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 cmitting a photon is unprcdictablc, 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 $\mathrm{Δ}t≈10$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 eV.

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

Step by step solution

01

Part (a) Step 1:Given Information

Emission of photon $=500\mathrm{η}\mathrm{m}$(wavelength )

Length of time $\mathrm{Δ}t=10ns$

02

Part (b) Step 1:soluction

It is clear that the energy of the photon cannot be determined accurately from result of part (a). The accuracy in measuring the energy uncertainty decides by the length of the$\mathrm{Δ}t$

Conclusion: Thus, from the explanation above states that, the energy of the photon cannot determine accurately.

03

Part (c) Step 1:soluction

Applying h=6.63 $×{10}^{-34}\mathrm{J},\mathrm{Δ}t=10\mathrm{ns}$,

$$\begin{gathered} \mathrm{Δ}Eâ‰\dots \frac{6.63×{10}^{-3/}\mathrm{J}}{10ns\left(\frac{{10}^{-9\mathrm{s}}}{10\mathrm{m}}\right)} \\ \mathrm{Δ}\mathrm{E}â‰\dots 6.63×{10}^{-26\mathrm{J}} \\ =6.63×{10}^{-26\mathrm{J}}\left(\frac{1.0\mathrm{V}}{1.6×{10}^{-{}^{-1}\mathrm{J}}}\right) \\ \mathrm{Δ}\mathrm{E}=4.14×{10}^{-7}\mathrm{cV} \end{gathered}$$

Conclusion:Thus, the uncertainty of energy of the photon is$∆\mathrm{E}=4.14×{10}^{-7}\mathrm{eV}$

04

Part (d) Step 1:Given Information

$$\begin{gathered} \mathrm{hc}=1240\mathrm{cVnm} \\ \mathrm{λ}=⊃00\mathrm{nm} \end{gathered}$$

05

Part (d) Step 2:soluction

Formula to be used:

$\mathrm{E}=\frac{\mathrm{hc}}{2}$

Calculation:Applying values,

$$\begin{gathered} \mathrm{E}=\frac{1240\mathrm{cVmm}}{50\mathrm{mm}} \\ \mathrm{E}=2.48\mathrm{eV} \end{gathered}$$

The fractional uncertainty, in the photon energy is,

$$\begin{gathered} \frac{\mathrm{Δ}\mathrm{E}}{\mathrm{E}}=\frac{4.14×{10}^{-7}\mathrm{eV}}{24\mathrm{seV}} \\ \frac{\mathrm{Δ}\mathrm{E}}{\mathrm{E}}=1.7×{10}^{-7}\mathrm{eV} \end{gathered}$$

Conclusion:Thus, the fractional uncertainty in the photon energy is$\left(\frac{∆E}{E}\right)1.7×{10}^{-7}\mathrm{cV}$

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