the block of copper affect the calculations at all? Or is it just a simple into plug and chug into c= λv? Also, I got 150pm, is this right?

Posts: 93
Joined: Wed Sep 30, 2020 10:00 pm

### Re: HW 1a 7b

Yes the answer is 150pm. You could check your answers on sapling btw. It's in the homepage of sapling titled Atkins_7e_SSM. Yeah, I would basically plug and chug into that equation.

Nathan Tong 3G
Posts: 91
Joined: Wed Sep 30, 2020 9:49 pm

### Re: HW 1a 7b

Yes, you are correct. All you need to do to solve the problem is plug in c = 3 x 108 and v = 2.0 x 1018 into λ = c/v, and then convert to picometers by multiplying by 1012.

Dylan_Nguyen_2C
Posts: 87
Joined: Wed Sep 30, 2020 9:41 pm

### Re: HW 1a 7b

For this one, I first plugged in the values for c = λv equation
2.99x10^8 = λ(2.0x10^18)
λ = (2.99x10^8)/(2.0x10^18)
λ = 1.495x10^-10
Then I multiplied the result by 1x10^12 to convert to picketers
(1.495x10^-10)(1x10^12) = 149.5 pm
You are correct; the answer is around 150 pm

rhettfarmer-3H
Posts: 101
Joined: Wed Sep 30, 2020 9:59 pm

### Re: HW 1a 7b

This problem is simple rearrangement of the equation c=wavelength times frequency. Thus, to relate frequency to wavelength we get wavelength=c/v. so now we just plug in the wave length we got and then we get it in meters. From here we must convert m to picom. Which is 1m is 10^12 pico meters. so multiply by 10^12. AS SHOWN IN WORK.
Attachments
IMG_2782.pdf