Problem 1A.7


Moderators: Chem_Mod, Chem_Admin

Yazmin Bocanegra 3L
Posts: 51
Joined: Thu Jul 25, 2019 12:17 am

Problem 1A.7

Postby Yazmin Bocanegra 3L » Thu Oct 17, 2019 9:14 pm

1A.7.
a) The frequency of violet light is 7.1 x 10^14 Hz. What is the wavelength (in nanometers) of violet light?
b) 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 wavelength (in picometers) of these x-rays?

I need help with this problem. Please and thank you!!

EvaLi_3J
Posts: 53
Joined: Wed Oct 02, 2019 12:16 am

Re: Problem 1A.7

Postby EvaLi_3J » Thu Oct 17, 2019 9:47 pm

A) Just use the equation c = wavelength * frequency. All light (including those that we cannot see like infrared and ultraviolet) have the same velosity, which is denoted by c, and it is equal to 3.0*10^8 m/s (this is not the more precise though, you might need to look up the precise value for precision). Now, you are given the frequency and you know the speed of light, you can easily calculate the wavelength. Just remember that 1 nanometer is 1*10^(-9) meter.

B) X-ray also has a velocity of c, so you can basically calculate this problem like a) above.


Hope that helps.

Daniel Honeychurch1C
Posts: 109
Joined: Thu Jul 11, 2019 12:15 am

Re: Problem 1A.7

Postby Daniel Honeychurch1C » Thu Oct 17, 2019 9:48 pm

Since c = λv, λ = c/v = (3.0 x 10^8 m/s) / frequency. For both problems, plug in the given frequencies to get the wavelength. Then convert to manometers ( 1 m = 10^9 nm, 1 m = 10^12 pm).

Suraj Doshi 2G
Posts: 100
Joined: Fri Aug 02, 2019 12:15 am

Re: Problem 1A.7

Postby Suraj Doshi 2G » Thu Oct 17, 2019 9:57 pm

For both parts we can use the equation: c = (lambda) * (v-frequency).

The equation for part a turns out to be:
3.00*10^8 = (lambda) * 7.1 x 10^14 Hz.

The equation for part b turns out to be:
3.00*10^8 = (lambda) * 2.0 x 10^18 Hz.
Just be sure to convert to picometers.

Tiffany Vo 3G
Posts: 52
Joined: Sat Aug 17, 2019 12:17 am

Re: Problem 1A.7

Postby Tiffany Vo 3G » Thu Oct 17, 2019 10:01 pm

For the first question, you'd have to use c = wave length x frequency, so you'd divide 3 x 10^8 meters per second by 7.1 x 10^14 Hz. If it helps, 1 Hz is equivalent to 1/second, so your units should cancel out and leave you just your wavelength. After that, it's just a matter of converting the meters to nanometers by multiplying by 10^9. The same process can be applied to the second question, but you'd have to multiply by 10^12 to convert from meters to picometers.

Aprice_1J
Posts: 55
Joined: Wed Sep 18, 2019 12:16 am

Re: Problem 1A.7

Postby Aprice_1J » Fri Oct 18, 2019 9:15 pm

When first looking at this problem something that is important to note is the change in units. To make sure that this doesn't get you confused because that is a little thing that is easy to overlook, I like to set it up as a fraction and make sure that all of the units cancel out. Some of the conversions will be given to you but make sure to memorize the SI untis. As for the problem you are looking to go from frequency to wavelength for both parts so you will be using c=v(frequency)x wavelength. C is a constant and then you plug in what you are given for frequency to find wavelength.


Return to “Properties of Light”

Who is online

Users browsing this forum: No registered users and 1 guest