The energy $E$ is given, from Einstein's formula, as: $E=h*nu$ Where: $h=6.63×10^-34Js$ is ; $nu$ is the frequency. You can relate frequency and wavelength $lambda$ through the speed of light $c$ as: $c=lambda*nu$ So finally you get: $E=hc/(lambda)$ And: $lambda=(hc)/E=(6.63×10^-34*3×10^8)/(3.097×10^4)=6.4×10^-30m$
You'd use the equation that establishes a relationship between energy and wavelength, which looks like this $E = h * c/(lamda)$, where $lamda$ - the wavelength of the photon;...
The inverse relationship that exists between wavelength and frequency is described by the equation $color(blue)( lamda * nu = c)" "$, where $lamda$ - the frequency of...
First find the frequency of the photon from the equation $c=flambda$ $therefore f=c/lambda=(3xx10^8m//s)/(0.050xx10^(-2)m)=6xx10^11Hz$. Energy of such photons is quantized and given by $E=hf$ $=(6.6xx10^(-34))(6xx10^11)$ $=3.96xx10^(-22)J$
The speed is given by: $c=fxxlambda$ A red photon has a longer wavelength ( $~700"nm"$ ) than a blue photon ($~400"nm"$). Consequently it has a lower frequency as does the...
Let 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...
Relativistic 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...
About $"9.78 keV"$, 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)...
