How Should You Round Photon Energy Changes in Physics Calculations?

[cupcake]

please help me,, Mastering physic.. rounding number :(

A photon of wavelength 4.38×10^−2 nm strikes a free electron and is scattered at an angle of 38.0 degree from its original direction.

Find the wavelength of the scattered light.
and my answer is this 4.43×10−2 nm (correct answer)

Find the change in energy of the photon.

E= hc / lambda1 = 1.24*10^-6 / (0.0438*10^-9) = 28310.50 eV

E= hc/ lambda2 = 1.24*10^-6 / (0.0443*10^-9) = 27990.97

and the change in energy is 319.529..
i have tried to enter 319 or 320 but mastering physics gave me

"Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

please help to round this... TT

***hate mastering physics...

 

[Redbelly98]

Since the wavelength was given to 3 significant figures, the energies are each known to 3 significant figures.

This makes the two energies accurate to the nearest ___ eV.

Does that help?

 

[cupcake]

This makes the two energies accurate to the nearest ___ eV.

Does that help?

what did you mean by the nearest ____ eV??

so, what number should i enter?
320 and 319 don't help much...

i already lost some points because of it.. :(

 

[Borek]

E= hc/ lambda2 = 1.24*10^-6 / (0.0443*10^-9) = 27990.97

If I understand correctly you have used rounded down 443 to calculate energy here - you should use whatever came out from the calculator. Use full precision when calculating, rounded down values are for reporting only.

 

[Redbelly98]

what did you mean by the nearest ____ eV??

I mean, is it accurate to the nearest 1eV, 10eV, 100eV, or what? Remember, we're using 3 significant figures here.