Quick question: Finding the wavelength?

[nukeman]

Quick question: Finding the wavelength??

Homework Statement

The energy of a photon is 5.87 x 10^-20 J. What is its wavelength in nanometers?

I use E = H(c/λ) Correct?

Since I am trying to find λ, I change it to this?

λ = H(c/E) ?

6.63x10^-34 x (3 x 10^8 / 5,87 x 10^-20)

= .000003

in nanometers, 3000

Is this correct? I don't think it is :(

Homework Equations

The Attempt at a Solution

 

[gneill]

You've lost a lot of precision through rounding. Why not stick to scientific notation for intermediate results? 3000 nm is in the ballpark, but you should have a couple more significant figures.

 

[nukeman]

You've lost a lot of precision thr9ough rounding. Why not stick to scientific notation for intermediate results? 3000 nm is in the ballpark, but you should have a couple more significant figures.

Ohh... So, I was correct? as in my method for solving this?

My damm $150 calculator seems to not want to give me all the sig figs lol.

 

[collinsmark]

Homework Statement

The energy of a photon is 5.87 x 10^-20 J. What is its wavelength in nanometers?

I use E = H(c/λ) Correct?

Since I am trying to find λ, I change it to this?

λ = H(c/E) ?

6.63x10^-34 x (3 x 10^8 / 5,87 x 10^-20)

= .000003

in nanometers, 3000

Is this correct? I don't think it is :(

It's pretty close to being correct. But not very precise.

I think maybe your calculator is truncating some of the significant figures perhaps. When you arrived at the answer "= .000003", did the result stop at the '3' because the calculator was unable to display more digits? If so, you might want to set up your calculator to display in scientific notation (or engineering notation) if you can. If your calculator doesn't have this capability, you might try a trick by setting c = 3 x 10⁸⁺⁹ nm/s.* The answer from that naturally comes out in nanometers.

By the way, Planck's constant is conventionally denoted by lower-case 'h'. It's not that big of deal; it's only a convention. But h is a very important physical constant, so it's a pretty standard convention.

*(Be careful when changing units like I described. It works in this case because h = 6.63 x 10^-34 has units of J·s, and your 5.97 x 10^-20 energy has units of Joules. The Joules unit from h cancel the Joule unit from E. In other physics problems that don't have this type of perfect cancellation, you would also have to convert Joules from units of kg·m²/s² to some other units like kg·nm²/s². You don't have to do that in this case because of the cancellations. But in general, just be careful.)

 

[gneill]

Ohh... So, I was correct? as in my method for solving this?

My damm $150 calculator seems to not want to give me all the sig figs lol.

Your method worked out fine. I usually remember $E = h \nu$ (where $\nu$ is the frequency) and $\nu = c/\lambda$.

 

[collinsmark]

Ohh... So, I was correct? as in my method for solving this?

Your method looks good to me.

My damm $150 calculator seems to not want to give me all the sig figs lol.

A $150 calculator should have to capability to display in scientific notation, or "engineering" notation that I like to use (in "engineering" notation exponents are always a multiple of 3, so it's a no-brainer to convert to mega, kilo, milli, micro, nano, etc.).