Answer
$2.65\times10^{-8}$
$2.65\times10^{-8}$
Work Step by Step
We know that the fractional energy uncertainty is given by $\dfrac{\Delta E}{E}$ where $\Delta E=\dfrac{\hslash}{\Delta t}=\dfrac{h}{2\pi \Delta t}$ and $E=hf=\dfrac{hc}{\lambda}$.
Thus,
$$\dfrac{\Delta E}{E}=\dfrac{\dfrac{h}{2\pi \Delta t}}{\dfrac{hc}{\lambda}}$$
$$\dfrac{\Delta E}{E}= \dfrac{\color{red}{\bf\not}h}{2\pi \Delta t} \dfrac{\lambda}{\color{red}{\bf\not}hc} $$
$$\dfrac{\Delta E}{E}= \dfrac{ \lambda}{2\pi c \Delta t} $$
Plugging the known;
$$\dfrac{\Delta E}{E}= \dfrac{ 500\times10^{-9}}{2\pi \times3\times10^8 \times 10\times10^{-9} } $$
$$\dfrac{\Delta E}{E}=\color{red}{\bf 2.65\times10^{-8}}$$
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We know that the wavelength’s fractional uncertainty is given by
$$ \lambda+\Delta \lambda=\dfrac{hc}{E+\Delta E}$$
$$ \lambda+\Delta \lambda=\dfrac{hc}{E\left(1+\frac{\Delta E}{E}\right)}$$
where $hc/E=\lambda$
$$ \lambda+\Delta \lambda=\lambda\left(1+\frac{\Delta E}{E}\right)^{-1}$$
Using the binomial expansion for $\left(1+\frac{\Delta E}{E}\right)^{-1}$
$$ \lambda+\Delta \lambda=\lambda\left(1-\frac{\Delta E}{E}\right) $$
$$ \color{red}{\bf\not}\lambda+\Delta \lambda=\color{red}{\bf\not}\lambda -\frac{\Delta E}{E}\lambda $$
$$ \Delta \lambda= -\frac{\Delta E}{E}\lambda $$
Thus,
$$\left|\dfrac{\Delta \lambda}{ \lambda}\right|=\frac{\Delta E}{E}$$
Plugging from above;
$$\dfrac{\Delta \lambda}{ \lambda}=\color{red}{\bf 2.65\times10^{-8}}$$
