Question

Question #db111

Answer 1

4.82 keV photon

Energy `E = (hc)/lambda implies lambda = (6.63 xx 10^(-34) cdot 3 xx 10^8)/(4.82 xx10^3 cdot color(red)(1.6 xx 10^(-19)) )approx 0.26 " nm"`

[Term in red: An eV is the energy required to increase the potential of an electron by 1 Volt so: `1 "eV" = e^(-) " J"`]

4.82 keV neutron

Here we do use the De Broglie relation `lambda = "h/p"`.

The neutron's energy is: `E = p^2/(2m) implies p = sqrt(2Em) implies lambda = h/sqrt(2Em_n)`

That is:

`(6.63xx10^(-34))/sqrt(2 ( 4.82 xx10^(3) cdot color(red)(1.6 xx 10^(-19)) ) 1.67 xx10^(-27) ) approx 0.4 xx 10^(-12) " m"`