The first thing to do here is to calculate the ionization energy for one atom of aluminium, $"Al"$, by suing the fact that you know the energy needed to ionize...
1 Answers 1 viewsThe energy of a photon is proportional to its frequency, as stated by the Planck - Einstein's equation $color(blue)(E = h * nu)" "$, where $E$ -...
1 Answers 1 viewsThe inverse relationship that exists between wavelength and frequency is described by the equation $color(blue)( lamda * nu = c)" "$, where $lamda$ - the frequency of...
1 Answers 1 viewsLet us calculate the frequency $nu$ of the photon. We have the expression between the velocity of light in vacuum $c=nuxxlambda$ or $nu=c/lambda$......(1) where $lambda$ is the wave-length of...
1 Answers 1 viewsFor electromagnetic radiation: $c=flambda$ where $c$ is the speed of light, $f$ is the frequency, $lambda$ is the wavelength. Rearranging: $lambda = c/f=(3.0xx10^8)/(3.0xx10^10)=10^-2$ $m$ The energy of a photon is...
1 Answers 1 viewsTo find the energy of a photon you use the Planck Expression: $sf(E=hf)$ $sf(h)$ is the Planck Constant which = $sf(6.63xx10^(-34)color(white)(x)J.s)$ $sf(f)$ is the frequency This becomes: $sf(E=(hc)/lambda)$ $sf(c)$ is...
1 Answers 1 viewsIt has no rest mass. It's a photon. It's light. You CANNOT logically use $lambda = h/(mv)$ on a photon. Doing so violates Einstein's theory of relativity...
1 Answers 1 viewsWe're asked to calculate the energy of one photon of a green light, given its wavelength of $5200$ $"Å"$. To do this, we can use the equation...
1 Answers 1 viewsRelativistic effects are present here... but I assume that the transition from level $a$ to level $g$: has $DeltaL = pm 1$ (the total change in angular momentum is...
1 Answers 1 viewsAbout $"9.78 keV"$, assuming $100%$ transfer of energy. The de Broglie wavelength $lambda$ is given by $lambda = h/p = h/(mv)$, where: $h = 6.626 xx 10^(-34)...
1 Answers 1 views