91Ó°ÊÓ

a. Starting with the expression$\mathrm{Δ}f\mathrm{Δ}t≈1$for a wave packet, find an expression for the product

$\mathrm{Δ}E\mathrm{Δ}t$for a photon.

b. Interpret your expression. What does it tell you?

c. The Bohr model of atomic quantization says that an atom in an excited state can jump to a lower-energy state by emitting a photon. The Bohr model says nothing about how long this process takes. You'll learn in Chapter 41 that the time any particular atom spends in the excited state before emitting a photon is unpredictable, but the average lifetime $\mathrm{Δ}t$of many atoms can be determined. You can think of $\mathrm{Δ}t$as being the uncertainty in your knowledge of how long the atom spends in the excited state. A typical value is $\mathrm{Δ}t≈10\mathrm{ns}$. Consider an atom that emits a photon with a $500\mathrm{nm}$wavelength as it jumps down from an excited state. What is the uncertainty in the energy of the photon? Give your answer in $\mathrm{eV}$.

d. What is the fractional uncertainty $\mathrm{Δ}E/E$in the photon's energy?

Step by step solution

01

a.) Given Information : Expression for a wave packet ΔfΔt≈1

a.) We know that the energy of a photon is E = hf, which means that $\mathrm{Δ}E=h\mathrm{Δ}f$. Now, we have been told in the question that the expression $\mathrm{Δ}f\mathrm{Δ}t≈1$is valid, meaning that we can treat the photon as a wave packet. Substitute $\mathrm{Δ}E/h$for $\mathrm{Δ}f$

$\frac{\mathrm{Δ}E}{h}×\mathrm{Δ}t≈1$

$\mathrm{Δ}E\mathrm{Δ}t≈h$

02

b.) Given Information : Interpretation on the basis of part a 

b.) We can't know the exact energy of a photon, where the uncertainty in the knowledge of a photon's energy depends on the time period taken to measure this energy.

03

c.) Given Information : an atom that emits a photon with a  500 nm wavelength as it jumps down from an excited state. 

We can simply use the equation obtained in part (a) to find $\mathrm{Δ}E$when $\mathrm{Δ}t=10\mathrm{ns}$

$\mathrm{Δ}E=\frac{h}{\mathrm{Δ}t}=\frac{6.626×{10}^{-34}\mathrm{J}·\mathrm{s}}{10×{10}^{-9}\mathrm{s}}=6.626×{10}^{-26}\mathrm{J}$

$\mathrm{Δ}E=6.626×{10}^{-26}\mathrm{J}×\frac{1\mathrm{eV}}{1.6×{10}^{-19}\mathrm{J}}=4.14×{10}^{-7}\mathrm{eV}$

04

d.) Given Information : The photon's energy is

$E=\frac{hc}{\mathrm{λ}}=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}{500×{10}^{-9}\mathrm{m}}=3.976×{10}^{-19}\mathrm{J}$

$E=3.976×{10}^{-19}\mathrm{J}×\frac{1\mathrm{eV}}{1.6×{10}^{-19}\mathrm{J}}=2.485\mathrm{eV}$

the fractional uncertainty is

$\frac{\mathrm{Δ}E}{E}=\frac{4.14×{10}^{-7}\mathrm{eV}}{2.485\mathrm{eV}}=1.67×{10}^{-7}$

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