Here is the formula for the deBroglie wavelength of an electron versus its kinetic energy: $$ \lambda(E_k) = h\left/\sqrt{\frac{(E_k+m_eC^2)^2-m_e^2C^4}{C^2}}\right.$$ and here is simple script that prints a CSV with plot data:

import csv
from scipy import constants as spc
import numpy as np

vev = np.arange(1,1e6,100);
vWavelength = map( lambda n: \
               spc.h/np.sqrt( (np.power(spc.eV*n+spc.m_e*np.power(spc.c,2),2)-np.power(spc.m_e,2)*np.power(spc.c,4))/np.power(spc.c,2) ), \
               vev );
np.savetxt('plotdata/1a.csv', np.transpose(np.array([vev, vWavelength])));

The vWavelength line is abysmal, in my opinion. How can I make this code better?


1 Answer 1


First, simplify: factor out \$C^4\$ out of square root:

\$\lambda(E_k) = h\left/C\sqrt{(\frac{E_k}{C^2}+m_e)^2-m_e^2}\right.\$

Simplify even more: factor out \$m_e^2\$:

\$\lambda(E_k) = h\left/C m_e\sqrt{(\frac{E_k}{(C m_e)^2}+1)^2-1}\right.\$

Do not put everything in lambda:

# These are all constants; no need to recompute them.
# Consider putting them into a namespace.
# I didn't think too hard to figure appropriate names; my physics is too rusty

CM = spc.c * spc.me
e1 = spc.ev/(CM * CM)
scale = spc.h / CM

# This is an actual computation.
# Can be converted to lambda, but I strongly advise against.

def deBroglieWaveLength(n):
    arg = e1 * n + 1.0
    return scale / np.sqrt(arg * arg - 1)
  • \$\begingroup\$ Why avoid lambdas? \$\endgroup\$ Commented Mar 24, 2017 at 21:10
  • \$\begingroup\$ @enthdegree Mostly to give it a clear and definite name. \$\endgroup\$
    – vnp
    Commented Mar 24, 2017 at 21:53

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