Find wavelength of a quantum of electromagnetic radiation

trivk96

1. The problem statement, all variables and given/known data
A quantum of electromagnetic radiation has
an energy of 0.877 keV.
What is its wavelength? The speed of light
is 2.99792 × 10
8 m/s, and Planck’s constant
is 6.62607 × 10−34J · s.
Answer in units of nm


2. Relevant equations
E=hf
v=fλ
... λ=v/(E/h)

3. The attempt at a solution

When i solved, i got 1.413728e-9 nm... I have checked my units. can some just help and point me in the right direction

Dickfore

v is usually written as c when one speaks of the speed of light in vacuum. Also, the double fraction reduces to:
[tex]
\frac{c}{\frac{E}{h}} = \frac{h \, c}{E}
[/tex]
For this answer, you need to know the conversion factor between an electron-volt (eV) and a joule as energy units. Do you know it?

trivk96

Yes i did convert it but i still got it wrong

Dickfore

how did you convert it, and what did you get?

trivk96

I did it again and i got 1.414E-8 ... and i think that is in meters. Am i right??

so that means that the answer is14.14nm
_______________________________________________________________________ _

I used plancks constant in eV's. Its on the ap equation sheet

Dickfore

I didn't get that. What did you get for the energy in joules?

trivk96

1.405109518e-16 J

Dickfore

This is correct. Now:
[tex]
\frac{h \, c}{E} = \frac{6.626 \times 10^{-34} \, \mathrm{J} \cdot \mathrm{s} \times 2.998 \times 10^8 \, \mathrm{m} \cdot \mathrm{s}^{-1}}{1.4051 \times 10^{-16} \, \mathrm{J}}
[/tex]

The product and ratio of the mantissas, gives:
[tex]
\frac{6.626 \times 2.998}{1.4051} = 14.14
[/tex]
The exponents sum up to [itex]-34 + 8 - (-16) = -10[/itex]. You may read off the units from the above fraction fairly easily.

What should the answer be in scientific form?

trivk96

so in nm, it would be 1.414

Dickfore

yes, except that you need to use as many significant figures, as there are in variable with the least number of significant figures given in the problem. Fundamental constants are usually known to a lot of significant figures.

trivk96

Thank You